567 nice and_hard_inequality

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567
INEQUALITY PROBLEMS WITH SOLUTIONS

IF
—A if a, b, c —re positive re—l num˜ersD then
a
b
+
b
c
+
c
a
≥
a2 + 1
b2 + 1
+
b2 + 1
c2 + 1
+
c2 + 1
a2 + 1
.
˜Avet a, b, c, d ˜e positive re—l num˜ersF€rove th—t
a2
− bd
b + 2c + d
+
b2
− ca
c + 2d + a
+
c2
− db
d + 2a + b
+
d2
− ac
a + 2b + c
≥ 0.
SolutionX
—Afy g—u™hyEƒ™hw—rz9s inequ—lityD ‡e h—veX
a2
+ b2
(a2 + 1) (b2 + 1) ≥ a2
+ b2
(ab + 1)
= ab a2
+ b2
+ a2
+ b2
≥ ab a2
+ b2
+ 2
⇒
a
b
+
b
a
=
a2
+ b2
ab
≥
a2
+ b2
+ 2
(a2 + 1) (b2 + 1)
=
a2 + 1
b2 + 1
+
b2 + 1
a2 + 1
fy ghe˜yshev9s inequ—lityD ‡e h—ve
a2
b2
=
a2
b2 + 1
+
a2
b2 (b2 + 1)
≥
a2
b2 + 1
+
b2
b2 (b2 + 1)
=
a2
+ 1
b2 + 1
.
„herefore
1 +
a
b
2
= 1 + 2
a
b
+
b
a
+
a2
b2
≥ 1 + 2
a2 + 1
b2 + 1
+
b2 + 1
a2 + 1
+
a2
+ 1
b2 + 1
= 1 +
a2 + 1
b2 + 1
2
.
„herefore
a
b
+
b
c
+
c
a
≥
a2 + 1
b2 + 1
+
b2 + 1
c2 + 1
+
c2 + 1
a2 + 1
—s requireF
˜Axoti™e th—t
2(a2
− bd)
b + 2c + d
+ b + d =
2a2
+ b2
+ d2
+ 2c(b + d)
b + 2c + d
=
(a − b)2
+ (a − d)2
+ 2(a + c)(b + d)
b + 2c + d
(1)
end simil—rlyD
2(c2
− db)
d + 2a + b
+ b + d =
(c − d)2
+ (c − b)2
+ 2(a + c)(b + d)
d + 2a + b
(2)
…sing g—u™hyEƒ™hw—rz9s inequ—lityDwe get
(a − d)2
b + 2c + d
+
(c − d)2
d + 2a + b
≥
[(a − b)2
+ (c − d)2
]
(b + 2c + d) + (d + 2a + b)
(3)
Q

(a − d)2
b + 2c + d
+
(c − b)2
d + 2a + b
≥
[(a − d)2
+ (c − b)2
]2
(b + 2c + d) + (d + 2a + b)
(4)
2(a + c)(b + d)
b + 2c + d
+
2(a + c)(b + d)
d + 2a + b
≥
8(a + c)(b + d)
(b + 2c + d) + (d + 2a + b)
(5)
prom @IAD@PAD@QAD@RA —nd @SAD we get
2(
a2
− bd
b + 2c + d
+
c2
− db
d + 2a + b
) + b + d ≥
(a + c − b − d)2
+ 4(a + c)(b + d)
a + b + c + d
= a + b + c + d.
or
a2
− bd
b + 2c + d
+
c2
− db
d + 2a + b
≥
a + c − b − d
2
sn the s—me m—nnerDwe ™—n —lso show th—t
b2
− ca
c + 2d + a
+
d2
− ac
a + 2b + c
≥
b + d − a − c
2
—nd ˜y —dding these two inequ—litiesDwe get the desired resultF
inqu—lity holds if —nd only if a = c —nd b = dF
PD
vet a, b, c ˜e positive re—l num˜ers su™h th—t
a + b + c = 1
€rove th—t the following inequ—lity holds
ab
1 − c2
+
bc
1 − a2
+
ca
1 − b2
≤
3
8
SolutionX prom the given ™ondition „he inequ—lity is equiv—lent to
4ab
a2 + b2 + 2(ab + bc + ca)
≤
3
2
˜ut from g—uhy ƒhw—rz inequ—lity
4ab
a2 + b2 + 2(ab + bc + ca)
≤
ab
a2 + ab + bc + ca
+
ab
b2 + ab + bc + ca
=
ab
(a + b)(a + c)
+
ab
(b + c)(a + b)
=
a(b + c)2
(a + b)(b + c)(c + a)
„hus ‡e need prove th—t
3(a + b)(b + c)(c + a) ≥ 2 a(b + c)2
whi™h redu™es to the o˜vious inequ—lity
ab(a + b) ≥ 6abc
„he Solution is ™ompletedFwith equ—lity if —nd only if
a = b = c =
1
3
R

yr ‡e ™—n use the f—™t th—t
4ab
a2 + b2 + 2(ab + bc + ca)
≤
4ab
(2ab + 2ac) + (2ab + 2bc)
≤
ab
2a(b + c)
+
ab
2b(a + c)
=
1
2
b
b + c
+
a
a + c
=
1
2
b
b + c
+
c
b + c
=
3
2
QD vet a, b, c ˜e the positive re—l num˜ersF €rove th—t
1 +
ab2
+ bc2
+ ca2
(ab + bc + ca)(a + b + c)
≥
4. 3
(a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)
(a + b + c)2
SolutionX wultiplying ˜oth sides of the —˜ove inequ—lity with (a + b + c)2
it9s equiv—lent to
prove th—t
(a + b + c)2
+
(a + b + c)(ab2
+ bc2
+ ca2
)
ab + bc + ca
≥ 4. 3
(a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)
‡e h—ve
(a + b + c)2
+
(a + b + c)(ab2
+ bc2
+ ca2
)
ab + bc + ca
=
(a2
+ ab + bc)(c + a)(c + b)
ab + bc + ca
fy using ewEqw inequ—lity ‡e get
(a2
+ ab + bc)(c + a)(c + b)
ab + bc + ca
≥ 3.
3
(a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)[(a + b)(b + c)(c + a)]2
ab + bc + ca
ƒin™e it9s suffi™es to show th—t
√
3. 3
(a + b)(b + c)(c + a) ≥ 2.
√
ab + bc + ca
whi™h is ™le—rly true ˜y ewEqw inequ—lity —g—inF „he Solution is ™ompletedF iqu—lity
holds for a = b = c
RD
vet a0, a1, . . . , an ˜e positive re—l num˜ers su™h th—t ak+1 −ak ≥ 1 for —ll k = 0, 1, . . . , n−1.
€rove th—t
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
an − a0
≤ 1 +
1
a0
1 +
1
a1
· · · 1 +
1
an
SolutionX ‡e will prove it ˜y indu™tionF
por n = 1 ‡e need to ™he™k th—t
1 +
1
a0
1 +
1
a1 − a0
≤ 1 +
1
a0
1 +
1
a1
whi™h is equiv—lent to a0(a1 − a0 − 1) ≥ 0, whi™h is true ˜y given ™onditionF
vet
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
≤ 1 +
1
a0
1 +
1
a1
· · · 1 +
1
ak
S

it rem—ins to prove th—tX
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak+1 − a0
≤
≤ 1 +
1
a0
1 +
1
a1
· · · 1 +
1
ak+1
fy our hypothesis
1 +
1
a0
1 +
1
a1
· · · 1 +
1
ak+1
≥
≥ 1 +
1
ak+1
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
id estD it rem—ins to prove th—tX
1 +
1
ak+1
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
≥
≥ 1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak+1 − a0
fut
1 +
1
ak+1
1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
≥
≥ 1 +
1
a0
1 +
1
a1 − a0
· · · 1 +
1
ak+1 − a0
⇔
⇔
1
ak+1
+
1
ak+1a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
≥
≥
1
(ak+1 − a0)a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
⇔
⇔ 1 ≥
1
ak+1 − a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
fut ˜y our ™onditions ‡e o˜t—inX
1
ak+1 − a0
1 +
1
a1 − a0
· · · 1 +
1
ak − a0
≤
≤
1
k
1 +
1
1
· · · 1 +
1
k − 1
= 1.
„husD the inequ—lity is provenF
SD
qiven a, b, c > 0F €rove th—t
3 a2 + bc
b2 + c2
≥ 9.
3
√
abc
(a + b + c)
Solution X „his ineq is equiv—lent toX
a2
+ bc
3
abc(a2 + bc)
2
(b2 + c2)
≥
9
(a + b + c)
3
fy ewEqw ineq D ‡e h—ve
a2
+ bc
3
abc(a2 + bc)
2
(b2 + c2)
=
T

=
a2
+ bc
3
(a2 + bc)c(a2 + bc)b(b2 + c2)a
≥
3(a2
+ bc)
sym
a2b
ƒimil—rlyD this ineq is true if ‡e prove th—tX
3(a2
+ b2
+ c2
+ ab + bc + ca)
sym
a2b
≥
9
(a + b + c)
3
a3
+ b3
+ c3
+ 3abc ≥
sym
a2
b
‡hi™h is true ˜y ƒ™hur ineqF iqu—lity holds when a = b = c
TD
vet a, b, c ˜e nonneg—tive re—l num˜ers su™h th—t ab + bc + ca > 0F €rove th—t
1
2a2 + bc
+
1
2b2 + ca
+
1
2c2 + ab
≥
2
ab + bc + ca
.
„he inequ—lity is equiv—lent to
ab + bc + ca
2a2 + bc
≥ 2, (1)
or
a(b + c)
2a2 + bc
+
bc
bc + 2a2
≥ 2.(2)
…sing the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
bc
bc + 2a2
≥
( bc)
2
bc(bc + 2a2)
= 1.(3)
„hereforeD it suffi™es to prove th—t
a(b + c)
2a2 + bc
≥ 1.(4)
ƒin™e
a(b + c)
2a2 + bc
≥
a(b + c)
2(a2 + bc)
it is enough to ™he™k th—t
a(b + c)
a2 + bc
≥ 2, (5)
whi™h is — known resultF
‚em—rkX
2ca + bc
2a2 + bc
+
2bc + ca
2b2 + ca
≥
4c
a + b + c
.
UD
vet a, b, c ˜e non neg—tive re—l num˜ers su™h th—t ab + bc + ca > 0F €rove th—t
1
2a2 + bc
+
1
2b2 + ca
+
1
2c2 + ab
+
1
ab + bc + ca
≥
12
(a + b + c)2
.
SolutionX IA ‡e ™—n prove this inequ—lity using the following —uxili—ry result
if 0 ≤ a ≤ min{a, b}D then
1
2a2 + bc
+
1
2b2 + ca
≥
4
(a + b)(a + b + c)
.
U

in f—™tD this is used to repl—™ed for 4no two of whi™h —re zero4D so th—t the fr—™tions
1
2a2 + bc
,
1
2b2 + ca
,
1
2c2 + ab
,
1
ab + bc + ca
h—ve me—ningsF
fesidesD the i—ker —lso works for itX
1
2a2 + bc
+
1
2b2 + ca
+
1
2c2 + ab
≥
2(ab + bc + ca)
a2b2 + abc(a + b + c)
fut our Solution for ˜oth of them is exp—nd
vet a, b, c ˜e non neg—tive re—l num˜ers su™h th—t ab + bc + ca > 0F €rove th—t
1
2a2 + bc
+
1
2b2 + ca
+
1
2c2 + ab
+
1
ab + bc + ca
≥
12
(a + b + c)2
.
PA gonsider ˜y ewEqw inequ—lityD ‡e h—ve
2 a2
+ ab + b2
(a + b + c)
= (2b + a) 2a2
+ bc + (2a + b) 2b2
+ ca
≥ 2 (2a + b)(2b + a) (2a2 + bc) (2b2 + ca).
end ˜y ewEqw inequ—lityD ‡e h—ve
c2
(2a + b)
2a2 + bc
+
c2
(2b + a)
2b2 + ca
≥ 2
c4(2a + b)(2b + a)
(2a2 + bc) (2b2 + ca)
≥
2c2
(2a + b)(2b + a)
(a2 + ab + b2) (a + b + c)
=
4c2
a + b + c
+
6abc
a + b + c
c
a2 + ab + b2
2c2
a + bc2
+ 2ab2
+ b2
c
2a2 + bc
=
c2
(2a + b)
2a2 + bc
+
c2
(2b + a)
2b2 + ca
≥
4c2
a + b + c
+
6abc
a + b + c
c
a2 + ab + b2
=
4 a2
+ b2
+ c2
a + b + c
+
6abc
a + b + c
c
a2 + ab + b2
≥
4 a2
+ b2
+ c2
a + b + c
+
6abc
a + b + c
(a + b + c)2
c (a2 + ab + b2)
=
4 a2
+ b2
+ c2
ab + c
+
6abc
ab + bc + ca
⇒
2a2
b + 2ab2
+ 2b2
c + 2bc2
+ 2c2
a + 2ca2
2a2 + bc
V

= (b + c) +
2c2
a + bc2
+ 2ab2
+ b2
c
2a2 + bc
≥ (b + c) +
4 a2
+ b2
+ c2
a + b + c
+
6abc
ab + bc + ca
=
8 a2
+ b2
+ c2
+ ab + bc + ca
a + b + c
−
2 a2
b + ab2
ab + bc + ca
⇒
1
2a2 + bc
+
1
ab + bc + ca
≥
4 a2
+ b2
+ c2
+ ab + bc + ca
(a + b + c) ( (a2b + ab2))
≥
12
(a + b + c)2
.
<=>
(a + b)(a + c)
2a2 + bc
+
a2
+ bc
2a2 + bc
− 2 ≥
12(ab + bc + ca)
(a + b + c)2
prom
2a2
+ 2bc
2a2 + bc
− 3 =
bc
2a2 + bc
≥ 1
‡e get
a2
+ bc
2a2 + bc
− 2 ≥ 0
xowD ‡e will prove the stronger
(a + b)(a + c)
2a2 + bc
≥
12(ab + bc + ca)
(a + b + c)2
prom ™—u™hyEs™h—rztD ‡e h—ve
(a + b)(a + c)
2a2 + bc
= (a+b)(b+c)(c+a)(
1
(2a2 + bc)(b + c)
≥
3(a + b)(b + c)(c + a)
ab(a + b) + bc(b + c) + ca(c + a)
pin—llyD ‡e only need to prove th—t
(a + b)(b + c)(c + a)
ab(a + b) + bc(b + c) + ca(c + a)
≥
4(ab + bc + ca)
(a + b + c)2
(a + b + c)2
ab + bc + ca
≥
4[ab(a + b) + bc(b + c) + ca(c + a)
(a + b)(b + c)(c + a)
= 4 −
8abc
(a + b)(b + c)(c + a)
a2
+ b2
+ c2
ab + bc + ca
+
8abc
(a + b)(b + c)(c + a)
≥ 2
whi™h is old pro˜lemF yur Solution —re ™ompleted equ—lity o™™ur if —nd if only
a = b = c, a = b, c = 0
or —ny ™y™li™ permutionF
VD vet a, b, c ˜e positive re—l num˜ers su™h th—t 16(a + b + c) ≥ 1
a + 1
b + 1
c F €rove th—t
1
a + b + 2(a + c)
3 ≤
8
9
.
SolutionX „his pro˜lem is r—ther e—syF …sing the ewEqw inequ—lityD ‡e h—veX
a + b + 2(c + a) = a + b +
c + a
2
+
c + a
2
≥ 3
3 (a + b)(c + a)
2
.
W

ƒo th—tX
1
a + b + 2(c + a)
3 ≤
2
27(a + b)(c + a)
.
„husD it9s enough to ™he™k th—tX
1
3(a + b)(c + a)
≤ 4 ⇐⇒ 6(a + b)(b + c)(c + a) ≥ a + b + c,
whi™h is true sin™e
9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca)
—nd
16abc(a + b + c) ≥ ab + bc + ca ⇒
16(ab + bc + ca)2
3
≥ ab + bc + ca ⇐⇒ ab + bc + ca ≥
3
16
.
„he Solution is ™ompletedF iqu—lity holds if —nd only if a = b = c = 1
4 F
WD vet x, y, z ˜e positive re—l num˜ers su™h th—t xyz = 1F €rove th—t
x3
+ 1
x4 + y + z
+
y3
+ 1
y4 + z + x
+
z3
+ 1
z4 + x + y
≥ 2
√
xy + yz + zx.
SolutionX …sing the ewEqw inequ—lityD ‡e h—ve
2 (x4 + y + z)(xy + yz + zx) = 2 [x4 + xyz(y + z)](xy + yz + zx)
= 2 (x3 + y2z + yz2)(x2y + x2z + xyz)
≤ (x3
+ y2
z + yz2
) + (x2
y + x2
z + xyz)
= (x + y + z)(x2
+ yz) =
(x + y + z)(x3
+ 1)
x
.
it follows th—t
x3
+ 1
x4 + y + z
≥
2x
√
xy + yz + zx
x + y + z
.
edding this —nd it —n—logous inequ—litiesD the result followsF
IHD vet a, b, c ˜e nonneg—tive re—l num˜ers s—tisfying a + b + c =
√
5F €rove th—t
(a2
− b2
)(b2
− c2
)(c2
− a2
) ≤
√
5
SolutionX por this oneD ‡e ™—n —ssume ‡vyq th—t c ≥ b ≥ a so th—t ‡e h—ve
P = (a2
− b2
)(b2
− c2
)(c2
− a2
) = (c2
− b2
)(c2
− a2
)(b2
− a2
) ≤ b2
c2
(c2
− b2
).
elso note th—t
√
5 = a + b + c ≥ b + c sin™e a ≥ 0F xowD using the ewEqw inequ—lity ‡e
h—ve
(c + b) ·
√
5
2
− 1 · c
2
·
√
5
2
+ 1 b
2
· (c − b)
≤ (c + b)
√
5(b + c)
5
5
≤
√
5;
ƒo th—t ‡e get P ≤
√
5F end hen™e ‡e —re doneF iqu—lity holds if —nd only if (a, b, c) =√
5
2 + 1;
√
5
2 − 1; 0 —nd —ll its ™y™li™ permut—tionsF 2
IH

IID vet a, b, c > 0 —nd a + b + c = 3F €rove th—t
1
3 + a2 + b2
+
1
3 + b2 + c2
+
1
3 + c2 + a2
≤
3
5
SolutionX ‡e h—veX
1
3 + a2 + b2
+
1
3 + b2 + c2
+
1
3 + c2 + a2
≤
3
5
<=>
3
3 + a2 + b2
+
3
3 + b2 + c2
+
3
3 + c2 + a2
≤
9
5
a2
+ b2
3 + a2 + b2
≥
6
5
…sing g—u™hyEƒ™hw—rz9s inequ—lityX
a2
+ b2
3 + a2 + b2
( 3 + a2
+ b2
) ≥ ( a2 + b2)2
„h—t me—ns ‡e h—ve to prove
( a2 + b2)2
≥
6
5
( (3 + a2
+ b2
))
(a2
+ b2
) + 2 (a2 + b2)(a2 + c2) ≥
54
5
+
12
5
a2
8 a2
+ 10 ab ≥ 54 <=> 5(a + b + c)2
+ 3 a2
≥ 54
it is true with a + b + c = 3F
IPD
qiven a, b, c > 0 su™h th—t ab + bc + ca = 1F €rove th—t
1
4a2 − bc + 1
+
1
4b2 − ca + 1
+
1
4c2 − ab + 1
≥ 1
SolutionX in f—™tD the sh—rper inequ—lity holds
1
4a2 − bc + 1
+
1
4b2 − ca + 1
+
1
4c2 − ab + 1
≥
3
2
.
1
a(4a + b + c)
+
1
b(4b + c + a)
+
1
c(4c + a + b)
≥
3
2
.
1
a(4a + b + c)
4a + b + c
a
≥
1
a
2
=
1
a2b2c2
.
2
3a2b2c2
≥
4a + b + c
a
+
4b + c + a
b
+
4c + a + b
c
.
ƒin™e
4a + b + c
a
= 3 +
a + b + c
a
= 9 +
(a + b + c)(ab + bc + ca)
abc
= 9 +
a + b + c
abc
,
II

this inequ—lity ™—n ˜e written —s
9a2
b2
c2
+ abc(a + b + c) ≤
2
3
,
whi™h is true ˜e™—use
a2
b2
c2
≤
ab + bc + ca
3
3
=
1
27
,
—nd
abc(a + b + c) ≤
(ab + bc + ca)2
3
=
1
3
.
IQD qiven a, b, c ≥ 0 su™h th—t ab + bc + ca = 1F €rove th—t
1
4a2 − bc + 2
+
1
4b2 − ca + 2
+
1
4c2 − ab + 2
≥ 1
SolutionX xoti™e th—t the ™—se abc = 0 is trivi—l so let us ™onsider now th—t abc > 0F …sing
the ewEqw inequ—lityD ‡e h—ve
4a2
− bc + 2(ab + bc + ca) = (2a + b)(2a + c) ≤
[c(2a + b) + b(2a + c)]2
4bc
=
(ab + bc + ca)2
bc
=
1
bc
.
it follows th—t
1
4a2 − bc + 2
≥ bc.
edding this —nd its —n—logous inequ—litiesD ‡e get the desired resultF
IRD qiven a, b, c —re positive re—l num˜ersF €rove th—t
(
1
a
+
1
b
+
1
c
)(
1
1 + a
+
1
1 + b
+
1
1 + c
) ≥
9
1 + abc
.
SolutionX „he origin—l inequ—lity is equiv—lent to
abc + 1
a
+
abc + 1
b
+
abc + 1
c
1
a + 1
+
1
b + 1
+
1
c + 1
≥ 9
or
cyc
1 + a2
c
a
1
a + 1
+
1
b + 1
+
1
c + 1
≥ 9
fy g—u™hy ƒ™hw—rz ineq —nd ewEqw ineqD
cyc
1 + a2
c
a
≥
cyc
c(1 + a)2
a(1 + c)
≥ 3 3
(1 + a)(1 + b)(1 + c)
—nd
1
a + 1
+
1
b + 1
+
1
c + 1
≥
3
3
(1 + a)(1 + b)(1 + c)
wultiplying these two inequ—litiesD the ™on™lusion followsF iqu—lity holds if —nd only if
a = b = c = 1F
ISF qiven a, b, c —re positive re—l num˜ersF €rove th—tX
a(b + 1) + b(c + 1) + c(a + 1) ≤
3
2
(a + 1)(b + 1)(c + 1)
IP

SolutionX g—seIFif a + b + c + ab + bc + ca ≤ 3abc + 3 <=> 4(ab + bc + ca + a + b + c) ≤
3(a + 1)(b + 1)(c + 1) …sing g—u™hyEƒ™h—wrz9s inequ—lity D‡e h—veX
( a(b + 1) + b(c + 1) + c(a + 1))2
≤ 3(ab + bc + ca + a + b + c) ≤
9(a + 1)(b + 1)(c + 1)
4
„he inequ—lity is trueF g—sePF ifa + b + c + ab + bc + ca ≤ 3abc + 3.
<=>
9(a + 1)(b + 1)(c + 1)
4
≥ 2(a + b + c + ab + bc + ca) + 3abc + 3
fy ewEqw9s inequ—lity X
2 ab(b + 1)(c + 1) ≤ [ab(c + 1) + (b + 1)] = a + b + c + ab + bc + ca + 3abc + 3
=> ab + bc + ca + a + b + c + 2 ab(b + 1)(c + 1) ≤
9
4(a + 1)(b + 1)(c + 1)
=> ( a(b + 1) + b(c + 1) + c(a + 1))2
≤ [
3
2
(a + 1)(b + 1)(c + 1)]2
=> Q.E.D
inqu—lity holds when a = b = c = 1.
ITD qiven a, b, c —re positive re—l num˜ersF €rove th—tX
1
a2 + b2
+
1
b2 + c2
+
1
c2 + a2
≥
10
(a + b + c)2
SolutionX essume c = min{a, b, c}F „hen
1
a2 + c2
+
1
b2 + c2
≥
2
ab + c2
⇐⇒ (ab − c2
)(a − b)2
≥ 0
end ˜y g—u™hyEs™hw—rz
((a2
+ b2
) + 8(ab + c2
))
1
a2 + b2
+
2
ab + c2
≥ 25
ren™e ‡e need only to proveX
5(a + b + c)2
≥ 2((a2
+ b2
) + 8(ab + c2
)) ⇐⇒
3(a − b)2
+ c(10b + 10a − 11c) ≥ 0
iqu—lity for a = b, c = 0 or permut—tionsF
IUD vet a, b —nd c —re nonEneg—tive num˜ers su™h th—t ab + ac + bc = 0F €rove th—t
a2
(b + c)2
a2 + 3bc
+
b2
(a + c)2
b2 + 3ac
+
c2
(a + b)2
c2 + 3ab
≤ a2
+ b2
+ c2
Solution:
fy g—u™hyEƒ™hw—rz ineq D ‡e h—ve
a2
(b + c)
2
a2 + bc
=
a2
(b + c)
3
(a2 + bc)(b + c)
=
a2
(b + c)
3
b(a2 + c2) + c(a2 + b2)
≤
a2
(b + c)
4
(
b2
b(a2 + c2)
+
c2
c(a2 + b2)
) =
a2
(b + c)
4
(
b
a2 + c2
+
c
a2 + b2
)
ƒimil—rlyD ‡e h—ve
LHS ≤ a2
(b + c)(
b
a2 + c2
+
c
a2 + b2
) =
c(a2
(b + c) + b2
(c + a))
a2 + b2
IQ

= a2
+ b2
+ c2
+
abc(a + b)
a2 + b2
≤ a2
+ b2
+ c2
+
abc(a + b)
a2 + b2
≤ a2
+ b2
+ c2
+ ab + bc + ca
whi™h is true ˜y ewEqw ineq
„he origin—l inequ—lity ™—n ˜e written —s
(a + b)2
(a + c)2
a2 + bc
≤
8
3
(a + b + c)2
.
ƒin™e (a + b)(a + c) = (a2
+ bc) + a(b + c) ‡e h—ve
(a + b)2
(a + c)2
a2 + bc
=
(a2
+ bc)2
+ 2a(b + c)(a2
+ bc) + a2
(b + c)2
a2 + bc
= a2
+ bc + 2a(b + c) +
a2
(b + c)2
a2 + bc
,
—nd thus the —˜ove inequ—lity is equiv—lent to
a2
(b + c)2
a2 + bc
≤
8
3
(a + b + c)2
− a2
− 5 ab,
or
a2
(b + c)2
a2 + bc
≤
5(a2
+ b2
+ c2
) + ab + bc + ca
3
.
ƒin™e
5(a2
+ b2
+ c2
) + ab + bc + ca
3
≥ a2
+ b2
+ c2
+ ab + bc + ca
it is enough show th—t
a2
(b + c)2
a2 + bc
≤ a2
+ b2
+ c2
+ ab + bc + ca.
FiFh
IVD qiven
a1 ≥ a2 ≥ . . . ≥ an ≥ 0, b1 ≥ b2 ≥ . . . ≥ bn ≥ 0
n
i=1
ai = 1 =
n
i=1
bi
pind the m—xmium of
n
i=1
(ai − bi)2
‡SolutionXithout loss of gener—lityD —ssume th—t
a1 ≥ b1
xoti™e th—t for
a ≥ x ≥ 0, b, y ≥ 0
‡e h—ve
(a − x)2
+ (b − y)2
− (a + b − x)2
− y2
= −2b(a − x + y) ≤ 0.
e™™ording to this inequ—lityD ‡e h—ve
(a1 − b1)2
+ (a2 − b2)2
≤ (a1 + a2 − b1)2
+ b2
2,
(a1 + a2 − b1)2
+ (a3 − b3)2
≤ (a1 + a2 + a3 − b1)2
+ b2
3, · · · · · ·
IR

(a1 + a2 + · · · + an−1 − b1)2
+ (an − bn)2
≤ (a1 + a2 + · · · + an − b1)2
+ b2
n.
edding these inequ—litiesD ‡e get
n
i=1
(ai − bi)2
≤ (1 − b1)2
+ b2
2 + b2
3 + · · · + b2
n
≤ (1 − b1)2
+ b1(b2 + b3 + · · · + bn)
= (1 − b1)2
+ b1(1 − b1) = 1 − b1 ≤ 1 −
1
n
.
iqu—lity holds for ex—mple when
a1 = 1, a2 = a3 = · · · = an = 0
—nd
b1 = b2 = · · · = bn =
1
n
IWD qiven
a, b, c ≥ 0
su™h th—t
a2
+ b2
+ c2
= 1
€rove th—t
1 − ab
7 − 3ac
+
1 − bc
7 − 3ba
+
1 − ca
7 − 3cb
≥
1
3
SolutionX pirstD ‡e will show th—t
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
≤
1
2
.
1
7 − 3ab
=
1
3(1 − ab) + 4
≤
1
9
1
3(1 − ab)
+ 1 .
it follows th—t
1
7 − 3ab
≤
1
27
1
1 − ab
+
1
3
,
—nd thusD it is enough to show th—t
1
1 − ab
+
1
1 − bc
+
1
1 − ca
≤
9
2
,
whi™h is †—s™9s inequ—lityF xowD ‡e write the origin—l inequ—lity —s
3 − 3ab
7 − 3ac
+
3 − 3bc
7 − 3ba
+
3 − 3ca
7 − 3cb
≥ 1,
or
7 − 3ab
7 − 3ac
+
7 − 3bc
7 − 3ba
+
7 − 3ca
7 − 3cb
≥ 1 + 4
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
.
ƒin™e
4
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
≤ 2
IS

it is enough to show th—t
7 − 3ab
7 − 3ac
+
7 − 3bc
7 − 3ba
+
7 − 3ca
7 − 3cb
≥ 3,
whi™h is true —™™ording to the ewEqw inequ—lityF
PID vet
a, b, c ≥ 0
su™h th—t
a + b + c > 0
—nd
b + c ≥ 2a
por
x, y, z > 0
su™h th—t
xyz = 1
€rove th—t the following inequ—lity holds
1
a + x2(by + cz)
+
1
a + y2(bz + cx)
+
1
a + z2(bx + cy)
≥
3
a + b + c
SolutionX ƒetting
u =
1
x
, v =
1
y
—nd
w =
1
z
—nd using the ™ondition
uvw = 1
the inequ—lity ™—n ˜e rewritten —s
u
au + cv + bw
=
u2
au2 + cuv + bwu
3
a + b + c
F
epplying g—u™hyD it suffi™es to prove
(u + v + w)
2
a u2 + (b + c) uv
3
a + b + c
1
2
· (b + c − 2a) (x − y)2
0D
whi™h is o˜vious due to the ™ondition for
a, b, c
PPD qiven
x, y, z > 0
su™h th—t
xyz = 1
IT

€rove th—t
1
(1 + x2)(1 + x7)
+
1
(1 + y2)(1 + y7)
+
1
(1 + z2)(1 + z7)
≥
3
4
SolutionX pirst ‡e prove this ineq e—sy
1
(1 + x2)(1 + x7)
≥
3
4(x9 + x
9
2 + 1)
end this ineq ˜e™—meX
1
x9 + x
9
2 + 1
+
1
y9 + y
9
2 + 1
+
1
z9 + z
9
2 + 1
≥ 1
with
xyz = 1
it9s —n old result
PQD vet
a, b, c
˜e positive re—l num˜ers su™h th—t
3(a2
+ b2
+ c2
) + ab + bc + ca = 12
€rove th—t
a
√
a + b
+
b
√
b + c
+
c
√
c + a
≤
3
√
2
.
SolutionX vet
A = a2
+ b2
+ c2
, B = ab + bc + ca
2A + B = 2 a2
+ ab ≤
3
4
3 a2
+ ab = 9.
fy g—u™hy ƒ™hw—rz inequ—lityD ‡e h—ve
a
√
a + b
=
√
a
a
a + b
≤
√
a + b + c
a
a + b
.
fy g—u™hy ƒ™hw—rz inequ—lity —g—inD ‡e h—ve
b
a + b
=
b2
b(a + b)
≥
(a + b + c)2
b(a + b)
=
A + 2B
A + B
a
a + b
= 3 −
b
a + b
≤ 3 −
A + 2B
A + B
=
2A + B
A + B
hen™eD it suffi™es to prove th—t
(a + b + c) ·
2A + B
A + B
≤
9
2
IU

gonsider
(a + b + c)
√
2A + B
= (A + 2B) (2A + B)
≤
(A + 2B) + (2A + B)
2
=
3
2
(A + B)
⇒ (a + b + c) ·
2A + B
A + B
≤
3
2
√
2A + B ≤
9
2
—s requireF
fy ewEqw ineq e—sy to see th—t
3 ≤ a2
+ b2
+ c2
≤ 4
fy g—u™hyEƒ™hw—rz ineqD ‡e h—ve
LHS2
= (
a
√
a + c
(a + b)(a + c)
) ≤ (a2
+ b2
+ c2
+ ab + bc + ca)(
a
(a + b)(a + c)
)
…sing the f—mili—r ineq
9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca)
‡e h—ve
a
(a + b)(a + c)
=
2(ab + bc + ca)
(a + b)(b + c)(c + a)
≤
9
4(a + b + c)
end ‡e need to prove th—t
9(a2
+ b2
+ c2
+ ab + bc + ca)
4(a + b + c)
≤
9
2
⇔
6 − (a2
+ b2
+ c2
)
24 − 5(a2 + b2 + c2)
≤ 1
⇔ (6 − (a2
+ b2
+ c2
))2
≤ 24 − 5(a2
+ b2
+ c2
)
⇔ (3 − (a2
+ b2
+ c2
))(4 − (a2
+ b2
+ c2
)) ≤ 0
‡hi™h is true ‡e —re done equ—lity holds when
a = b = c = 1
PRF
qiven
a, b, c ≥ 0
€rove th—t
1
(a2 + bc)(b + c)2
≤
8(a + b + c)2
3(a + b)2(b + c)2(c + a)2
SolutionX in f—™tD the sh—rper —nd ni™er inequ—lity holdsX
a2
(b + c)2
a2 + bc
+
b2
(c + a)2
b2 + ca
+
c2
(a + b)2
c2 + ab
≤ a2
+ b2
+ c2
+ ab + bc + ca.
a2
(b + c)2
a2 + bc
+
b2
(c + a)2
b2 + ca
+
c2
(a + b)2
c2 + ab
≤ a2
+ b2
+ c2
+ ab + bc + ca
IV

PSF
qiven
a, b, c ≥ 0
su™h th—t
ab + bc + ca = 1
€rove th—t
1
8
5 a2 + bc
+
1
8
5 b2 + ca
+
1
8
5 c2 + ab
≥
9
4
essume ‡vyq
a ≥ b ≥ c
this ineq
1
8
5 a2 + bc
−
5
8
+
1
8
5 b2 + ca
−
5
8
+
1
8
5 c2 + ab
− 1 ≥ 0
8 − 8a2
− 5bc
8a2 + 5bc
+
8 − 8b2
− 5ca
8b2 + 5ca
+
1 − 8
5 c2
− ab
c2 + 8
5 ab
≥ 0
8a(b + c − a) + 3bc
8a2 + 5bc
+
8b(a + c − b) + 5ac
8b2 + 5ca
+
c(a + b − 8
5 c)
c2 + 8
5 ab
≥ 0
xoti™e th—t ‡e only need to prove this ineq when
a ≥ b + c
˜y the w—y ‡e need to prove th—t
8b
8b2 + 5ca
≥
8a
8a2 + 5bc
(a − b)(8ab − 5ac − 5bc) ≥ 0
i—sy to see th—tX if
a ≥ b + c
then
8ab = 5ab + 3ab ≥ 5ac + 6bc ≥ 5ac + 5ac
ƒo this ineq is trueD ‡e h—ve qFdFe D equ—lity hold when
(a, b, c) = (1, 1, 0)
PTD qive
a, b, c ≥ 0
€rove th—tX
a
b2 + c2
+
b
a2 + c2
+
c
a2 + b2
≥
a + b + c
ab + bc + ca
+
abc(a + b + c)
(a3 + b3 + c3)(ab + bc + ca)
a
b2 + c2
=
a2
ab2 + c2a
≥
(a + b + c)2
(ab2 + c2a)
,
it suffi™es to prove th—t
a + b + c
(ab2 + c2a)
≥
1
ab + bc + ca
+
abc
(ab + bc + ca) (a3 + b3 + c3)
,
IW

˜e™—use
a + b + c
(ab2 + c2a)
−
1
ab + bc + ca
=
3abc
(ab + bc + ca) (ab2 + ca2)
,
3 a3
+ b3
+ c3
≥ ab2
+ c2
a ,
2 a3
+ b3
+ c3
≥ ab2
+ c2
a .
‚em—rkX
a
b2 + c2
+
b
c2 + a2
+
c
a2 + b2
≥
a + b + c
ab + bc + ca
+
3abc(a + b + c)
2(a3 + b3 + c3)(ab + bc + ca)
.
qive
a, b, c ≥ 0
€rove th—t
1
a2 + bc
+
1
b2 + ca
+
1
c2 + ab
≥
3
ab + bc + ca
+
81a2
b2
c2
2(a2 + b2 + c2)4
iqu—lity o™™ur if —nd if only
a = b = c, a = b, c = 0
it is true ˜e™—use
(1)
1
a2 + bc
+
1
b2 + ca
+
1
c2 + ab
≥
3 a2
+ b2
+ c2
a3b + ab3 + b3c + bc3 + c3a + ca3
—nd
(2)
3 a2
+ b2
+ c2
a3b + ab3 + b3c + bc3 + c3a + ca3
≥
3
ab + bc + ca
+
81a2
b2
c2
2(a2 + b2 + c2)4
.
fe™—use
a2
(a3b + ab3)
−
1
ab + bc + ca
=
abc(a + b + c)
(ab + bc + ca) ( (a3b + ab3))
,
2(a + b + c) a2
+ b2
+ c2 4
≥ 27abc(ab + bc + ca) a3
b + ab3
,
(a) (a + b + c) a2
+ b2
+ c2
≥ 9abc,
(b) a2
+ b2
+ c2
≥ ab + bc + ca,
(c) 2 a2
+ b2
+ c2 2
≥ 3 a3
b + ab3
,
whi™h
(c)
PH

is equiv—lent to
a2
− ab + b2
(a − b)2
≥ 0,
whi™h is trueF
PUD vet
a, b, c
˜e nonneg—tive num˜ersD no two of whi™h —re zeroF €rove th—t
a2
(b + c)
b2 + bc + c2
+
b2
(c + a)
c2 + ca + a2
+
c2
(a + b)
a2 + ab + b2
2(a2
+ b2
+ c2
)
a + b + c
.
SolutionX
a2
(b + c)
b2 + bc + c2
=
4a2
(b + c)(ab + bc + ca)
(b2 + bc + c2) (ab + bc + ca)
≥
4a2
(b + c)(ab + bc + ca)
(b2 + bc + c2 + ab + bc + ca)
2
=
4a2
(ab + bc + ca)
(b + c)(a + b + c)2
,
it suffi™es to prove
a2
b + c
≥
(a + b + c) a2
+ b2
+ c2
2(ab + bc + ca)
,
or
a2
b + c
+ a ≥
(a + b + c)3
2(ab + bc + ca)
,
or
a
b + c
≥
(a + b + c)2
2(ab + bc + ca)
,
whi™h is true ˜y g—u™hyEƒ™hw—rz inequ—lity
a
b + c
=
a2
a(b + c)
≥
(a + b + c)2
2(ab + bc + ca)
.
‡e just w—nt to give — little note hereF xoti™e th—t
a2
(b + c)
b2 + bc + c2
+
a(b + c)
a + b + c
=
a(b + c)(a2
+ b2
+ c2
+ ab + bc + ca)
(b2 + bc + c2)(a + b + c)
,
—nd
2(a2
+ b2
+ c2
)
a + b + c
+
a(b + c)
a + b + c
=
2(a2
+ b2
+ c2
+ ab + bc + ca)
a + b + c
.
„hereforeD the inequ—lity ™—n ˜e written in the form
a(b + c)
b2 + bc + c2
+
b(c + a)
c2 + ca + a2
+
c(a + b)
a2 + ab + b2
≥ 2,
xote th—t
cyc
a(b + c)
b2 + bc + c2
=
cyc
4a(b + c)(ab + bc + ca)
4(b2 + bc + c2)(ab + bc + ca) cyc
4a(ab + bc + ca)
(b + c)(a + b + c)2
.
PI

ƒo th—t ‡e h—ve to proveX
cyc
4a(ab + bc + ca)
(b + c)(a + b + c)2
2,
or
cyc
a
b + c
(a + b + c)2
2(ab + bc + ca)
,
whi™h is o˜viously true due to the g—u™hyEƒ™hw—rz inequ—lityF
„his is —nother new SolutionF pirstD ‡e will prove th—t
(a2 + ac + c2)(b2 + bc + c2) ≤
ab(a + b) + bc(b + c) + ca(c + a)
a + b
.(1)
indeedD using the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
√
ac ·
√
bc + a2 + ac + c2 · b2 + bc + c2 ≤ (ac + a2 + ac + c2)(bc + b2 + bc + c2)
= (a + c)(b + c).
it follows th—t
(a2 + ac + c2)(b2 + bc + c2) ≤ ab + c2
+ c a + b −
√
ab ≤ ab + c2
+ c a + b −
2ab
a + b
=
ab(a + b) + bc(b + c) + ca(c + a)
a + b
.
xowD from @IAD using the ewEqw inequ—lityD ‡e get
1
a2 + ac + c2
+
1
b2 + bc + c2
≥
2
(a2 + ac + c2)(b2 + bc + c2)
≥
2(a + b)
ab(a + b) + bc(b + c) + ca(c + a)
.
(2)
prom
(2)
‡e h—ve
a(b + c)
b2 + bc + c2
= ab
1
a2 + ac + c2
+
1
b2 + bc + c2
≥
2ab(a + b)
ab(a + b) + bc(b + c) + ca(c + a)
= 2.
PWD if
a, b, c > 0
then the following inequ—lity holdsX
a2
(b + c)
b2 + bc + c2
+
b2
(c + a)
c2 + ca + a2
+
c2
(a + b)
a2 + ab + b2
≥ 2
a3 + b3 + c3
a + b + c
„his inequ—lity is equiv—lent to
a2
(b + c)(a + b + c)
b2 + bc + c2
≥ 2 (a3 + b3 + c3) (a + b + c)
or
a2
+
a2
(ab + bc + ca)
b2 + bc + c2
≥ 2 (a3 + b3 + c3) (a + b + c),
PP

˜e™—use
2 (a3 + b3 + c3) (a + b + c) ≤ a2
+ b2
+ c2
+
a3
+ b3
+ c3
(a + b + c)
a2 + b2 + c2
,
a2
b2 + bc + c2
≥
a3
+ b3
+ c3
(a + b + c)
(a2 + b2 + c2) (ab + bc + ca)
,
˜y g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
a2
b2 + bc + c2
≥
a2
+ b2
+ c2 2
a2 (b2 + bc + c2)
=
a2
+ b2
+ c2 2
2 a2b2 + a2bc
,
a2
+ b2
+ c2 3
(ab + bc + ca) ≥ a3
+ b3
+ c3
(a + b + c) 2 a2
b2
+ a2
bc .
vet
A = a4
, B =
1
2
a3
b + ab3
, C = a2
b2
, D = a2
bc,
‡e h—ve
a2
+ b2
+ c2 2
= A + 2C,
a2
+ b2
+ c2
(ab + bc + ca) = 2B + D,
a3
+ b3
+ c3
(a + b + c) = A + 2B,
—nd
2 a2
b2
+ a2
bc = 2C + D.
(A + 2C) (2B + D) ≥ (A + 2B) (2C + D) ,
or
2 (A − D) (B − C) ≥ 0,
A ≥ D
—nd
B ≥ C
QHD qiven
a, b, c ≥ 0
su™h th—t
a + b + c = 1
€rove th—t
2 a2b + b2c + c2a + ab + bc + ca ≤ 1
‚ewrite the inform inequ—lity —s
2 a2b + b2c + c2a + ab + bc + ca ≤ (a + b + c)2
PQ

2 (a2b + b2c + c2a) (a + b + c) ≤ a2
+ b2
+ c2
+ ab + bc + ca
essume th—t ˜ is the num˜er ˜etien — —nd ™F „henD ˜y —pplying the ewEqw inequ—lityD ‡e
get
2 (a2b + b2c + c2a) (a + b + c) ≤
a2
b + b2
c + c2
a
b
+ b(a + b + c)
it is thus suffi™ient to prove the stronger inequ—lity
a2
+ b2
+ c2
+ ab + bc + ca ≥
a2
b + b2
c + c2
a
b
+ b(a + b + c)
„his inequ—lity is equiv—lent to
c(a − b)(b − c)
b
≥ 0,
whi™h is o˜viously true —™™ording to the —ssumption of
b
row to prove
a4
+ 2 a3
c ≥ a2
b2
+ 2 a3
b
only ˜y ewEqw iquiv—lent to prove
(a − b)2
(a + b)2
≥ 4(a − b)(b − c)(a − c)(a + b + c)
‡vyq ‡e ™—n —ssume th—t
a ≥ b ≥ c, a − b = x, b − c = y
then ‡e need to prove th—t
x2
(2c + 2y + x)2
+ y2
(2c + y)2
+ (x + y)2
(2c + x + y)2
≥ xy(x + y)(3c + 2x + y)
˜y
(x + y)4
≥ xy(x + y)(x + 2y)
—nd
(x + y)3
≥ 3xy(x + y)
‡e h—ve ™ompleted the Solution
QID vet
a, b, c
˜e positive num˜ers su™h th—t
a2
b2
+ b2
c2
+ c2
a2
≥ a2
b2
c2
pind the minimum of e
A =
a2
b2
c3(a2 + b2)
+
b2
c2
a3(b2 + c2)
+
c2
a2
b3(c2 + a2)
xo one like this pro˜lemc ƒetting
x =
1
a
, y =
1
b
, z =
1
c
PR

‡e h—ve
x2
+ y2
+ z2
≥ 1
‡e will prove th—t
x3
y2 + z2
+
y3
x2 + z2
+
z3
x2 + y2
≥
√
3
2
…sing g—u™hyEƒ™hw—rzX
LHS ≥
(x2
+ y2
+ z2
)2
x(y2 + z2) + y(x2 + z2) + z(x2 + y2)
fy ewEqw ‡e h—veX
x(y2
+z2
)+y(x2
+z2
)+z(x2
+y2
) ≤
2
3
(x2
+y2
+z2
)(x+y+z) ≤
2
√
3
(x2
+y2
+z2
) x2 + y2 + z2
fe™—use
x2
+ y2
+ z2
≥ 1
ƒo
(x2
+ y2
+ z2
)2
2√
3
(x2 + y2 + z2) x2 + y2 + z2
≥
√
3
2
‡e done3
QPF
vet xDyDz ˜e non neg—tive re—l num˜ers su™h th—t x2
+ y2
+ z2
= 1
F find the minimum —nd m—ximum of f = x + y + z − xyz.
Solution IF
pirst ‡e fix z —nd let m = x+y = x+
√
1 − x2 − z2 = g(x)(0 ≤ x ≤
√
1 − z2), then ‡e h—ve
g (x) = 1 −
x
√
1 − x2 − z2
,
‡e get
g (x) > 0 ⇔ 0 ≤ x <
1 − z2
2
—nd
g (x) < 0 ⇔
1 − z2
2
< x ≤ 1 − z2,
so ‡e h—ve
mmin = min{g(0), g( 1 − z2)} = 1 − z2
—nd
mmax = g
1 − z2
2
= 2 − 2z2.
e™tu—llyD f —nd written —s
f = f(m) = −
z
2
m2
+ m + 1 − z2 z
2
+ z,
e—sy to prove th—t the —xis of symmetry
m =
1
z
> 2 − 2z2
PS

so f(m) is in™re—sing in the interv—l of mD thusD ‡e h—ve
f(m) ≥ f( 1 − z2) = 1 − z2 + z
—nd
f(m) ≤ f( 2 − 2z2) =
z3
2
+
z
2
+ 2 − 2z2.
ƒin™e
( 1 − z2 + z)2
= 1 + 2z 1 − z2 ≥ 1
‡e get f(m) ≥ 1 —nd when two of xDyDz —re zero ‡e h—ve f = 1, soWegetfmin = 1.
vet
h(z) =
z3
2
+
z
2
+ 2 − 2z2,
e—sy to prove th—t
h (z) > 0 ⇔ 0 ≤ z <
1
√
3
andh (z) < 0 ⇔
1
√
3
< z ≤ 1
then ‡e get
f(m) ≤ h
1
√
3
=
8
√
3
9
,
when x = y = z =
1
√
3
Wehavef =
8
√
3
9
D so ‡e getfmax =
8
√
3
9
.
honeF
Solution PF
‡hen two of xDyDz —re zero ‡e h—vef = 1D —nd ‡e will prove th—t f ≥ 1 then ‡e ™—n get
fmin = 1F e™tu—llyD ‡e h—ve
f ≥ 1 ⇔ x + y + z − xyz ≥ 1 ⇔ (x + y + z) x2
+ y2
+ z2
− xyz ≥
x2 + y2 + z2
3
⇔ (x + y + z) x2
+ y2
+ z2
− xyz
2
≥
x2
+ y2
+ z2 3
⇔ x2
y2
z2
+ 2
sym
x5
y + x3
y3
+ x3
y2
z ≥ 0,
the l—st inequ—lity is o˜vious trueD so ‡e got f ≥ 1; ‡henx = y = z =
1
√
3
‡e h—ve
f =
8
√
3
9
,
—nd ‡e will prove th—t
f ≤
8
√
3
9
then ‡e ™—n get
fmax =
8
√
3
9
e™tu—llyD ‡e h—ve
f ≤
8
√
3
9
⇔ x + y + z − xyz ≤
8
√
3
9
⇔ (x + y + z) x2
+ y2
+ z2
− xyz ≤
8
√
3
9
x2 + y2 + z2
3
⇔ 27 (x + y + z) x2
+ y2
+ z2
− xyz
2
≤ 64 x2
+ y2
+ z2 3
⇔
1
4 cyc
S (x, y, z) (y − z)
2
≥ 0,
PT

where
S(x, y, z) = 17y2
(2y−x)2
+17z2
(2z−x)2
+56y2
(z−x)2
++56z2
(y−x)2
+24x4
+6y4
+6z4
+57x2
(y2
+z2
)+104y2
z2
is o˜vious positiveD so the l—st inequ—lity is o˜vious trueD so ‡e gotfmax =
8
√
3
9
.
QQD por positive re—l num˜ersD show th—t
a3
(b + c − a)
a2 + bc
+
b3
(c + a − b)
b2 + ca
+
c3
(a + b − c)
c2 + ab
≤
ab + bc + ca
2
ineq
a2
+ b2
+ c2
+
ab + bc + ca
2
≥
a3
(b + c − a)
a2 + bc
+ a2
a2
+ b2
+ c2
+
ab + bc + ca
2
≥ (ab + bc + ca)(
a2
a2 + bc
)
a2
+ b2
+ c2
+ (ab + bc + ca)(
bc
a2 + bc
) ≥
5
2
(ab + bc + ca)
a2
+ b2
+ c2
ab + bc + ca
+
bc
a2 + bc
≥
5
2
…se two ineq
bc
a2 + bc
+
ab
c2 + ab
+
ac
b2 + ac
≥
4abc
(a + b)(b + c)(c + a)
+ 1(1)
it is e—sy to proveF
a2
+ b2
+ c2
ab + bc + ca
+
8abc
(a + b)(b + c)(c + a)
≥ 2(2)
ƒo e—sy to see th—t
a2
+ b2
+ c2
ab + bc + ca
+
bc
a2 + bc
≥
a2
+ b2
+ c2
ab + bc + ca
+
4abc
(a + b)(b + c)(c + a)
+ 1
≥
a2
+ b2
+ c2
2(ab + bc + ca)
+ 2 ≥
5
2
‡e h—ve done 3
a3
(b + c − a)
a2 + bc
+
b3
(c + a − b)
b2 + ca
+
c3
(a + b − c)
c2 + ab
≤
3abc(a + b + c)
2(ab + bc + ca)
Solution
a2
+ b2
+ c2
ab + bc + ca
+
bc
a2 + bc
+
3abc(a + b + c)
2(ab + bc + ca)
2 ≥ 3
end ‡e prove th—t
3abc(a + b + c)
2(ab + bc + ca)
2 ≥
4abc
(a + b)(b + c)(c + a)
3(a + b + c)(a + b)(b + c)(c + a) ≥ 8(ab + bc + ca)2
„his ineq is true ˜e™—use
3(a + b + c)(a + b)(b + c)(c + a) ≥
8
3
(a + b + c)2
(ab + bc + ca) ≥ 8(ab + bc + ca)2
PU

ƒo
LHS ≥
a2
+ b2
+ c2
ab + bc + ca
+
4abc
(a + b)(b + c)(c + a)
+
4abc
(a + b)(b + c)(c + a)
+ 1 ≥ 3
vet
a, b, c > 0
ƒhow th—t
a3
(b + c − a)
a2 + bc
+
b3
(c + a − b)
b2 + ca
+
c3
(a + b − c)
c2 + ab
≤
9abc
2(a + b + c)
pirstD‡e prove this lenm—X
a2
a2 + bc
+
b2
b2 + ca
+
c2
c2 + ab
≤
(a + b + c)2
2(ab + bc + ca)
bc
a2 + bc
+
ac
b2 + ac
+
ab
c2 + ab
+
a2
+ b2
+ c2
2(ab + bc + ca)
≥ 2
whi™h is true from
bc
a2 + bc
+
ac
b2 + ac
+
ab
c2 + ab
≥ 1 +
4abc
(a + b)(b + c)(c + a)
a2
+ b2
+ c2
2(ab + bc + ca)
+
4abc
(a + b)(b + c)(c + a)
≥ 1
equ—lity o™™ur if —nd if only
a = b = c
or
a = b, c = 0
‚eturn to your inequ—lityD‡e h—ve
(
a3
(b + c − a)
a2 + bc
+ a2
) ≤ a2
+ b2
+ c2
+
9abc
2(a + b + c)
or
(ab + bc + ca)
a2
a2 + bc
≤ a2
+ b2
+ c2
+
9abc
2(a + b + c)
prom
a2
a2 + bc
+
b2
b2 + ca
+
c2
c2 + ab
≤
(a + b + c)2
2(ab + bc + ca)
‡e only need to prove th—t
(a + b + c)2
2
≤ a2
+ b2
+ c2
+
9abc
2(a + b + c)
or
a2
+ b2
+ c2
+
9abc
a + b + c
≥ 2(ab + bc + ca)
‡hi™h is s™hur inequ—lityF yur Solution —re ™ompleted equ—lity o™™ur if —nd if only
a = b = c, a = b, c = 0
QQD vet
a, b, c > 0
PV

su™h th—t
a + b + c = 1
„hen
a3
+ bc
a2 + bc
+
b3
+ ca
b2 + ca
+
c3
+ ab
c2 + ab
≥ 2
prom the ™ondition
a − 1 = −(b + c)
it follows th—t
a3
+ bc
a2 + bc
= −
a2
(b + c)
a2 + bc
+ 1
„hus it suffi™es to prove th—t
a + b + c ≥
a2
(b + c)
a2 + bc
por
a, b, c
positive re—ls prove th—t
ab(a + b)
c2 + ab
≥ a
<=> a4
+ b4
+ c4
− b2
c2
− c2
a2
− a2
b2
≥ 0
ab(a + b)
c2 + ab
+
c2
(a + b)
c2 + ab
= 2 a
—nd our inequ—lity ˜e™omes
c2
(a + b)
(c2 + ab)
≤ a
˜ut
c2
(a + b)
(c2 + ab)
=
c2
(a + b)2
(c2 + ab)(a + b)
=
(ca + cb)2
a(b2 + c2) + b(a2 + c2)
≤
c2
a2
a(b2 + c2)
+
c2
b2
b(a2 + c2)
= a
QR
vet
a, b, c ≥ 0
su™h th—t
a + b + c = 1
„hen
6(a2
+ b2
+ c2
) ≥
a3
+ bc
a2 + bc
+
b3
+ ca
b2 + ca
+
c3
+ ab
c2 + ab
Solution
6(a2
+ b2
+ c2
) +
a2
(b + c)
a2 + bc
≥ 3
6(a2
+ b2
+ c2
) − 2(a + b + c)2
≥ (a−
a2
(b + c)
a2 + bc
)
4 (a − b)(a − c) ≥
a(a − b)(a − c)
a2 + bc
(a − b)(a − c)(4 −
a
a2 + bc
) ≥ 0
PW

essuming ‡vyq
a ≥ b ≥ c
then e—sy to see th—t
4 −
a
a2 + bc
≥ 0
—nd
4 −
c
c2 + ab
≥ 0
(c − a)(c − b)(4 −
c
c2 + ab
) ≥ 0and(a − b)(a − c)(4 −
a
a2 + bc
) ≥ 0
‡e h—ve two ™—ses
g—se I
4 −
b
b2 + ac
≤ 0
then
(b − c)(b − a)(4 −
b
b2 + ac
) ≥ 0
so this ineq is true
g—se P
4 −
b
b2 + ac
≤ 0
e—sy to see th—t
4 −
c
c2 + ab
≥ 4 −
b
b2 + ac
ƒo
LHS ≥ (c − b)2
(4 −
c
c2 + ab
) + (a − b)(b − c)(
b
b2 + ac
−
c
c2 + ab
) ≥ 0
FiFh
QSF vet
x, y, z
˜e re—l num˜ers s—tisfyX
x2
y2
+ 2yx2
+ 1 = 0
pind the m—ximum —nd minimum v—lues ofX
f(x, y) =
2
x2
+
1
x
+ y(y +
1
x
+ 2)
SolutionX €ut
t =
1
x
; k = y + 1
D ‡e h—veX
t2
+ k2
= 1
f(x, y) = t2
+ tk
€ut
t = cos α; k = sin α
then
f(x, y) = cos α2
+ cos α sinα =
sin 2α2 =
1
2
+
1
√
2
cos (2α −
π
4
)
max f(x, y) =
1
2
+
1
√
2
QH

min f(x, y) =
1
2
−
1
√
2
FiFh
F
QTF
ƒuppose —D˜D™Dd —re positive integers with ab + cd = 1.
„henD por We = 1, 2, 3, 4,let (xi)2
+ (yi)2
= 1, where xi —nd yi —re re—l num˜ersF
ƒhow th—t
(ay1 + by2 + cy3 + dy4)2
+ (ax4 + bx3 + cx2 + dx1)2
≤ 2(
a
b
+
b
a
+
c
d
+
d
c
).
ƒolitionX
…se g—u™hyEƒ™hw—rtz D ‡e h—ve
(ay1 + by2 + cy3 + dy4)2
≤
(ab + cd)(
(ay1 + by2)2
ab
+
(cy3 + dy4)2
cd
) =
(ay1 + by2)2
ab
+
(cy3 + dy4)2
cd
ƒimil—rX
(ax4 + bx3 + cx2 + dx1)2
≤
(ab + cd)(
(ax4 + bx3)2
ab
+
(cx2 + dx1)2
cd
)
=
(ax4 + bx3)2
ab
+
(cx2 + dx1)2
cd
futX
(ay1 + by2)2
≤ (ay1 + by2)2
+ (ax1 − bx2)2
= a2
+ b2
+ 2ab(y1y2 − x1x2)
ƒimil—rF
(cx2 + dx1)2
≤ c2
+ d2
+ 2cd(x1x2 − y1y2)
D then ‡e getX
(ay1 + by2)2
ab
+
(cx2 + dx1)2
cd
≤
a
b
+
b
a
+
c
d
+
d
c
@IA
„he s—me —rgument show th—tX
(cy3 + dy4)2
cd
+
(ax4 + bx3)2
ab
≤
a
b
+
b
a
+
c
d
+
d
c
@PA
gom˜ining @IAY@PA ‡e get F FiFh
QUF
in —ny ™onvex qu—dril—ter—l with sides
a ≤ b ≤ c ≤ d
QI

—nd —re— F
€rove th—t X
F ≤
3
√
3
4
c2
SolutionX
„he inequ—lity is rewritten —sX
(−a + b + c + d)(a − b + c + d)(a + b − c + d)(a + b + c − d) ≤ 27c4
.
‡e su˜stitute x = −a + b + c + dD y = a − b + c + dD z = a + b − c + dD t = a + b + c − dF
„hen
x + y − z + t
4
= c —nd x ≥ y ≥ z ≥ t.
„hus ‡e h—veX xyzt ≤ 27(
x + y − z + t
4
)4
.
„he left side of the inequ—lity is m—ximum when z = y
while the right side of the inequ—lity is minimum @‡e h—ve fixed xDy —nd tAF
„hen ‡e just prove th—t xy2
t ≤ 27(
x + t
4
)4
.
fe™—use xy2
t ≤ x3
tD ‡e just h—ve to prove
x3
t ≤ (
x + t
4
)4
end then it follows th—t the —˜ove inequ—lity is —lso trueF
x
3
+
x
3
+
x
3
+ t ≥
4 4
x
3
·
x
3
·
x
3
· t
hen™e
27(
x + y
4
)4
≥ x3
t
QVF
vet efg ˜e — tri—ngleF €rove th—tX
1
a
+
1
b
+
1
c
≤
1
a + b − c
+
1
b + c − a
+
1
c + a − b
SolutionX
IF
1
a + b − c
+
1
b + c − a
=
2b
(a + b − c)(b + c − a)
=
2b
b2 − (c − a)2
≥
2b
b2
=
2
b
ƒimil—rlyD ‡e h—ve
1
b + c − a
+
1
c + a − b
≥
2
c
1
c + a − b
+
1
a + b − c
≥
2
a
edd three inequ—lities together —nd divide ˜y P to get the desired resultF
PF
use u—r—m—t— for the num˜er —rr—ys (b + c − a; c + a − b; a + b − c) (a; b; c)
—nd the ™onvex fun™tion
f (x) =
1
x
QP

yr m—ke the su˜stitution x = 1
2 (b + c − a)D y = 1
2 (c + a − b)D
z = 1
2 (a + b − c) —nd get
a = y + z, b = z + x, c = x + y,
so th—t the inequ—lity in question ™—n ˜e rewritten —s
1
y + z
+
1
z + x
+
1
x + y
≤
1
2x
+
1
2y
+
1
2z
D
wh—t dire™tly follows from ewErwX
2
y + z
≤
1
2y
+
1
2z
,
2
z + x
≤
1
2z
+
1
2x
,
2
x + y
≤
1
2x
+
1
2y
QWF
vet a, b, c ˜e nonneg—tive re—l num˜ersF €rove th—t
a3
+ b3
+ c3
+ 3abc ≥
(a2
b + b2
c + c2
a)2
ab2 + bc2 + ca2
+
(ab2
+ bc2
+ ca)2
a2b + b2c + c2a
SolutionX
if a = 0 or b = 0 or c = 0 Dit9s trueFif
abc > 0
€ut x = a
b , y = b
c , z = c
a F‡e need prove
x
z
+
y
x
+
z
y
+ 3 ≥
(xy + yz + zx)2
xyz(x + y + z)
+
(x + y + z)2
xy + yz + zx
x
z
+
y
x
+
z
y
≥
x2
y2
+ y2
z2
+ z2
x2
xyz(x + y + z)
+
x2
+ y2
+ z2
xy + yz + zx
x2
z
+
z2
y
+
y2
x
≥
(x2
+ y2
+ z2
)(x + y + z)
xy + yz + zx
x3
y
z
+
y3
z
x
+
z3
x
y
≥ x2
y + y2
z + z2
x
fy using ew qw9inequ—lityD ‡e h—veX
x3
y
z
+ xyz ≥ 2x2
y,
y3
z
x
+ xyz ≥ 2y2
z
z3
x
y
+ xyz ≥ 2z2
y, x2
y + y2
z + z2
x ≥ 3xyz
‡e h—ve done
RHF
vet x, y, z ˜e positive re—l num˜ersF €rove th—tX
x +
1
y
− 1 y +
1
z
− 1 ≥ 3.
SolutionF
‡e rewrite the inequ—lity —s
y
x
+
1
xyz
− 2 x + 1 −
2
xyz
xy + 3 ≥ 0.
QQ

€utting xyz = k3
D then there exist a, b, c > 0 su™h th—t x = ka
b , y = kb
c , z = kc
a .
„he inequ—lity ˜e™omes
a2
bc
+
1
k2
− 2k
a
b
+ k2
−
2
k
b
a
+ 3 ≥ 0
f(k) = a3
+ k2
−
2
k
a2
b +
1
k2
− 2k ab2
+ 3abc ≥ 0
‡e h—ve th—t
f (k) =
2(k3
+ 1)
k3
k a2
b − ab2
f (k) = 0 ⇔ k =
ab2
a2b
.
prom nowD —™™ording to the †—ri—tion fo—rdD ‡e ™—n dedu™e our inequ—lity to show th—t
f
ab2
a2b
≥ 0
or equiv—lentlyD
a3
+ b3
+ c3
+ 3abc ≥
(a2
b + b2
c + c2
a)2
ab2 + bc2 + ca2
+
(ab2
+ bc2
+ ca2
)2
a2b + b2c + c2a
.
FiFh
RIF
qiven a, b, c ≥ 0F€rove th—tX
(a + b + c)2
2(ab + bc + ca)
≥
a2
a2 + bc
+
b2
b2 + ca
+
c2
c2 + ab
SolutionX
‡e h—ve
2a2
(a + b)(a + c)
−
a2
a2 + bc
=
a2
(a − b)(a − c)
(a + b)(a + c)(a2 + bc)
≥ 0
@e—sy to ™he™k ˜y †orni™u ƒ™hurA it suffi™es to prove th—t
(a + b + c)2
2(ab + bc + ca)
≥
2a2
(a + b)(a + c)
=
2 ab(a + b)
(a + b)(b + c)(c + a)
essume th—t a + b + c = 1 —nd put q = ab + bc + ca, r = abcD then the inequ—lity ˜e™omes
1
4q
≥
q − 3r
q − r
⇔
q − r
q − 3r
≥ 4q
⇔
2r
q − 3r
≥ 4q − 1
fy ƒ™hur9s inequ—lity for third degreeD ‡e h—ve r ≥ 4q−1
9 D then
2r
q − 3r
≥
2r
q − 4q−1
3
=
6r
1 − q
QR

it suffi™es to show th—t
6r ≥ (4q − 1)(1 − q)
fut this is just ƒ™hur9s inequ—lity for fourth degree
a4
+ abc a ≥ ab(a2
+ b2
)
‡e h—ve doneF
PF
ƒuppose a + b + c = 3F ‡e need to proveX
f(r) = 4q4
− 9q3
+ 24qr2
− 54q2
r − 72r2
− 243r + 216qr ≤ 0
f (r) = 48qr − 54q2
− 144r − 243 + 216q
f (r) = 48(q − 3) ≤ 0, sof (r) ≤ f (0) = −54q2
− 144 + 216q ≤ 0
ƒoD with q ≤ 9
4 , f(r) ≤ f(0) = q3
(4q − 9) ≤ 0
‡ith q ≥ 9
4 D ‡e h—veX f(r) ≤ f(4q−9
3 ) ≤ 0 @trues with q ≥ 9
4 A
RPF
vet a, b, c ˜e nonneg—tive re—l num˜ers su™h th—t a2
+ b2
+ c2
= 1F €rove th—t
a3
b2 − bc + c2
+
b3
+ c3
a2
≥
√
2
SolutionX
a3
b2 − bc + c2
+
b3
+ c3
a2
≥
(a2
+ b2
+ c2
)2
a[b2 − bc + c2 + a(b + c)]
≥
1
2.a[3−2a2
4 ]
=
1
2 a2(
3
2 −a2
2 )2
≥
√
2
RQF
vet ∆ABC —nd max(A, B, C) ≤ 90F €rove th—t X
cosAcosB
sin2C
+
cosBcosC
sin2A
+
cosCcosA
sin2B
≥
√
3
2
SolutionX
fut if A = 90◦
the left side does not existF
if max{A, B, C} < 90◦
F vet a2
+ b2
− c2
= z, a2
+ c2
− b2
= y —nd b2
+ c2
− a2
= x.
ren™eD x, y —nd z —re positive —nd
cyc
cos A cos B
sin 2C
=
cyc
b2
+c2
−a2
2bc · a2
+c2
−b2
2ac
2 · 2S
ab · a2+b2−c2
2ab
=
=
cyc
ab(b2
+ c2
− a2
)(a2
+ c2
− b2
)
8c2S(a2 + b2 − c2)
=
cyc
xy
(x + y)z
a2b2
2(a2b2 + a2c2 + b2c2) − a4 − b4 − c4
=
=
cyc
xy
(x + y)z
(x + z)(y + z)
4(xy + xz + yz)
.
„husD it rem—ins to prove th—t
cyc
xy
(x + y)z
(x + z)(y + z)
xy + xz + yz
≥
√
3,
QS

whi™h is equiv—lent to
cyc
x2
y2
(x + z)(y + z)
x + y
≥ xyz 3(xy + xz + yz).
fy g—u™hyEƒ™hw—rtz ‡e o˜t—inX
cyc
x2
y2
(x + z)(y + z)
x + y
·
cyc
x + y
(x + z)(y + z)
≥ (xy + xz + yz)2
.
ren™eD ‡e need to prove th—t
(xy + xz + yz)2
≥
cyc
x + y
(x + z)(y + z)
· xyz 3(xy + xz + yz).
‡e o˜t—inX
(xy + xz + yz)2
≥
cyc
x + y
(x + z)(y + z)
· xyz 3(xy + xz + yz) ⇔
⇔
√
xy + xz + yz
3
(x + y)(x + z)(y + z) ≥
cyc
xyz(x + y) 3(x + y) ⇔
⇔ (xy + xz + yz)3
(x + y)(x + z)(y + z) ≥ 3x2
y2
z2
cyc
(x + y)3
+
+6x2
y2
z2
cyc
(x + y)(x + z) (x + y)(x + z) ⇔
⇔ (xy + xz + yz)3
(x + y)(x + z)(y + z) ≥ 3x2
y2
z2
cyc
(2x3
+ 3x2
y + 3x2
z)+
+3x2
yz
cyc
(x + y)(x + z)2 z2(x + y)y2(x + z).
fy ewEqw 2 z2(x + y)y2(x + z) ≤ y2
x + y2
z + z2
x + z2
y.
ren™eD it rem—ins to prove th—t
⇔ (xy + xz + yz)3
(x + y)(x + z)(y + z) ≥ 3x2
y2
z2
cyc
(2x3
+ 3x2
y + 3x2
z)+
+3x2
yz cyc(x + y)(x + z)(y2
x + y2
z + z2
x + z2
y), whi™h is equiv—lent to
sym(x5
y4
+ x5
y3
z − 5x4
y3
z2
+ x4
y4
z + 2x3
y3
z3
) ≥ 0, whi™h is true ˜y ewEqw ˜e™—use
sym
(x5
y4
+ x5
y3
z − 5x4
y3
z2
+ x4
y4
z + 2x3
y3
z3
) ≥ 0 ⇔
⇔
sym
(x5
y4
+ x5
y3
z +
1
3
x4
y4
z +
2
3
x4
z4
y + 2x3
y3
z3
) ≥
sym
5x4
y3
z2
.
FiFh
RRF
vet a, b, c ˜e nonneg—tive re—l num˜ersD no two of whi™h —re zeroF €rove th—t
a
b + c
+
b
c + a
+
c
a + b
+
a2
b + b2
c + c2
a
ab2 + bc2 + ca2
≥
5
2
QT

Solution
IFFF—sume p = 1 —nd vemm— ab2
+ bc2
+ ca2
≤ 4
27 − r
⇔
a
b + c
+
b
c + a
+
c
a + b
+
ab(a + b)
ab2 + bc2 + ca2
≥
7
2
‡e h—ve
a
b + c
+
b
c + a
+
c
a + b
+
ab(a + b)
ab2 + bc2 + ca2
≥
1 − 2q + 3r
q − r
+
27q − 81r
4 − 27r
‡e need prove th—t
1 − 2q + 3r
q − r
+
27q − 81r
4 − 27r
≥
7
2
⇔
1 + r
q − r
− 2 +
27q − 12
4 − 27r
+ 3 ≥
7
2
⇔
1 + r
q − r
+
27q − 12
4 − 27r
≥
5
2
⇔ −135r2
+ r(81q + 2) + (54q2
+ 8 − 44q) ≥ 0
f(r) = −135r2
+ r(81q + 2) + (54q2
+ 8 − 44q)
f (r) = −270r + 81q + 2 ≥ 0 @˜e™—use q ≥ 9rA
⇒ f(r) ≥ f(
4q − 1
9
) =
570q2
− 349q + 55
9
≥ 0
PFFFFFFFF
a
b + c
+
b
c + a
+
c
a + b
+
a2
b + b2
c + c2
a
ab2 + bc2 + ca2
≥
5
2
⇔
⇔
cyc
a
b + c
−
1
2
≥
(a2
c − a2
b)
a2c
⇔
⇔
cyc
a − b − (c − a)
2(b + c)
≥
(a − b)(b − c)(c − a)
a2c + b2a + c2b
⇔
⇔
cyc
a − b
2
1
b + c
−
1
a + c
≥
(a − b)(b − c)(c − a)
a2c + b2a + c2b
⇔
⇔
cyc
(a − b)2
(a + c)(b + c)
≥
2(a − b)(b − c)(c − a)
a2c + b2a + c2b
.
if (a − b)(b − c)(c − a) ≤ 0 then the inequ—lity holdsF
vet (a − b)(b − c)(c − a) > 0 —nd a2
c+b2
a+c2
b
a2b+b2c+c2a = t. „hen t > 1.
fy ewEqw ‡e o˜t—inX
cyc
(a − b)2
(a + c)(b + c)
≥ 3 3 (a − b)2(b − c)2(c − a)2
(a + b)2(a + c)2(b + c)2
.
27(a2
c + b2
a + c2
b)3
≥ 8(a + b)2
(a + c)2
(b + c)2
(a − b)(b − c)(c − a).
QU

fut
(a + b)(a + c)(b + c) =
cyc
(a2
b + a2
c) + 2abc ≤
4
3 cyc
(a2
b + a2
c).
id estD it rem—ins to prove th—t 27t3
≥ 8 · 16
9 (t + 1)2
(t − 1), whi™h o˜viousF
RSF
por —ll nonneg—tive re—l num˜ers a, b —nd cD no two of whi™h —re zeroD
1
(a + b)2
+
1
(b + c)2
+
1
(c + a)2
≥
3 3abc(a + b + c)(a + b + c)2
4(ab + bc + ca)3
Solution ‚epl—™ing a, b, c ˜y 1
a , 1
b , 1
c respe™tivelyD ‡e h—ve to prove th—t
a2
b2
(a + b)2
≥
3 3(ab + bc + ca)(ab + bc + ca)2
4(a + b + c)3
.
xowD using g—u™hy ƒ™hw—rz inequ—lityD ‡e h—ve
a2
b2
(a + b)2
≥
(ab + bc + ca)2
(a + b)2 + (b + c)2 + (c + a)2
=
(ab + bc + ca)2
2(a2 + b2 + c2 + ab + bc + ca)
.
(ab + bc + ca)2
2(a2 + b2 + c2 + ab + bc + ca)
≥
3 3(ab + bc + ca)(ab + bc + ca)2
4(a + b + c)3
or equiv—lentlyD
2(a + b + c)3
≥ 3 3(ab + bc + ca)(a2
+ b2
+ c2
+ ab + bc + ca),
th—t is
4(a + b + c)6
≥ 27(ab + bc + ca)(a2
+ b2
+ c2
+ ab + bc + ca)2
fy ewEqwD ‡e see th—t
27(ab + bc + ca)(a2
+ b2
+ c2
+ ab + bc + ca)2
≤
1
2
2(ab + bc + ca) + (a2
+ b2
+ c2
+ ab + bc + ca) + (a2
+ b2
+ c2
+ ab + bc + ca)
3
= 4(a+b+c)6
.
„hereforeD our Solution is ™ompleted
RTF
2
3
(
1
a2 + bc
+
1
b2 + ca
+
1
c2 + ab
) ≥
1
ab + bc + ca
+
2
a2 + b2 + c2
SolutionX
‚ewrite our inequ—lity —sX
1
a2 + bc
≥
3(a + b + c)2
2(a2 + b2 + c2)(ab + bc + ca)
.
‡e will ™onsider P ™—sesX g—se IF a2
+b2
+c2
≤ 2(ab+bc+ca), then —pplying g—u™hy ƒ™hw—rz
inequ—lityD ‡e ™—n redu™e our inequ—lity to
6
a2 + b2 + c2 + ab + bc + ca
≥
(a + b + c)2
(a2 + b2 + c2)(ab + bc + ca)
,
(a2
+ b2
+ c2
− ab − bc − ca)(2ab + 2bc + 2ca − a2
− b2
− c2
) ≥ 0, whi™h is trueF
g—se PF a2
+ b2
+ c2
≥ 2(ab + bc + ca), then (a + b + c)2
≤ 2(a2
+ b2
+ c2
), whi™h yields th—t
3(a + b + c)2
2(a2 + b2 + c2)(ab + bc + ca)
≤
3
ab + bc + ca
,
QV

—nd ‡e just need to prove th—t
1
a2 + bc
+
1
b2 + ca
+
1
c2 + ab
≥
3
ab + bc + ca
,
whi™h is just your very known @—nd ni™eA inequ—lityF
RUF
vet a, b, c ˜e nonneg—tive re—l num˜ersD no two of whi™h —re zeroF €rove th—t
(a)
1
2a2 + bc
+
1
2b2 + ca
+
1
2c2 + ab
≥
1
ab + bc + ca
+
2
a2 + b2 + c2
SolutionX
Ist Solution
fy g—u™hy inequ—lityD
cyc
(b + c)2
(2a2
+ bc)
cyc
1
2a2 + bc
≥ 4(a + b + c)2
it rem—ins to show th—t
cyc
(b + c)2
(2a2
+ bc) ≤ 4(a2
+ b2
+ c2
)(ab + bc + ca)
whi™h is e—syF Pnd SolutionF
ƒin™e
cyc
ab + bc + ca
2a2 + bc
=
cyc
bc
2a2 + bc
+
cyc
a(b + c)
2a2 + bc
‡e need to show
cyc
bc
2a2 + bc
≥ 1
—nd
cyc
a(b + c)
2a2 + bc
≥
2(ab + bc + ca)
a2 + b2 + c2
„he former is illEknownX if x, y, z ≥ 0 su™h th—t xyz = 1D then
1
2x + 1
+
1
2y + 1
+
1
2z + 1
≥ 1
„he l—terX ˜y g—u™hy inequ—lityD
cyc
a(b + c)(2a2
+ bc)
cyc
a(b + c)
2a2 + bc
≥ 4(ab + bc + ca)2
„he result then follows from the following identity
cyc
a(b + c)(2a2
+ bc) = 2(ab + bc + ca)(a2
+ b2
+ c2
)
Qrd SolutionF
LHS − RHS
a + b + c
=
2(a + b + c)(a − b)2
(b − c)2
(c − a)2
+ 3abc cyc(a2
+ ab + b2
)(a − b)2
(2a2 + bc)(2b2 + ca)(2c2 + ab)(ab + bc + ca)(a2 + b2 + c2)
Qrd SolutionF
essume th—t c = min{a, b, c}D then the g—u™hy ƒ™hw—rz inequ—lity yields
1
2a2 + bc
+
1
2b2 + ca
≥
4
2(a2 + b2) + c(a + b)
,
QW

then ‡e just need to prove th—t
4
2(a2 + b2) + c(a + b)
+
1
ab + 2c2
≥
1
ab + bc + ca
+
2
a2 + b2 + c2
,
or equiv—lently
c(a + b − 2c)
(ab + 2c2)(ab + bc + ca)
≥
2c(a + b − 2c)
(a2 + b2 + c2)(2a2 + 2b2 + ac + bc)
,
th—t is
(a2
+ b2
+ c2
)(2a2
+ 2b2
+ ac + bc) ≥ 2(ab + 2c2
)(ab + bc + ca),
whi™h is true sin™e a2
+ b2
+ c2
≥ ab + bc + ca —nd 2a2
+ 2b2
+ ac + bc ≥ 2(ab + 2c2
). honeF
RVF
vet a, b, c ˜e positive re—l num˜er F €rove th—tX
(c) 2(
1
a2 + 8bc
+
1
b2 + 8ca
+
1
c2 + 8ab
) ≥
1
ab + bc + ca
+
1
a2 + b2 + c2
SolutionX
‚epl—™ing a, b, c ˜y 1
a , 1
b , 1
c respe™tivelyD ‡e ™—n rewrite our inequ—lity —s
4(a + b + c)
a
8a2 + bc
+
b
8b2 + ca
+
c
8c2 + ab
≥ 2 +
2abc(a + b + c)
a2b2 + b2c2 + c2a2
.
xowD —ssume th—t c = mina, b, cD then ‡e h—ve the following estim—tionsX
a(4a + 4b + c)
8a2 + bc
+
b(4a + 4b + c)
8b2 + ca
− 2 =
(a − b)2
(32ab − 12ac − 12bc + c2
)
(8a2 + bc)(8b2 + ca)
≥ 0,
—nd
2abc(a + b + c)
a2b2 + b2c2 + c2a2
≤
2c(a + b + c)
ab + 2c2
.
‡ith these estim—tionsD ‡e ™—n redu™e our inequ—lity to
3ac
8a2 + bc
+
3bc
8b2 + ca
+
4c(a + b + c)
8c2 + ab
≥
2c(a + b + c)
ab + 2c2
,
or
3a
8a2 + bc
+
3b
8b2 + ca
≥
2(a + b + c)(4c2
− ab)
(ab + 2c2)(ab + 8c2)
.
e™™ording to g—u™hy ƒ™hw—rz inequ—lityD ‡e h—ve
3a
8a2 + bc
+
3b
8b2 + ca
≥
12
8(a + b) + c a
b + b
a
.
6
8(a + b) + c a
b + b
a
≥
(a + b + c)(4c2
− ab)
(ab + 2c2)(ab + 8c2)
.
if 4c2
≤ abD then it is trivi—lF ytherwiseD ‡e h—ve
a + b ≤ c + ab
c , —nd a
b + b
a ≤ ab
c2 + c2
ab . ‡e need to prove
6
8 c + ab
c + c ab
c2 + c2
ab
≥
2c + ab
c (4c2
− ab)
(ab + 2c2)(ab + 8c2)
,
RH

whi™h isD —fter exp—ndingD equiv—lent to
(9ab − 4c2
)(ab − c2
)2
c(ab + 8c2)(c4 + 8abc2 + 9a2b2)
≥ 0,
whi™h is true —s c = mina, b, c.
yur Solution is ™ompletedF
RWF
vet a, b, c > 0F€rove th—tX
5
3
(
1
4a2 + bc
+
1
4b2 + ca
+
1
4c2 + ab
) ≥
2
ab + bc + ca
+
1
a2 + b2 + c2
SolutionX
essume th—t c = min(a, b, c)D then ‡e h—ve the following estim—tionsX
1
4a2 + bc
+
1
4b2 + ca
−
4
8ab + ac + bc
=
(a − b)2
(32ab − 12ac − 12bc + c2
)
(4a2 + bc)(4b2 + ca)(8ab + ac + bc)
≥ 0,
—nd
1
a2 + b2 + c2
≤
1
2ab + c2
.
…sing theseD ‡e ™—n redu™e our inequ—lity to
20
8ab + ac + bc
+
5
ab + 4c2
≥
6
ab + ac + bc
+
3
2ab + c2
.
henote x = a + b ≥ 2
√
ab then this inequ—lity ™—n ˜e rewritten —s
f(x) =
20
cx + 8ab
−
6
cx + ab
+
5
ab + 4c2
−
3
2ab + c2
≥ 0.
‡e h—ve
f (x) =
6c
(cx + ab)2
−
20
(cx + 8ab)2
≥
20c
(cx + ab)(cx + 8ab)
−
20c
(cx + 8ab)2
=
140abc
(cx + ab)(cx + 8ab)2
≥ 0.
„his shows th—t f(x) is in™re—singD —nd ‡e just need to prove th—t f(2
√
ab) ≥ 0, whi™h is
equiv—lent to
7c(13t2
+ 6tc + 8c2
)(t − c)2
t(t + 2c)(4t + c)(2t2 + c2)(t2 + 4c2)
≥ 0,
‡here
t =
√
ab
„his is o˜viously nonneg—tiveD so our Solution is ™ompletedF
SHF
vet a, b —nd c re—l num˜ers su™h th—t a + b + c + d = e = 0F €rove th—tX
30(a4
+ b4
+ c4
+ d4
+ e4
) ≥ 7(a2
+ b2
+ c2
)2
SolutionX
xoti™e th—t there exitst three num˜ers —mong a, b, c, d, e h—vinh the s—me singF vet these
num˜er ˜e a, b, c, d, e F‡ithout loss of gener—lityD‡e m—y —ssume th—t a, b, c ≥ 0@it notD‡e
™—n t—ke −1, −b, −cAF
xow Dusing the g—u™hyEƒ™h—wrz inequ—lityD‡e h—veX
[(9(a4
+b4
+c4
)+2(d4
+e4
))+7d4
+7e4
)](84+63+63) ≥ [2 21(9(a4 + b4 + c4) + 2(d4 + e4))+21d2
+21e2
]2
.
RI

end thusDit suffi™es to prove th—tX
2 9(a4 + b4 + c4) + 2(d4 + e4)) ≥
√
21(a2
+ b2
+ c2
).
yr
36(a4
+ b4
+ c4
) + 8(d4
+ e4
) ≥ 21(a2
+ b2
+ c2
)2
.
ƒin™e
d4
+ e4
≥
(d2
+ e2
)2
2
≥
(d + e)4
8
=
(a + b + c)4
8
,
it is enough to show th—t
36(a4
+ b+
c4
) + (a + b + c + d)4
≥ 21(a2
+ b2
+ c2
)2
‡hi™h is true —nd it is e—sy to proveF
SIF
vet a, b, c > 0F€rove th—tX
a(b + c)
a2 + bc
+
b(c + a)
b2 + ca
+
c(a + b)
c2 + ab
≤
1
2
27 + (a + b + c)
1
a
+
1
b
+
1
c
Solution
a2
(b + c)2
(a2 + bc)2
+ 2
ab(b + c)(c + a)
(a2 + bc)(b2 + ca)
≤
15
2
+
1
4
b + c
a
xoti™e th—t
(a2
+ bc)(b2
+ ca) − ab(b + c)(c + a) = c(a + b)(a − b)2
then
2
ab(b + c)(c + a)
(a2 + bc)(b2 + ca)
≤ 6
@IA yther h—ndD
a2
(b + c)2
(a2 + bc)2
≤
a2
(b + c)2
4a2bc
=
1
4
b
c
+
c
b
+ 2
@PA prom @IA —nd @PA ‡e h—ve done3
fesidesD ˜y the s—m w—ysD ‡e h—ve — ni™e Solution for —n old pro˜lemX
a(b + c)
a2 + bc
≤
√
a
1
√
a
SPF
por —ny positive re—l num˜ers a, b —nd cD
a(b + c)
a2 + bc
+
b(c + a)
b2 + ca
+
c(a + b)
c2 + ab
≤
√
a +
√
b +
√
c
1
√
a
+
1
√
b
+
1
√
c
SolutionX
‡e h—ve the inequ—lity is equiv—lent to
a(b + c)
a2 + bc
2
≤
√
a
1
√
a
RP

<=>
a(b + c)
a2 + bc
+ 2
ab(a + c)(b + c)
(a2 + bc)(b2 + ca)
≤
√
a
1
√
a
‡e ™—n e—sily prove th—t
ab(a + c)(b + c)
(a2 + bc)(b2 + ca)
≤ 3
ƒoD it suffi™es to prove th—t
<=>
a(b + c)
a2 + bc
+ 6 ≤
√
a
1
√
a
„o prove this ineqD ‡e only need to prove th—t
a + b
√
ab
−
c(a + b)
c2 + ab
− 1 ≥ 0
fut this is trivi—lD ˜e™—use
a + b
√
ab
−
c(a + b)
c2 + ab
− 1 = (a + b)
1
√
ab
−
c
c2 + ab
− 1
≥ 2
√
ab
1
√
ab
−
c
c2 + ab
− 1 =
c −
√
ab
2
c2 + ab
≥ 0
‡e —re doneF
SQF
vet a, b, c > 0F €rove th—tX
(a + b + c)3
3abc
+
ab2
+ bc2
+ ca2
a3 + b3 + c3
≥ 10
Solution
a3
+ b3
+ c3
3abc
+
ab2
+ bc2
+ ca2
a3 + b3 + c3
+
(a + b)(b + c)(c + a)
abc
≥ 10
…sing ewEqw9s inequ—lity D‡e h—veX
a3
+ b3
+ c3
3abc
+
ab2
+ bc2
+ ca2
a3 + b3 + c3
≥ 2
ab2 + bc2 + ca2
3abc
≥ 2
(a + b)(b + c)(c + a)
abc
≥ 8
SRF
in —ny tri—ngle efg show th—t
ama + bmb + cmc ≤
√
bcma +
√
camb +
√
abmc
SolutionX
‡e h—ve to prove the inequ—lity
ama + bmb + cmc ≤
√
bcma +
√
camb +
√
abmc
D where maD mbD mc —re the medi—ns of — tri—ngle efgF
ƒin™e 2bc
b+c ≤
√
bcD 2ca
c+a ≤
√
ca —nd 2ab
a+b ≤
√
ab
RQ

˜y the rwEqw inequ—lityD it will ˜e enough to show the stronger inequ—lity
ama + bmb + cmc ≤
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc
D
sin™e then ‡e will h—ve
ama + bmb + cmc ≤
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc
≤
√
bcma +
√
camb +
√
abmc
—nd the initi—l inequ—lity will ˜e provenF
ƒo in the followingD ‡e will ™on™entr—te on proving this stronger inequ—lityF
fe™—use the inequ—lity ‡e h—ve to prove is symmetri™D ‡e ™—n ‡vyq —ssume th—t a ≥ b ≥ cF
„henD ™le—rlyD bc ≤ ca ≤ abF
yn the other h—ndD using the formul—s m2
a = 1
4 2b2
+ 2c2
− a2
—nd m2
b = 1
4 2c2
+ 2a2
− b2
D
‡e ™—n get —s — result of — str—ightforw—rd ™omput—tionF
ma
b + c
2
−
mb
c + a
2
=
3ac + 3bc + a2
+ b2
+ 4c2
(a + b − c) (b − a)
4 (b + c)
2
(c + a)
2
xowD the fr—™tion on the right h—nd side is ≤ 0D sin™e 3ac + 3bc + a2
+ b2
+ 4c2
≥ 0 @this
is trivi—lAD
a + b − c > 0 @in f—™tD a + b > c ˜e™—use of the tri—ngle inequ—lityA —nd b − a ≤ 0 @sin™e
a ≥ bAF ren™eD
ma
b + c
2
−
mb
c + a
2
≤ 0
wh—t yields ma
b+c
2
≤ mb
c+a
2
—nd thus ma
b+c ≤ mb
c+a F ƒimil—rlyD using b ≥ cD ‡e ™—n find
mb
c+a ≤ mc
a+b F
„husD ‡e h—ve
ma
b + c
≤
mb
c + a
≤
mc
a + b
ƒin™e ‡e h—ve —lso bc ≤ ca ≤ abD the sequen™es
ma
b + c
;
mb
c + a
;
mc
a + b
—nd (bc; ca; ab) —re equ—lly sortedF „husD the ‚e—rr—ngement inequ—lity yields
ma
b + c
· bc +
mb
c + a
· ca +
mc
a + b
· ab ≥
ma
b + c
· ca +
mb
c + a
· ab +
mc
a + b
· bc
—nd
ma
b + c
· bc +
mb
c + a
· ca +
mc
a + b
· ab ≥
ma
b + c
· ab +
mb
c + a
· bc +
mc
a + b
· ca
ƒumming up these two inequ—litiesD ‡e get
2
ma
b + c
· bc + 2
mb
c + a
· ca + 2
mc
a + b
· ab
RR

≥
ma
b + c
· (ca + ab) +
mb
c + a
· (ab + bc) +
mc
a + b
· (bc + ca)
„his simplifies to
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc
≥
ma
b + c
· a (b + c) +
mb
c + a
· b (c + a) +
mc
a + b
· c (a + b)
iF eF to
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc ≥ ama + bmb + cmc
„husD ‡e h—ve
ama + bmb + cmc ≤
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc
—nd the Solution is ™ompleteF xote th—t in e—™h of the inequ—lities
ama + bmb + cmc ≤
√
bcma +
√
camb +
√
abmc
—nd
ama + bmb + cmc ≤
2bc
b + c
ma +
2ca
c + a
mb +
2ab
a + b
mc
equ—lity holds only if the tri—ngle efg is equil—ter—lF
SSF
por aD bD c positive re—ls prove th—t
a2
+ 3 b2
+ 3 c2
+ 3 ≥
4
3
3 √
abc (ab + bc + ca)
SolutionX
hivide abc for ˜oth term —nd t—ke x = bc
a ; y = ac
b ; z = ab
c —nd ‡e must prove th—tX
(xy + 3
xy ) ≥ (4
3 )3
(x + y + z) xote th—tX
LHS ≥ 3(x2
+y2
+z2
)+x2
y2
z2
+
4
x2y2z2
≥ (x+y+z)2
+4 ≥ 4(x+y+z) ≥ (
4
3
)3
(x+y+z).
STF veta, b, c > 0 F€rove th—tX
a + b
c
√
a2 + b2
+
b + c
a
√
b2 + c2
+
c + a
b
√
c2 + a2
≥
3
√
6
√
a2 + b2 + c2
.
Solution
IFFFeltern—tivelyD using ghe˜yshev —nd g—u™hyD
cycl
a + b
c
√
a2 + b2
≥
2(a + b + c)
3
·
9
cycl c
√
a2 + b2
=
6(a + b + c)
cycl c
√
a2 + b2
—nd
cycl
c a2 + b2 ≤
a + b + c
3
cycl
a2 + b2 ≤
a + b + c
3
6(a2 + b2 + c2)
gom˜ining ‡e get the desired resultF
SUF
vet a, b, c > 0 su™h th—t a2
+ b2
+ c2
+ abc = 4
€rove th—t
a2
b2
+ b2
c2
+ c2
a2
≤ a2
+ b2
+ c2
RS

SolutionX
vet a = 2 yz
(x+y)(x+z) , b = 2 xz
(x+y)(y+z) —nd c = 2 xy
(x+z)(y+z) ,
where x, y —nd z —re positive num˜ers @ e—sy to ™he™k th—t it exists AF
cyc
xy
(x + z)(y + z)
≥
cyc
4x2
yz
(x + y)(x + z)(y + z)2
,
whi™h equiv—lent to cyc(x4
y2
+x4
z2
−2x4
yz+2x3
y3
−2x2
y2
z2
) ≥ 0, whi™h true ˜y ewEqwF
SVF
vet a, b, c > 0 su™h th—t a + b + c = 1F€rove th—t
b2
a + b2
+
c2
b + c2
+
a2
c + a2
≥
3
4
Solution
‡e h—ve
b2
a + b2
+
c2
b + c2
+
a2
c + a2
≥
a2
+ b2
+ c2 2
(a4 + b4 + c4) + (ab2 + bc2 + ca2)
ren™e it suffi™es to prove th—t
a2
+ b2
+ c2 2
(a4 + b4 + c4) + (ab2 + bc2 + ca2)
≥
3
4
⇔ 4 a2
2
≥ 3 ab2
a + 3 a4
⇔ 4 a4
+ 8 a2
b2
≥ 3 a4
+ 3 a2
b2
+ abc2
+ a3
c
⇔ a4
+ 5 a2
b2
≥ 3abc a + 3 a3
c
ƒin™e ‡e —lw—ys h—ve
3 a3
c + b3
a + c3
b ≤ a2
+ b2
+ c2 2
= a4
+ b4
+ c4
+ 2 a2
b2
+ b2
c2
+ c2
a2
„herefor it suffi™es to prove th—t
3 a2
b2
+ b2
c2
+ c2
a2
≥ 3abc (a + b + c)
whi™h o˜viously trueF
SWF
vet a; b; c > 0F €rove th—t
ab+c
+ ba+c
+ ca+b
≥ 1
Solution
if a ≥ 1 or b ≥ 1 or c ≥ 1 then the inequ—lity is true
if 0 ≤ a, b, c ≤ 1 then suppose c = mina, b, c
C if a + b < 1 ‡e h—ve b + c < 1 Dc + a < 1
RT

epply fernoull‡e 9 inequ—lity
(
1
a
)b+c
) = (1 +
1 − a
a
)b+c
< 1 +
(b + c)(1 − a)
a
<
a + b + c
a
„herefore ab+c
> a
a+b+c
ƒimil—r for bc+a
—nd ca+b
dedu™e ab+c
+ ba+c
+ ca+b
> 1
C if a + b > 1 then ab+c
+ ba+c
+ ca+b
> ab+c
+ ba+c
≥ aa+b
+ ba+b
epply fernoull‡e 9 inequ—lity ‡e h—ve X haa+b
= (1 + (a − 1))a+b
> 1 + (a + b)(a − 1)
ƒimil—r forba+b
when™e ab+c
+ ba+c
+ ca+b
> 2 + (a + b)(a + b − 2) = (a + b − 1)2
+ 1 ≥ 1
THF
vet a, b, c ˜e the sidelengths of tri—ngle with perimeter 2 (⇒ a + b + c = 2). €rove th—t
a3
b
+
b3
c
+
c3
a
−
a3
c
−
b3
a
−
c3
b
< 3
SolutionX
„his ineq is equiv—lent toX
|a4
c + c4
b + b4
a − a4
b − bc
− c4
a| ≤ 3abc
<=> |(a − b)(b − c)(c − a)(a2
+ b2
+ c2
+ ab + bc + ca)| ≤ 3abc
fy ‚—v‡e ƒu˜stitution D denoteX a = x + y, b = y + z, c = z + xD so x + y + z = 1D this ineq
˜e™omesX
|(x − y)(y − z)(z − x)(3(x2
+ y2
+ z2
) + 5(xy + yz + zx)| ≤ 3(x + y)(y + z)(z + x)
i—sy to see th—t |(x − y)(y − z)(z − x)| ≤ (x + y)(y + z)(z + x)
ƒo ‡e need to prove (3(x2
+ y2
+ z2
) + 5(xy + yz + zx) ≤ 3 = 3(x + y + z)2
<=> xy + yz + zx ≥ 0
whi™h is o˜vious true
FiFh
F
TIF
qiven xDyDzbHF€rove th—t
x(y + z)2
2x + y + z
+
y(x + z)2
x + 2y + z
+
z(x + y)2
x + y + 2z
= (3xyz(x + y + z))
SolutionX
cyc
x(y + z)2
2x + y + z
− 3xyz(x + y + z) =
=
cyc
x(y + z)2
2x + y + z
− yz + xy + xz + yz − 3xyz(x + y + z) =
=
cyc
z2
(x − y) − y2
(z − x)
2x + y + z
+
cyc
z2
(x − y)2
2 xy + xz + yz + 3xyz(x + y + z)
=
RU

=
cyc
(x − y)
z2
2x + y + z
−
z2
2y + x + z
+
+
cyc
z2
(x − y)2
2 xy + xz + yz + 3xyz(x + y + z)
=
=
cyc
(x − y)2

 z2
2 xy + xz + yz + 3xyz(x + y + z)
−
z2
(2x + y + z)(2y + x + z)

 .
(2x + y + z)(2y + x + z) ≥ 2 xy + xz + yz + 3xyz(x + y + z) .
fut
(2x + y + z)(2y + x + z) ≥ 2 xy + xz + yz + 3xyz(x + y + z) ⇔
⇔ 2x2
+ 2y2
+ z2
+ 3xy + xz + yz ≥ 2 3xyz(x + y + z),
x2
+ y2
+ z2
≥ xy + xz + yz ≥ 3xyz(x + y + z).
it seems th—t the following inequ—lity is true tooF
vet x, y —nd z —re positive num˜ersF €rove th—tX
x(y + z)2
3x + 2y + 2z
+
y(x + z)2
2x + 3y + 2z
+
z(x + y)2
2x + 2y + 3z
≥
4
7
3xyz(x + y + z)
TPF
vet a, b ∈ R su™h th—t 9a2
+ 8ab + 7b2
≤ 6 €rove th—t X7a + 5b + 12ab ≤ 9
SolutionX
IFFF
fy ewEqw inequ—lityD ‡e see th—t
7a + 5b + 12ab ≤ 7 a2
+
1
4
+ 5 b2
+
1
4
+ 12ab
= (9a2
+ 8ab + 7b2
) − 2(a − b)2
+ 3 ≤ (9a2
+ 8ab + 7b2
) + 3 ≤ 6 + 3 = 9.
iqu—lity holds if —nd only if a = b = 1
2 .
TQF
vet x, y —nd z —re positive num˜ers su™h th—t x + y + z = 1
x + 1
y + 1
z .
€rove th—t xyz + yz + zx + xy ≥ 4.
SolutionX
xyz(x + y + z)
yz + zx + xy
+ x + y + z −
4xyz(x + y + z)2
(yz + zx + xy)2
≥
3xyz
yz + zx + xy
+ x + y + z −
4xyz(x + y + z)2
(yz + zx + xy)2
=
x3
(y − z)2
+ y3
(z − x)2
+ z3
(x − y)2
(yz + zx + xy)2
≥ 0 =⇒
1 +
1
x
+
1
y
+
1
z
≥
4
xyz
⇐⇒ xyz + yz + zx + xy ≥ 4
RV

TRF
vet a, b, ca0 s—tisfy a + b + c = 1
€rove th—t (a2
+ b2
)(b2
+ c2
)(c2
+ a2
)a 1
32
SolutionX
let f(a, b, c) = (a2
+ b2
)(b2
+ c2
)(c2
+ a2
)
let c = max(a, b, c);
‡e h—ve
f(a, b, c) ≤ f(a + b, 0, c)@whi™h is equiv—lent ab(−4abc2
+ a3
b + ab3
− 4a2
c2
− 4b2
c2
− 2c4
) ≤
0@trueA ‡e will prove th—t f(a + b, 0, c) = f(1 − c, 0, c) ≤ 1
2 whi™h is equiv—lent to
1
32
∗ (16c4
− 32c3
+ 20c2
− 4c − 1))(−1 + 2c)2
≤ 0
remem˜er th—t
16c4
− 32c3
+ 20c2
− 4c − 1 = 4(2c2
− 2c + 1−
√
5
4 )(2c2
− 2c + 1+
√
5
4 ) ≥ 0 for every c ∈ [0, 1]
TSF
vet a, b, c ˜e the sides of tri—ngleF €rove th—tX
a
2a − b + c
+
b
2b − c + a
+
c
2c − a + b
≥
3
2
SolutionX
the inequ—lity is equiv—lent to
1
1 + a
a+c−b
≤
3
2
fy g—u™hy ‡e h—ve X
a
a + c − b
+ 1 ≥ 2
a
a + c − b
ƒo ‡e need to prove
a + c − b
a
≤ 3
fe™—use a, b, c ˜e the sides of — tri—ngle so ‡e h—veX
a + c − b
a
=
sin A + sin C − sin B
sin A
=
2 sin C
2 cos B
2
cos A
2
it9s following th—t
a + c − b
a
=
cos B
2
√
sin C + cos C
2
√
sin A + cos A
2
√
sin B
cos A
2 cos B
2 cos C
2
≤
cos2 B
2 + cos2 C
2 + cos2 A
2 (sin C + sin A + sin B)
cos A
2 cos B
2 cos C
2
= 2 (cos B + cos C + cos A) + 6 ≤ 2.
3
2
+ 6 = 3
TTF
vet a, b, c, d > 0F€rove the following inequ—lityF‡hen does the equ—lity holdc
3
1
+
5
1 + a
+
7
1 + a + b
+
9
1 + a + b + c
+
36
1 + a + b + c + d
4 1 +
1
a
+
1
b
+
1
c
+
1
d
RW

SolutionX
‡e ™—n h—ve
(1 + a + b + c + d)(
4
25
+
16
25a
+
36
25b
+
64
25c
+
4
d
) ≥ (
2
5
+
4
5
+
6
5
+
8
5
+ 2)2
= 9
so
(
4
25
+
16
25a
+
36
25b
+
64
25c
+
4
d
) ≥ 36
1
1 + a + b + c + d
—nd
(1 + a + b + c)(
9
100
+
9
25a
+
81
100b
+
36
25c
) ≥ (
3
10
+
3
5
+
9
10
+
6
5
)
so ‡e h—ve
(
9
100
+
9
25a
+
81
100b
+
36
25c
) ≥ 9
1
1 + a + b + c
—nd
(1 + a + b)(
7
36
+
7
9a
+
7
4b
) ≥ 7
‡e get
(
7
36
+
7
9a
+
7
4b
) ≥
7
1 + a + b
—nd
(1 + a)(
5
9
+
20
9a
) ≥ 5
then
5
9
+
20
9a
≥
5
1 + a
—nd —dd these inequ—lity up ‡e ™—n solve the pro˜lemF
TUF
vet —D˜D™ ˜e positive re—l num˜er su™h th—t 9 + 3abc = 4(ab + bc + ca)
€rove th—t a + b + c ≥ 3
SolutionX
„—ke a = x + 1; b = y + 1; c = z + 1Dthen ‡e must prove th—tX
x + y + z ≥ 0 when 5(x + y + z) + xy + xz + yz = 3xyz
‡e ™onsider three ™—seX
g—se IXxyz ≥ 0 ⇒ (x+y+z)2
3 + 5(x + y + z)
≥ 5(x + y + z) + xy + xz + yz = 3xyz ≥ 0 ⇒ x + y + z ≥ 0
g—se PX x ≥ 0; y ≥ 0; z ≤ 0
essume th—t x + y + z ≤ 0 ⇒ −x ≥ y + zF
5(x + y + z) = yz(3x − 1) − x(y + z) ≥ −4yz + (y + z)2
= (y − z)2
≥ 0
g—se QX x ≤ 0; y ≤ 0; z ≤ 0Fy˜serve th—t −x, −y, −z ∈ [0, 1] thenX
0 = 5(x+y+z)+xy+yz+xz−3xyz ≤ 5(x+y+z)+2(xy+xz+yz) ≤ 5(x+y+z)+
2(x + y + z)2
3
⇒ x+y+z ≥ 0
ƒo ‡e h—ve doneF
TVF
if a, b, c, d —re nonEneg—tive re—l num˜ers su™h th—t a + b + c + d = 4D then
a2
b2 + 3
+
b2
c2 + 3
+
c2
d2 + 3
+
d2
a2 + 3
≥ 1.
SH

ƒyv…„iyxX
fy g—u™hyEƒ™hw—rz
[a2
(b2
+ 3) + b2
(c2
+ 3) + c2
(d2
+ 3)](
a2
b2 + 3
+
b2
c2 + 3
+
c2
d2 + 3
+
d2
a2 + 3
) ≥ (a2
+ b2
+ c2
)2
.
prom g—u™hy ‡e see th—t it is suffi™ient to prove th—t
(a2
+ b2
+ c2
+ d2
)2
≥ 3(a2
+ b2
+ c2
+ d2
) + a2
b2
+ b2
c2
+ c2
d2
+ d2
a2
whi™h ™—n ˜e rewritten —s
(a2
+ b2
+ c2
+ d2
)(a2
+ b2
+ c2
+ d2
− 3) ≥ a2
b2
+ b2
c2
+ c2
d2
+ d2
a2
xow you must homogeneize to h—ve r—r—z‡e form
(a2
+ b2
+ c2
+ d2
)(a2
+ b2
+ c2
+ d2
− 3(
a + b + c + d
4
)2
) ≥ a2
b2
+ b2
c2
+ c2
d2
+ d2
a2
whi™h follows from
a2
+ b2
+ c2
+ d2
≥ 4
—nd
(x + y + z + t)2
≥ 4(xy + yz + zt + tx)
with x = a2
—nd simil—rF
TWF
if a ≥ 2D b ≥ 2D c ≥ 2 —re re—lsD then prove th—t
8 a3
+ b b3
+ c c3
+ a ≥ 125 (a + b) (b + c) (c + a)
ƒyv…„iyxX
vets write LHS —s
8 ∗ (a3
b3
c3
+ abc + a4
b3
+ b4
c3
+ c4
a3
+ a4
c + b4
a + c4
b)
prom the wuirhe—ds inequ—lity ‡e h—ve th—t
a4
b3
+ b4
c3
+ c4
a3
≥ a3
b2
c2
= a2
b2
c2
(a + b + c)
—nd
a4
c + b4
a + c4
b ≥ a3
bc = abc(a2
+ b2
+ c2
)
F(∗∗)Nowletsseethat
a2
b2
c2
= (ab)(bc)(ca) ≥ (a + b)(b + c)(c + a)Thisiseasytoprove.
Fromthe@BBAWegetthat
LHS ≥ 8(a3
b3
c3
+ abc + a2
b2
c2
(a + b + c) + abc(a2
+ b2
+ c2
))
= 8a2
b2
c2
(abc +
1
abc
+ a + b + c +
a2
+ b2
+ c2
abc
)
so ‡e h—ve to prove th—t X
abc +
1
abc
+ a + b + c +
a2
+ b2
+ c2
abc
≥
125
8
SI

or
8a2
b2
c2
+ 8 + 8abc(a + b + c) + 8(a2
+ b2
+ c2
) ≥ 125abc
‡vyq let a ≥ b ≥ c
xow let abc = P ‡e will m—ke th—t following ™h—nge
a →
a
—nd b → b where ≥ 1 —nd a F
„he RHS doesnt ™h—ngeF in the LHS the first p—rt —lso doesn9t ™h—ngeF
a + b ≥
a
+ b
equiv—lent to ( − 1)(a − b ) whi™h is trueF elso ‡e get th—t
a2
+ b2
≥
a2
2
+ b2 2
ƒo —s ‡e get the num˜ers ™loser to e—™h other the LHS de™re—ses while the RHS rem—ins
the s—me so it is enough to prove the inequ—lity for the ™—se a = b = c whi™h is equiv—lent
to X
8a6
+ 8 + 24a4
+ 24a2
≥ 125a3
‡hi™h is pretty e—syF
UHD
vet a, b, c ≥ 0F €rove th—tX
1
√
2a2 + ab + bc
+
1
√
2b2 + bc + ca
+
1
√
2c2 + ca + ab
≥
9
2(a + b + c)
.
SolutionX
‡e h—veX
cyc
1
√
2a2 + ab + bc
=
cyc
2a + b + c
2 · 2a+b+c
2
√
2a2 + ab + bc
≥
≥
cyc
2a + b + c
2a+b+c
2
2
+ 2a2 + ab + bc
.
fut
cyc
2a + b + c
2a+b+c
2
2
+ 2a2 + ab + bc
≥
9
2(a + b + c)
⇔
⇔
cyc
(100a6
+ 600a5
b + 588a5
c + 1123a4
b2
− 357a4
c2
− 1842a3
b3
+
+1090a4
bc − 1414a3
b2
c + 1330a3
c2
b − 1218a2
b2
c2
) ≥ 0,
whi™h is e—syF
UIF vet —D˜D™ b H F€rove th—t X
a
a + 2b
+
b
b + 2c
+
c
c + 2a
≤
3(a2
+ b2
+ c2
)
(a + b + c)2
SP

SolutionX
⇐⇒ 1 −
a
a + 2b
≥ 3 −
3(a2
+ b2
+ c2
)
(a + b + c)2
=
6(ab + bc + ca)
(a + b + c)2
⇐⇒
b
a + 2b
+
c
b + 2c
+
a
c + 2a
≥
3(ab + bc + ca)
(a + b + c)2
fy g—u™hyEƒ™hw—rz ‡e get
b
a + 2b
b(a + 2b) ≥ (a + b + c)2
it suffi™e to show th—t
(a + b + c)4
≥ 3(ab + bc + ca)(ab + bc + ca + 2a2
+ 2b2
+ 2c2
)
‡ithout loss of generosityD—ssume th—t ab + bc + ca = 3Dthen it ˜e™omes
(a + b + c)2
− 9
2
≥ 0
whi™h is o˜viousF
UPF
vet a, b, c, d ˜e positive re—l num˜ersF €rove th—t the following inequ—lity holds
a4
+ b4
(a + b)(a2 + ab + b2)
+
b4
+ c4
(b + c)(b2 + bc + c2)
+
c4
+ d4
(c + d)(c2 + cd + d2)
+
d4
+ a4
(d + a)(d2 + da + a2)
≥
a2
+ b2
+ c2
+ d2
a + b + c + d
SolutionX
a4
+ b4
(a + b)(a2 + ab + b2)
≥
1
2 (a2
+ b2
)2
(a + b)(a2 + ab + b2)
„hus D it rem—ins to prove th—t
cyc
(a2
+ b2
)2
(a + b)(a2 + ab + b2)
≥
2(a2
+ b2
+ c2
+ d2
)
a + b + c + d
e™ording to g—u™hyEƒhw—rz inequ—lity ‡e h—ve X
LHS ≥
4(a2
+ b2
+ c2
+ d2
)2
A
where A = cyc(a + b)(a2
+ ab + b2
) = 2(a3
+ b3
+ c3
+ d3
) + 2 cyc ab(a + b)
2(a2
+ b2
+ c2
+ d2
)(a + b + c + d) ≥ 2(a3
+ b3
+ c3
+ d3
) + 2
cyc
ab(a + b)
⇔
cyc
2a3
+ 2a2
b + 2a2
c + 2a2
d − 2a3
− 2ab(a + b) =
cyc
2a2
c ≥ 0
UQF
if a, b, c —re nonneg—tive re—l num˜ersD then
a a2 + 4b2 + 4c2 ≥ (a + b + c)2
.
SolutionF
in the nontrivi—l ™—se when two of —D˜D™ —re nonzeroD ‡e t—ke squ—re ˜oth sides —nd write
SQ

the inequ—lity —s
a2
(a2
+ 4b2
+ 4c2
) + 2 ab (a2 + 4b2 + 4c2)(4a2 + b2 + 4c2) ≥≥ ( a)
4
.
epplying the g—u™hyEƒ™hw—rz inequ—lity in ™om˜in—tion with the trivi—l inequ—lity
(u + 4v)(v + 4u) ≥ 2u + 2v +
2uv
u + v
∀u, v ≥ 0, u + v > 0,
‡e get
ab (a2 + 4b2 + 4c2)(4a2 + b2 + 4c2) ≥ ab (a2 + 4b2)(b2 + 4a2) + 4c2
≥ ab 2a2
+ 2b2
+
2a2
b2
a2 + b2
+ 4c2
.
a4
+ 8 a2
b2
+ 4 ab(a2
+ b2
) + 8abc a + 4 a3
b3
a2+b2 ≥
≥ ( a)
4
.
„his inequ—lity redu™es toF
URF
vet —D˜D™ ˜e nonneg—tive re—l num˜ersD no two of whi™h —re zeroF €rove th—t
b2
+ c2
a2 + bc
+
c2
+ a2
b2 + ca
+
a2
+ b2
c2 + ab
≥
2a
b + c
+
2b
c + a
+
2c
a + b
.
SolutionF
fy the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
b2
+ c2
a2 + bc
≥
(b2
+ c2
)
2
(b2 + c2)(a2 + bc)
=
4(a2
+ b2
+ c2
)2
ab(a2 + b2) + 2 a2b2
.
2(a2
+ b2
+ c2
)2
≥
a ab(a2
+ b2
) + 2 a2
b2
b + c
.4
ƒin™e
ab(a2
+ b2
) + 2 a2
b2
=
= (b + c)[a3
+ 2a2
(b + c) + bc(b + c) + a(b2
− bc + c2
)] − 4a2
bc,
this inequ—lity ™—n ˜e written —s
2 a2
2
+ 4abc
a2
b + c
≥
≥ a[a3
+ 2a2
(b + c) + bc(b + c) + a(b2
− bc + c2
)],
or equiv—lentlyD
a4
+ 2 a2
b2
+ 4abc
a2
b + c
≥ abc a + 2 a3
(b + c).
xowD ˜y ghe˜yshev9s inequ—lityD ‡e h—ve
a2
b + c
≥
3(a2
+ b2
+ c2
)
2(a + b + c)
,
SR

—nd thusD it suffi™es to show th—t
a4
+ 2 a2
b2
+
6abc a2
a
≥ abc a + 2 a3
(b + c).
efter some simple ™omput—tionsD ‡e ™—n write this inequ—lity —s
a3
(a − b)(a − c) + b3
(b − c)(b − a) + c3
(c − a)(c − b) ≥ 0,
whi™h is ƒ™hur9s inequ—lityF „he Solution is ™ompletedF iqu—lity holds if —nd only if a =
b = c, ora = b —nd c = 0, or —ny ™y™li™ permut—tionFF
a2
b2
+ 2
a3
b3
a2 + b2
≥ 2abc a,
or
a2
b2
(a + b)2
a2 + b2
≥ 2abc a.
fy the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
a2
b2
(a + b)2
a2 + b2
≥
[ab(a + b) + bc(b + c) + ca(c + a)]2
(a2 + b2) + (b2 + c2) + (c2 + a2)
,
—nd thusD it is enough to to ™he™k th—t
[ab(a + b) + bc(b + c) + ca(c + a)]2
≥ 4abc(a + b + c)(a2
+ b2
+ c2
).
‡ithout loss of gener—lityD —ssume th—t b is ˜etien a—nd cF prom the ewEqw inequ—lityD ‡e
h—ve
4abc(a + b + c)(a2
+ b2
+ c2
) ≤ [ac(a + b + c) + b(a2
+ b2
+ c2
)]2
.
yn the other h—ndD one h—s
ac(a + b + c) + b(a2
+ b2
+ c2
) − [ab(a + b) + bc(b + c) + ca(c + a)] =
= −b(a − b)(b − c) ≤ 0.
gom˜ining these two inequ—litiesD the ™on™lusion followsF iqu—lity o™™urs if —nd only if
—a˜a™D or a = b = 0, orb = c = 0, orc = a = 0.
USF vet a, b, c ˜e positive re—l num˜erF €rove th—tX
2(b + c)
a
≥
27(a + b)(b + c)(c + a)
4(a + b + c)(ab + bc + ca)
SolutionF
fy ewEqw inequ—lity ‡e h—ve
1
a 2(a2 + bc)
=
√
b + c
√
2a. (ab + ac)(a2 + bc)
≥
2(b + c)
√
a.(a + b)(a + c)
(b + c)
2(b + c)
a
≥
9(a + b)(b + c)(c + a)
2(ab + bc + ca)
fy ghe˜yselv inequ—lity ‡e h—ve
(b + c)
2(b + c)
a
≥
2
3
(a + b + c).
2(b + c)
a
SS

ren™eD it suffi™es to show th—t
2(b + c)
a
≥
27(a + b)(b + c)(c + a)
4(a + b + c)(ab + bc + ca)
fy g—u™hyEƒ™hw—rz inequ—lityD ‡e get
2(b + c)
a
2
a(b + c)2
≥ 16(a + b + c)3
end ˜y ewEqw inequ—lityD
27(a + b)(b + c)(c + a) ≤ 8(a + b + c)3
pin—llyD ‡e need to show th—t
16(a + b + c)3
a(b + c)2
≥
27.8(a + b + c)3
(a + b)(b + c)(c + a)
16(a + b + c)2(ab + bc + ca)2
or
32(a + b + c)2
(ab + bc + ca)2
≥ 27(a + b)(b + c)(c + a) ((a + b)(b + c)(c + a) + 4abc)
or
5x2
+ 32y2
≥ 44xy
where ‡e setting x = (a + b)(b + c)(c + a)D y = abc —nd using the equ—lity (a + b + c)(ab +
bc + ca) = x + y
„he l—st inequ—lity is true ˜e™—use it equiv—lent (x − 8y)(5x − 4y) ≥ 0D o˜viouslyF
UTF
if —D˜D™ —re positive re—l num˜ersD then
1
a 2(a2 + bc)
≥
9
2(ab + bc + ca)
.
pirst SolutionF
fy rolder9s inequ—lityD ‡e h—ve
1
a
√
a2 + bc
2
a2
+ bc
a
≥
1
a
3
.
it follows th—t
1
a
√
a2 + bc
2
≥
(ab + bc + ca)3
a2
b2
c2
a2
b2
+ a2
bc
,
—nd hen™eD it suffi™es to prove th—t
2(ab + bc + ca)5
≥ 81a2
b2
c2
a2
b2
+ a2
bc .
ƒetting x = bc, y = ca —nd z = abD this inequ—lity ˜e™omes
2(x + y + z)5
≥ 81xyz(x2
+ y2
+ z2
+ xy + yz + zx).
…sing the illEknown inequ—lity
xyz ≤
(x + y + z)(xy + yz + zx)
9
,
ST

‡e see th—t it is enough to ™he™k th—t
2(x + y + z)4
≥ 9(xy + yz + zx)(x2
+ y2
+ z2
+ xy + yz + zx),
whi™h is equiv—lent to the o˜vious inequ—lity
(x2
+ y2
+ z2
− xy − yz − zx)(2x2
+ 2y2
+ 2z2
+ xy + yz + zx) ≥ 0.
„he Solution is ™ompletedF iqu—lity holds if —nd only if a = b = c.
ƒe™ond SolutionF fy the ewEqw inequ—lityD ‡e h—ve
1
a 2(a2 + bc)
=
√
b + c
√
2a (ab + ac)(a2 + bc)
≥
2(b + c)
√
a(a + b)(a + c)
.
b + c
2a
·
1
(a + b)(a + c)
≥
9
4(ab + bc + ca)
.
‡ithout loss of gener—lityD —ssume th—t a ≥ b ≥ cF ƒin™e
b + c
2a
≤
c + a
2b
≤
a + b
2c
—nd
1
(a + b)(a + c)
≤
1
(b + c)(b + a)
≤
1
(c + a)(c + b)
,
˜y ghe˜yshev9s inequ—lityD ‡e get
b + c
2a
·
1
(a + b)(a + c)
≥
1
3
b + c
2a
1
(a + b)(a + c)
=
2(a + b + c)
3(a + b)(b + c)(c + a)
b + c
2a
.
ƒoD it suffi™es to show th—t
b + c
2a
≥
27(a + b)(b + c)(c + a)
8(a + b + c)(ab + bc + ca)
.
ƒetting
t =
6 (a + b)(b + c)(c + a)
8abc
t ≥ 1. fy the ewEqw inequ—lityD ‡e h—ve
b + c
2a
≥ 3t.
elsoD it is e—sy to verify th—t
27(a + b)(b + c)(c + a)
8(a + b + c)(ab + bc + ca)
=
27t6
8t6 + 1
.
ƒoD it is enough to ™he™k th—t
3t ≥
27t6
8t6 + 1
,
SU

or
8t6
− 9t5
+ 1 ≥ 0.
ƒin™e t ≥ 1D this inequ—lity is true —nd the Solution is ™ompletedF
UTF
qive a1, a2, ..., an ≥ 0—re num˜ers h—ve sum is IF €rove th—t if n > 3 so
a1a2 + a2a3 + ... + ana1 ≤
1
4
SolutionX
vet n = 2kD where k ∈ N —nd a1 + a3 + ... + a2k−1 = xF
ren™eD
a1a2 + a2a3 + ... + ana1 ≤ (a1 + a3 + ... + a2k−1) (a2 + a4 + ... + a2k) =
= x(1 − x) ≤
1
4
vet n = 2k − 1 —nd a1 = mini{ai}F
ren™eD
a1a2 + a2a3 + ... + ana1 ≤
≤ a1a2 + a2a3 + ... + ana2 ≤ (a1 + a3 + ... + a2k−1) (a2 + a4 + ... + a2k−2) =
= x(1 − x) ≤
1
4
UUF
vet a, b, c ˜e nonEneg—tive re—l num˜ersF €rove th—t
a3
+ 2abc
a3 + (b + c)3
+
b3
+ 2abc
b3 + (c + a)3
+
c3
+ 2abc
c3 + (a + b)3
≥ 1
Solution
‡e h—ve
1−
a3
a3 + (b + c)3
=
a3
a3 + b3 + c3
−
a3
a3 + (b + c)3
=
3a3
bc(b + c)
(a3 + b3 + c3)(a3 + (b + c)3)
ren™eD it suffi™es to show th—t
2
a3
+ b3
+ c3
a3 + (b + c)3
≥
3a2
(b + c)
a3 + (b + c)3
⇔
2a3
− 3a2
(b + c) + 2b3
+ 2c3
a3 + (b + c)3
≥ 0
⇔
(a − b)(a2
− 2ab − 2b2
) + (a − c)(a2
− 2ac − 2c2
)
a3 + (b + c)3
≥ 0
⇔
(a − b)3
− (c − a)3
+ 3c2
(c − a) − 3b2
(a − b)
a3 + (b + c)3
≥ 0
c2
(c − a) − b2
(a − b)
a3 + (b + c)3
≥ 0
—nd
(a − b)3
− (c − a)3
a3 + (b + c)3
≥ 0
SV

„he first inequ—lity is equiv—lent to
⇔ (a − b)
a2
b3 + (c + a)3
−
b2
a3 + (b + c)3
≥ 0
pin—llyD to finish the SolutionD ‡e will show th—t if a ≥ bD then
a2
b3 + (c + a)3
≥
b2
a3 + (b + c)3
⇔ a5
− b5
≥ b2
(c + a)3
− a2
(b + c)3
⇔ a5
− b5
≥ a2
b2
(a − b) + c3
(b2
− a2
) + 3c2
ab(b − a)
whi™h is o˜viously true sin™e a ≥ b —nd c ≥ 0F
end the se™ond inequ—lity is equiv—lent to
(a − b)3 1
a3 + (b + c)3
−
1
b3 + (c + a)3
≥ 0
⇔
(a − b)4
(3c(a + b) + 3c2
)
(a3 + (b + c)3)(b3 + (c + a)3)
≥ 0
whi™h is o˜viously trueF
iqu—lity holds for a = b = c or abc = 0
UVF
vet a, b, c —re positive re—l num˜ersD prove th—t
3
a
b
+
b
c
+
c
a
+ 2
ab + bc + ca
a2 + b2 + c2
≥ 5
SolutionX
fy using the ill known
2
a
b
+
b
c
+
c
a
+ 1 ≥
21 a2
+ b2
+ c2
(a + b + c)
2
ƒetting x = ab+bc+ca
a2+b2+c2 ≤ 1F it suffi™es to show th—t
3 (10 − x2)
2x2 + 1
+ 2x ≥ 5
⇔
3 10 − x2
2x2 + 1
≥ 4x2
− 20x + 25
⇔
−8x4
+ 40x3
− 57x2
+ 20x + 5
2x2 + 1
≥ 0
⇔
(x − 1) −8x3
+ 32x2
− 25x − 5
2x2 + 1
≥ 0
whi™h is ™le—rly trueF
a
b + b
c + c
a
3
a
b + b
c + c
a
3
ab + bc + ca
a2 + b2 + c2
≥ 1
end ‡e —lso note th—t the folloid is not true
a
b + b
c + c
a
3
ab + bc + ca
a2 + b2 + c2
≥ 1
SW

UWF
vet a, b, c ˜e the sideElengths of — tri—ngle su™h th—t a2
+ b2
+ c2
= 3. €rove th—t
bc
1 + a2
+
ca
1 + b2
+
ab
1 + c2
≥
3
2
.
SolutionF
‡rite the inequ—lity —s
2bc
4a2 + b2 + c2
≥ 1.
ƒin™e 1 = b2
c2
a2b2+b2c2+c2a2 , this inequ—lity is equiv—lent to
2bc
4a2 + b2 + c2
−
b2
c2
a2b2 + b2c2 + c2a2
≥ 0,
or
abc
a2b2 + b2c2 + c2a2
(2a2
− bc)(b − c)2
a(4a2 + b2 + c2)
≥ 0.
‡ithout loss of gener—lityD —ssume th—t a ≥ b ≥ c. ƒin™e
(2a2
− bc)(b − c)2
a(4a2 + b2 + c2)
≥ 0,
(2b2
− ca)(c − a)2
b(4b2 + c2 + a2)
+
(2c2
− ab)(a − b)2
c(4c2 + a2 + b2)
≥ 0.
ƒin™e a, b, c —re the sideElengths of — tri—ngle —nd a ≥ b ≥ c, ‡e h—ve
2b2
− ca ≥ c(b + c) − ca = c(b + c − a) ≥ 0,
—nd
a − c −
b
c
(a − b) =
(b − c)(b + c − a)
c
≥ 0.
„hereforeD
(2b2
− ca)(c − a)2
b(4b2 + c2 + a2)
≥
b(2b2
− ca)(a − b)2
c2(4b2 + c2 + a2)
.
b(2b2
− ca)
4b2 + c2 + a2
+
c(2c2
− ab)
4c2 + a2 + b2
≥ 0,
or
b(2b2
− ca)
4b2 + c2 + a2
≥
c(ab − 2c2
)
4c2 + a2 + b2
.
ƒin™e ab − 2c2
− (2b2
− ca) = a(b + c) − 2(b2
+ c2
) ≤ a(b + c) − (b + c)2
≤ 0, it is enough to
™he™k th—t
b
4b2 + c2 + a2
≥
c
4c2 + a2 + b2
,
b(4c2
+ a2
+ b2
) − c(4b2
+ c2
+ a2
) = (b − c)[(b − c)2
+ (a2
− bc)] ≥ 0.
TH

„he Solution is ™ompletedF
VHF
vet a, b, c, d > 0 su™h th—t a2
+ b2
+ c2
+ d2
= 4Dthen
1
a
+
1
b
+
1
c
+
1
d
≤ 2 +
2
abcd
Solution
write the inequ—lity —s
abc + bcd + cda + dab ≤ 2abcd + 2.
‡ithout loss of gener—lityD —ssume th—t a ≥ b ≥ c ≥ d.
vet
t =
a2 + b2
2
,
=> 1 ≤ t ≤
√
2. ƒin™e
2
1
c
+
1
d
≥
1
a
+
1
b
+
1
c
+
1
d
≥
16
a + b + c + d
≥
16
4(a2 + b2 + c2 + d2)
= 4,
‡e h—vec + d ≥ 2cd.
„hereforeD
abc + bcd + cda + dab − 2abcd = ab(c + d − 2cd) + cd(a + b)
≤
a2
+ b2
2
(c + d − 2cd) + cd 2(a2 + b2)
= t2
(c + d − 2cd) + 2tcd.
t2
(c + d − 2cd) + 2tcd ≤ 2,
or
2tcd(1 − t) + t2
(c + d) ≤ 2.
…sing the ewEqw inequ—lityD ‡e get
c + d ≤
(c + d)2
+ 4
4
=
(4 − 2t2
+ 2cd) + 4
4
=
4 − t2
+ cd
2
.
ƒoD it is enough to ™he™k th—t
4tcd(1 − t) + t2
(4 − t2
+ cd) ≤ 4,
or
tcd(4 − 3t) ≤ (2 − t2
)2
.
ƒin™e 2 − t2
= c2
+d2
2 ≥ cd, ‡e h—ve
(2 − t2
)2
− tcd(4 − 3t) ≥ cd(2 − t2
) − tcd(4 − 3t) = 2cd(t − 1)2
≥ 0.
TI

„he Solution is ™ompletedF
VIF
vet a, b, c ˜e positive re—l num˜er F€roveX
cyc
3
(a2 + ab + b2)2 ≤ 3
3
cyc
(2a2 + bc)
2
SolutionX
cyc
3
(a2 + ab + b2)2 ≤ 3
3
cyc
(2a2 + bc)
2
↔
cyc
3
3(a2 + ab + b2)2 ≤ 3
9
cyc
(2a2 + bc)
2
fy holder9s inequ—lityX
cyc
3
3(a2 + ab + b2)2 ≤ 3 9(
cyc
(a2 + ab + b2))2
ƒo ‡e must proveX
3 9(
cyc
(a2 + ab + b2))2 ≤ 3 (
cyc
(2a2 + bc))
2
<=> (
cyc
(a2
+ ab + b2
))2
≤ 3
cyc
(2a2
+ bc)2
…sing g—u™hyEƒ™h—wrz9s inequ—lityD
3
cyc
(2a2
+ bc)2
≥ (
cyc
(2a2
+ bc))2
=
cyc
(a2
+ ab + b2
)2
FiFh
F
VPF
vet a, b, c > 0Fprove th—tX
cyc
1
a2 + bc
≥
cyc
1
a2 + 2bc
+
ab + bc + ca
2(a2b2 + b2c2 + c2a2)
SolutionX
IFFF
(a2
b2
+ b2
c2
+ c2
a2
)
cyc
1
a2 + bc
=
cyc
bc +
a2
(b2
+ c2
− bc)
a2 + bc
<=> 2
cyc
a2
(b2
− bc + c2
)
a2 + bc
≥ ab + bc + ca
<=> 2
cyc
a2
1 +
b2
− bc + c2
a2 + bc
≥ 2(a2
+ b2
+ c2
) + ab + bc + ca
<=>
cyc
a2
a2 + bc
≥ 1 +
ab + bc + ca
2(a2 + b2 + c2)
TP

fy g—u™hyEƒ™hw—rzD
cyc
a2
a2 + bc
≥
(a + b + c)2
cyc a2 + cyc bc
= 1 +
ab + bc + ca
cyc a2 + cyc bc
≥ 1 +
ab + bc + ca
2(a2 + b2 + c2)
FiFh
PFF
„— ™âX
(a2
b2
+ b2
c2
+ c2
a2
)
cyc
1
a2 + bc
=
cyc
b2
+ c2
− bc
b2
+ c2
− bc
a2 + bc
<=> 2(a2
+ b2
+ c2
) −
cyc
bc
b2
+ c2
− bc
a2 + bc
≤
3
2
(a2
+ b2
+ c2
)
r—y l
2
cyc
bc
b2
+ c2
− bc
a2 + bc
≥ a2
+ b2
+ c2
fy ewEqw9s inequ—lityX
2(b2
+ c2
− bc) ≥ b2
+ c2
end ‡e will proveX
cyc
bc(b2
+ c2
)
a2 + bc
≥ a2
+ b2
+ c2
fy g—u™hyEƒ™hw—rz9s inequ—lityX
LHS ≥
( ab
√
a2 + b2)2
bc(a2 + bc)
( bc b2 + c2)2
≥ ( a2
)(abc a + a2
b2
)
…sing g—u™hyEƒ™hw—rzD
a2 + b2 a2 + c2 ≥ a2
+ bc
<=> abc(a3
+ b3
+ c3
+ 3abc − a2
b ≥ 0)
it is trueF
VQF
if —D˜D™ —re positive re—l num˜ersD then
a2
(b + c)
b2 + c2
+
b2
(c + a)
c2 + a2
+
c2
(a + b)
a2 + b2
≥ a + b + c.
pirst SolutionF
‡e h—ve
a2
(b + c)
b2 + c2
− a =
ab(a − b) − ca(c − a)
b2 + c2
= ab(a − b)
1
b2 + c2
−
1
c2 + a2
=
ab(a + b)(a − b)2
(a2 + c2)(b2 + c2)
≥ 0.
TQ

„husD it follows th—t
a2
(b + c)
b2 + c2
− a ≥ 0,
or
a2
(b + c)
b2 + c2
+
b2
(c + a)
c2 + a2
+
c2
(a + b)
a2 + b2
≥ a + b + c,
whi™h is just the desired inequ—lityF iqu—lity holds if —nd only if a = b = c.
ƒe™ond SolutionF
r—ving in view of the identity
a2
(b + c)
b2 + c2
=
(b + c)(a2
+ b2
+ c2
)
b2 + c2
− b − c,
‡e ™—n write the desired inequ—lity —s
b + c
b2 + c2
+
c + a
c2 + a2
+
a + b
a2 + b2
≥
3(a + b + c)
a2 + b2 + c2
.
‡ithout loss of gener—lityD —ssume th—t a ≥ b ≥ cF ƒin™e a2
+ c2
≥ b2
+ c2
—nd
b + c
b2 + c2
−
a + c
a2 + c2
=
(a − b)(ab + bc + ca − c2
)
(a2 + c2)(b2 + c2)
≥ 0,
˜y ghe˜yshev9s inequ—lityD ‡e h—ve
[(b2
+ c2
) + (a2
+ c2
)]
b + c
b2 + c2
+
a + c
a2 + c2
≥ 2[(b + c) + (a + c)],
or
b + c
b2 + c2
+
a + c
a2 + c2
≥
2(a + b + 2c)
a2 + b2 + 2c2
.
2(a + b + 2c)
a2 + b2 + 2c2
+
a + b
a2 + b2
≥
3(a + b + c)
a2 + b2 + c2
,
whi™h is equiv—lent to the o˜vious inequ—lity
c(a2
+ b2
− 2c2
)(a2
+ b2
− ac − bc)
(a2 + b2)(a2 + b2 + c2)(a2 + b2 + 2c2)
≥ 0.
Solution Q
xote th—t from g—u™hyEƒ™hw—rtz inequ—lity ‡e h—ve
cyc
a2
(b + c)
(b2 + c2)
≥
cyc
cyc a2
(b + c)
2
cyc a2(b + c)(b2 + c2)
„herefore it suffi™es to show th—t
cyc
a2
(b + c)
2
≥ (a + b + c)
cyc
a2
(b + c)(b2
+ c2
)
efter exp—nsion —nd using the ™onvention p = a + b + c; q = ab + bc + ca; r = abc this is
equiv—lent to withX
⇔ r(2p3
+ 9r − 7pq) ≥ 0
TR

futD sin™e @from trivi—l inequ—lityA ‡e h—vep2
− 3q ≥ 0D hen™e it suffi™es to show th—t
p3
+ 9r ≥ 4pq, whi™h follows from ƒ™hur9s inequ—lityF
iqu—lity o™™urs if —nd only if —a˜a™ or when a = b; c = 0 —nd its ™y™li™ permut—tionsF
VRF
if —D˜D™ —re positive num˜ers su™h th—t a + b + c = 3 then
a
3a + b2
+
b
3b + c2
+
c
3c + a2
≤
3
4
.
SolutionX
is equiv—lent to
a
3a + b2
≤
3
4
3a
3a + b2
− 1 ≤ −
3
4
or
b2
b2 + 3a
≥
3
4
fy g—u™hy ƒ™hw—rz inequ—lityD ‡e h—ve
LHS ≥
(a2
+ b2
+ c2
)2
a4 + (a + b + c) ab2
it suffi™es to prove
4(a2
+ b2
+ c2
)2
≥ 3 a4
+ 3 a2
b2
+ 3 ab3
+ 3 a2
bc
⇔ (a2
+ b2
+ c2
)2
− 3 ab3
+ 3( a2
b2
− a2
bc) ≥ 0
fy †—sg9s inequ—lityD ‡e h—ve
(a2
+ b2
+ c2
)2
− 3 ab3
≥ 0
fy em Eqw inequ—lityD
a2
b2
− a2
bc ≥ 0
VSF
if a ≥ b ≥ c ≥ d ≥ 0 —nd a + b + c + d = 2, then
ab(b + c) + bc(c + d) + cd(d + a) + da(a + b) ≤ 1.
SolutionX
pirstDlet us prove — lemm—X
vemm—X
por —ny a + b + c + d = 2 —nd a ≥ b ≥ c ≥ d ≥ 0
a2
b + b2
c + c2
d + d2
a ≥ ab2
+ bc2
+ cd2
+ da2
Solution of lemm—X
vet
F(a) = (b − d)a2
+ (d2
− b2
)a + b2
c + c2
d − bc2
− cd2
F (a) = (b − d)(2a − b − d) ≥ 0 <=> F(a) ≥ F(b) = (c − d)b2
+ (d2
− c2
)b + cd(c − d)
TS

F (b) = (c − d)(2b − c − d) ≥ 0 <=> F(b) ≥ F(c) = 0
xowDlet us turn ˜—™k the Solution of the pro˜lemF prom lemm— ‡e h—veX
a2
b+b2
c+c2
d+d2
a+ab2
+bc2
+cd2
+da2
+2(abc+bcd+cda+dab) ≥ 2(ab2
+bc2
+cd2
+da2
+abc+bcd+cda+dab
it follows th—tY
(a + b + c + d)(a + c)(b + d) ≥ 2(ab(b + c) + bc(c + d) + cd(d + a) + da(a + b))
futD ˜y ewEqw inequ—lityX
(a + c)(b + d) = (a + c)(b + d) ≤
(a + b + c + d)2
4
= 1
VTF
if x, y, z, p, q ˜e nonneg—tive re—l num˜ers su™h th—t
(p + q)(yz + zx + xy) > 0
€rove th—tX
2(p + q)
(y + z)(py + qz)
+
2(p + q)
(z + x)(pz + qx)
+
2(p + q)
(x + y)(px + qy)
≥
9
yz + xy + zx
SolutionX
cyc
2(p + q)
(y + z)(py + qz)
−
9
yz + zx + xy
≡
F(x, y, z)
(y + z)(z + x)(x + y)(py + qz)(pz + qx)(px + qy)(yz + zx + xy)
.
F(x, y, z) = F(x, x + s, x + t)
= 16x4
p3
+ q3
s2
− st + t2
+x3
16(p + q) p2
+ q2
(s + t)(s − t)2
+ (p − 2q)2
(15p + 7q)
+5q2
(p + q)s2
t + (q − 2p)2
(15q + 7p) + 5p2
(q + p) st2
+x2 2(s − t)2
[p2
(p + 47q)s2
+ 51pq(p + q)st + q2
(q + 47p)t2
]
3
+
2[(5p + q)(5p − 12q)2
+ 6q3
]s4
75
+
[(77p − 145q)2
(7918p + 6699q) + 3003p3
+ 14297pq2
]s3
t
5633859
+
34(p + q)(p − q)2
s2
t2
3
+
[(77q − 145p)2
(7918q + 6699p) + 3003q3
+ 14297qp2
]st3
5633859
+
2[(5q + p)(5q − 12p)2
+ 6p3
]t4
75
+x [(2p + 5q)s − (2q + 5p)t]
2 4pq(p + q)s3
(2p + 5q)2
+
(4p4
+ 40p3
q + 65p2
q2
+ 113pq3
+ 30q4
)s2
t
(2p + 5q)3
+
(4q4
+ 40q3
p + 65q2
p2
+ 113qp3
+ 30p4
)st2
(2q + 5p)3
+
4qp(q + p)t3
(2q + 5p)2
+
s3
t2
(2p + 5q)2(2q + 5p)3
(2p − 3q)2
1505p6
+ 9948p5
q + 19439p4
q2
TT

+16869p3
q3
+ 6709p2
q4
+ 852pq5
+ 117q6
+4960p7
q + 6800p6
q2
+ p5
q3
+ 4p4
q4
+ 5p3
q5
+ 8q6
p2
+ 4pq7
+ 7q8
+
s2
t3
(2p + 5q)3(2q + 5p)2
(2q−3p)2
1505q6
+ 9948q5
p + 19439q4
p2
+ 16869q3
p3
+ 6709q2
p4
+ 852qp5
+ 117p6
+4960q7
p + 6800q6
p2
+ q5
p3
+ 4q4
p4
+ 5q3
p5
+ 8p6
q2
+ 4qp7
+ 7p8
+st [(13p + 47q)s − (13q + 47p)t]2 2pq(p + q)s2
(13p + 47q)2
+
49pq(p + q)st
6(743p2 + 6914pq + 743q2)
+
2qp(q + p)t2
(13q + 47p)2
+(p − q)2
st
q(324773p3
+ 3233274p2
q + 836101pq2
+ 419052q3
)s2
6(13p + 47q)(743p2 + 6914pq + 743q2)
+
2(p + q)(p − q)2
(832132509p4
+ 9284734492p3
q + 9070265998p2
q2
+ 9284734492pq3
+ 832132509q4
)st
3(13p + 47q)2(13q + 47p)2(743p2 + 6914pq + 743q2)
+
p(324773q3
+ 3233274q2
p + 836101qp2
+ 419052p3
)t2
6(13q + 47p)(743p2 + 6914pq + 743q2)
≥ 0,
whi™h is ™le—rly true for x = min{x, y, z}.
VUF
vet —D˜D™ ˜e positive num˜ers su™h th—t ab + bc + ca = 3. €rove th—t
1
a + b
+
1
b + c
+
1
c + a
≥ 1 +
3
2(a + b + c)
.
SolutionX
IFFFFFvet f(a, b, c) = 1
a+b + 1
b+c + 1
c+a − 1 − 3
2(a+b+c) —nd a = min{a, b, c}. „hen
f(a, b, c) − f a, (a + b)(a + c) − a, (a + b)(a + c) − a =
=
√
a + b −
√
a + c
2 1
(a + b)(a + c)
−
1
2(b + c) (a + b)(a + c) − a
+
3
2(a + b + c) 2 · (a + b)(a + c) − a
≥
≥
√
a + b −
√
a + c
2 1
4bc
−
1
2(b + c) ·
√
bc
+
2
3(b + c)2
≥ 0
sin™eD (a + b)(a + c) ≥ a +
√
bc —nd 2 · (a + b)(a + c) − a ≤ a + b + c ≤ 3(b+c)
2 . „husD
rem—in to prove th—t f(a, b, b) ≥ 0, whi™h equiv—lent to
(a − b)2
(2a3
+ 9a2
b + 12ab2
+ b3
) ≥ 0.
PFFFFFFF
„he inequ—lity is equiv—lent toX
1
a + b
+
1
b + c
+
1
c + a
≥
3
3(ab + bc + ca)
+
3
2(a + b + c)
↔
1
a + b
+
1
b + c
+
1
c + a
−
9
2(a + b + c)
≥
3
3(ab + bc + ca)
−
3
a + b + c
TU

↔
(a − b)2
2(a + c)(b + c)(a + b + c)
+
(b − c)2
2(a + b)(a + c)(a + b + c)
+
(c − a)2
2(b + c)(a + b)(a + b + c)
≥
3[(a − b)2
+ (b − c)2
+ (c − a)2
]
2(a + b + c + 3(ab + bc + ca))(a + b + c)( 3(ab + bc + ca))
↔ (a − b)2
.M + (b − c)2
.N + (c − a)2
.P ≥ 0
with X
M = (a + b)[(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca) − 3(a + b)(b + c)(c + a)
N = (b + c)[(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca) − 3(a + b)(b + c)(c + a)
P = (c + a)[(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca) − 3(a + b)(b + c)(c + a)
ƒuppose th—tXa ≥ b ≥ cF
ƒo ‡e h—veX
M = (a + b)([(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca) − 3(b + c)(c + a)) ≥ 0
fe™—use (a + b + c) 3(ab + bc + ca) ≥ 3c2
P = (a + c)([(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca) − 3(a + b)(b + c)) ≥ 0
fe™—use (a + b + c) 3(ab + bc + ca) ≥ 3b2
ƒo ‡e must proveX
N + P ≥ 0
it –s equiv—lent toX
X = [(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca)(a + b + 2c) − 6(a + b)(b + c)(c + a) ≥ 0
€ut
x = a + b + c; y = 3(ab + bc + ca)
X ≥ [(a + b + c) + 3(ab + bc + ca)] 3(ab + bc + ca)(a + b + c) − 6(a + b + c)(ab + bc + ca)
↔ x2
y ≥ xy2
↔ x ≥ y
@it –s true for —ll positive num˜ers —D˜D™AF
VVF
vet a, b, c˜e positive re—l num˜er F €rove th—tX
1
a + b
+
1
b + c
+
1
c + a
≥
a + b + c
2(ab + bc + ca)
+
3
a + b + c
SolutionX vet put p = a + b + c, q = ab + bc + ca, r = abcD „his inequ—lity is equiv—lent toX
p2
+ q
pq − r
≥
p
2q
+
3
p
⇐⇒
p2
+ 3
3p − r
≥
p
6
+
3
p
fy exp—nding expression ‡e h—veX
(p2
+ 3)6p − p2
(3p − r) − 18(3p − r) ≥ 0
TV

⇐⇒ 3p3
+ p2
r − 36p + 18r ≥ 0
prom the illEknown inequ—lityD the third degree ƒ™hur9s inequ—lity st—tesX
p3
− 4pq + 9r ≥ 0 ⇐⇒ p3
− 12p + 9r ≥ 0
‡e h—veX
⇐⇒ 3p3
+ p2
r − 36p + 18r ≥ 0
⇐⇒ 3(p3
− 12p + 9r) + r(p2
− 9) ≥ 0
yn the other h—ndD ‡e h—veX
r(p2
− 9) ≥ 0 ⇐⇒ (a − b)2
+ (b − c)2
+ (c − a)2
≥ 0
VWF if x, y, z —re nonneg—tive re—l num˜ers su™h th—t x + y + z = 3D then
4(
√
x +
√
y +
√
z) + 15 ≤ 9(
x + y
2
+
y + z
2
+
z + x
2
)
SolutionX „he inequ—lity9s true when x = 3, y = z = 0F if no two of x, y, z —re HD set x = a2
et™F it ˜e™omes
8(a + b + c) + 10 3a2 + 3b2 + 3c2 ≤ 9 2a2 + 2b2 + 2b2 + 2c2 + 2c2 + 2a2
⇐⇒ 10 3a2 + 3b2 + 3c2 − (a + b + c) ≤ 9
cyc
2a2 + 2b2 − (a + b)
⇐⇒ 10
cyc
(a − b)2
a + b + c +
√
3a2 + 3b2 + 3c2
≤ 9
cyc
(a − b)2
a + b +
√
2a2 + 2b2
⇐⇒
cyc
9
a + b +
√
2a2 + 2b2
−
10
a + b + c +
√
3a2 + 3b2 + 3c2
(a − b)2
≥ 0
xow e—™h term is nonneg—tiveD for in f—™tD
9(a + b + 3a2 + 3b2) ≥ 10(a + b + 2a2 + 2b2)
˜e™—use
9
√
3
√
2
− 10 2a2 + 2b2 > 2a2 + 2b2 ≥ a + b
WHF
if a, b, c —re nonneg—tive re—l num˜ers su™h th—t a + b + c = 3D then
1
4a2 + b2 + c2
+
1
4b2 + c2 + a2
+
1
4c2 + a2 + b2
≤
1
2
SolutionX
1
4a2 + b2 + c2
+
1
4b2 + c2 + a2
+
1
4c2 + a2 + b2
≤
1
2
⇔
⇔
sym
(a6
− 4a5
b + 13a4
b2
− 2a4
bc − 6a3
b3
− 12a3
b2
c + 10a2
b2
c2
) ≥ 0 ⇔
⇔
cyc
(a − b)2
(2c4
+ 2(a2
− 4ab + b2
)c2
+ a4
− 2a3
b + 4a2
b2
− 2ab3
+ b4
) ≥ 0,
whi™h true ˜e™—use
(a2
− 4ab + b2
)2
− 2(a4
− 2a3
b + 4a2
b2
− 2ab3
+ b4
) =
TW

= −(a − b)2
(a2
+ 6ab + b2
) ≤ 0.
WIF
vet —D˜D™ ˜e nonneg—tive re—l num˜ers su™h th—ta + b + c = 3. €rove th—t
a2
b
4 − bc
+
b2
c
4 − ca
+
c2
a
4 − ab
≤ 1.
SolutionF
ƒin™e
4a2
b
4 − bc
= a2
b +
a2
b2
c
4 − bc
the inequ—lity ™—n ˜e written —s
abc
ab
4 − bc
≤ 4 − a2
b.
…sing the illEknown inequ—lity a2
b + b2
c + c2
a + abc ≤ 4D ‡e get
4 − (a2
b + b2
c + c2
a) ≥ abc,
—nd hen™eD it suffi™es to prove th—t
abc
ab
4 − bc
≤ abc,
or equiv—lentlyD
ab
4 − bc
+
bc
4 − ca
+
ca
4 − ab
≤ 1.
ƒin™e
ab + bc + ca ≤
(a + b + c)2
3
= 3,
‡e get
ab
4 − bc
≤
ab
4
3
(ab + bc + ca) − bc
=
3ab
4ab + bc + 4ca
.
„hereforeD it is enough to ™he™k th—t
x
4x + 4y + z
+
y
4y + 4z + x
+
z
4z + 4x + y
≤
1
3
,
where x = ab, y = ca —nd z = bcF „his is — illEknown inequ—lityF WPF
if a, b, c —re nonneg—tive re—l num˜ersD then
a3
+ b3
+ c3
+ 12abc ≤ a2
a2 + 24bc + b2
b2 + 24ca + c2
c2 + 24ab
SolutionX
a2
+ 24bc 7 a2
+ b2
+ c2
+ 8(bc + ca + ab)
2
− 7a3
+ 8a2
(b + c) + 7a b2
+ c2
+ 92abc + 48bc(b + c)
2
= 24bc (b − c)2
109a2
+ 77ab + 77ac + 49b2
+ 89bc + 49c2
+(b + c − 2a)2
(25bc + 7ab + 7ca) ≥ 0 =⇒
a2
a2 + 24bc
≥
a2
[7a3
+ 8a2
(b + c) + 7a(b2
+ c2
) + 92abc + 48bc(b + c)]
7(a2 + b2 + c2) + 8(bc + ca + ab)
UH

= a3
+ b3
+ c3
+ 12abc.
WQF where a, b, c, d —re nonneg—tive re—l num˜ersF€rove the inequ—lityX
a 9a2 + 7b2 + b 9b2 + 7c2 + c 9c2 + 7d2 + d 9d2 + 7a2 ≥ (a + b + c + d)2
SolutionX
fy g—u™hyƒ™hw—rzD ‡e h—ve
4 a 9a2 + 7b2 ≥ a(9a + 7b)
9(a2
+b2
+c2
+d2
)+7(a+c)(b+d) ≥ 4(a+b+c+d)2
= 4(a+c)2
+4(b+d)2
+8(a+c)(b+d)
⇔ 9(a2
+ b2
+ c2
+ d2
) ≥ 4(a + c)2
+ 4(b + d)2
+ (a + c)(b + d)
a2
+ b2
+ c2
+ d2
≥ (a + c)(b + d)
2(a2
+ b2
+ c2
+ d2
) = 2(a2
+ c2
) + 2(b2
+ d2
) ≥ (a + c)2
+ (b + d)2
WRF vet a, b, c ˜e positive F €rove th—tF
3a2
7a2 + 5(b + c)2
≤ 1 ≤
a2
a2 + 2(b + c)2
SolutionX
„he right h—nd is trivi—l ˜y the rolder inequ—lity sin™e

 a
a2 + 2 (b + c)
2


2
a a2
+ 2 (b + c)
2
≥ a
3
end ( a)
3
≥ a a2
+ 2 (b + c)
2
⇔ ab (a + b) ≥ 6abcF por the left h—nd ˜y the
g—u™hy ƒ™hw—rz inequ—lity ‡e h—ve

 a
7a2 + 5 (b + c)
2


2
≤ a
a
7a2 + 5 (b + c)
2
essume a + b + c = 3D denote ab + bc + ca = 9−q2
3 , r = abc then ‡e will prove
a
12a2 − 30a + 45
≤
1
9
⇔ f (r) = 48r2
+ 222 + 52q2
r + 20q4
+ 75q2
− 270 ≥ 0
‡e h—ve
r ≥ max 0,
(3 + q)
2
(3 − 2q)
27
„hereforD if
q ≥
3
2
UI

then get r ≥ 0 —nd
f (r) ≥ f (0) = 20 q −
3
2
q +
3
2
q2
+ 6 ≥ 0
if q ≤ 3
2 then get r ≥ (3+q)2
(3−2q)
27 —nd
f (r) ≥ f
(3 + q)
2
(3 − 2q)
27
=
q2
(2q − 3) 96q3
− 396q2
+ 2322q − 5103
729
≥ 0
‡e h—ve doneF iqu—lity holds if —n only if a = b = c or a = b, c = 0 or —ny ™y™li™ permut—E
tionsF
WSF
if a, b, c, d, e —re positive re—l num˜ers su™h th—t a + b + c + d + e = 5D then
1
a
+
1
b
+
1
c
+
1
d
+
1
e
+
20
a2 + b2 + c2 + d2 + e2
≥ 9
SolutionD sym f(a, b) me—ns f(a, b)+f(a, c)+f(a, d)+f(a, e)+f(b, c)+f(b, d) +f(b, e)+
f(c, d) + f(c, e) + f(d, e)F ‡e will firstly rewrite the inequ—lity —s
1
a
+
1
b
+
1
c
+
1
d
+
1
e
−
25
a + b + c + d + e
≥ 4 −
4(a + b + c + d + e)2
5(a2 + b2 + c2 + d2 + e2)
.
…sing the identities
(a + b + c + d + e)
1
a
+
1
b
+
1
c
+
1
d
+
1
e
− 25 =
sym
(a − b)2
ab
—nd 5(a2
+ b2
+ c2
+ d2
+ e2
) − (a + b + c + d + e)2
= sym(a − b)2
‡e ™—n rewrite —g—in
the inequ—lity —s
1
a + b + c + d + e sym
(a − b)2
ab
≥
4
5
×
sym(a − b)2
a2 + b2 + c2 + d2 + e2
or sym Sab(a − b)2
≥ 0 where
Sxy =
1
xy
−
4
a2 + b2 + c2 + d2 + e2
for —ll x, y ∈ {a, b, c, d, e}F essume th—t a ≥ b ≥ c ≥ d ≥ e > 0F ‡e will show th—t
Sbc + Sbd ≥ 0 —nd Sab + Sac + Sad + Sae ≥ 0F indeedD ‡e h—ve
Sbc + Sbd =
1
bc
+
1
bd
−
8
a2 + b2 + c2 + d2 + e2
>
1
bc
+
1
bd
−
8
b2 + b2 + c2 + d2
≥
1
bc
+
1
bd
−
8
2bc + 2bd
≥ 0
—nd
Sab+Sac+Sad+Sae =
1
ab
+
1
ac
+
1
ad
+
1
ae
−
16
a2 + b2 + c2 + d2 + e2
≥
16
a(b + c + d + e)
−
16
a2 + 1
4 (b + c + d + e)2
≥ 0.
ren™eD with noti™e th—t
Sbd ≥ Sbc —nd Sae ≥ Sad ≥ Sac ≥ Sab
UP

‡e h—ve Sbd ≥ 0 —nd Sae ≥ 0, Sae + Sad ≥ 0, Sae + Sad + Sac ≥ 0F
„husD
Sbd(b − d)2
+ Sbc(b − c)2
≥ (Sbd + Sbc)(b − c)2
≥ 0(1)
—nd
Sae(a−e)2
+Sad(a−d)2
+Sac(a−c)2
+Sab(a−b)2
≥ (Sae+Sad)(a−d)2
+Sac(a−c)2
+Sab(a−b)2
≥ (Sae + Sad + Sac)(a − c)2
+ Sab(a − b)2
≥ (Sae + Sad + Sac + Sab)(a − b)2
≥ 0(2)
yn the other h—ndD Sbe ≥ Sbd ≥ 0 —nd Sde ≥ Sce ≥ Scd ≥ Sbd ≥ 0(3)F
„hereforeD from @IAD @PA —nd @QA ‡e get sym Sab(a − b)2
≥ 0.
iqu—lity o™™urs when a = b = c = d = e or a = 2b = 2c = 2d = 2eF
WTF
vet a, b, c ˜e nonneg—tive re—l num˜ersF €rove th—t
1 +
3abc
a2b + b2c + c2a
≥
2(ab + bc + ca)
a2 + b2 + c2
SolutionX
‡e ™—n prove it —s followX
‚ewriting the inequ—lity —s
3abc
a2b + b2c + c2a
≥
2(ab + bc + ca) − a2
− b2
− c2
a2 + b2 + c2
if 2(ab + bc + ca) ≤ a2
+ b2
+ c2
D it is trivi—lF
if 2(ab + bc + ca) ≥ a2
+ b2
+ c2
D —pplying ƒ™hur9s inequ—lityX
3abc ≥
(a + b + c)[2(ab + bc + ca) − a2
− b2
− c2
]
3
(a + b + c)[2(ab + bc + ca) − a2
− b2
− c2
]
3(a2b + b2c + c2a)
≥
2(ab + bc + ca) − a2
− b2
− c2
a2 + b2 + c2
(a + b + c)(a2
+ b2
+ c2
) ≥ 3(a2
b + b2
c + c2
a)
b(a − b)2
+ c(b − c)2
+ a(c − a)2
≥ 0
@„rueA
WUF
vet a, b, c ˜e positive re—l num˜ers su™h th—t abc = 1F €rove th—t
1
a
+
1
b
+
1
c
+
6
a + b + c
≥ 5.
SolutionX
IFFFF‡vyq —ssume a ≥ b ≥ cF
vet
f(a, b, c) =
1
a
+
1
b
+
1
c
+
6
a + b + c
f(a, b, c) ≥ f(a,
√
bc,
√
bc)
<=>
(
√
b −
√
c)2
bc(a + b + c)(a + 2
√
bc)
((a + b + c)(a + 2
√
bc) − 6bc) ≥ 0 ⇔
UQ

(a + b + c)(a + 2
√
bc) ≥ 6bc.
es
a ≥
b + c
2
≥
√
bc
so
(a + b + c)(a + 2
√
bc) ≥ 9bc ≥ 6bc
hen™e the —õve inequ—lity is trueF
f(
1
x2
, x, x) ≥ 5 ⇔
(x − 1)2
(2x4
+ 4x3
− 4x2
− x + 2) ≥ 0.
es 2x4
+ 4x3
− 4x2
− x + 2 > 0 if x > 0D so the —õve inequ—lity is trueF
„herefore
f(a, b, c) ≥ f(a,
√
bc,
√
bc) = f(
1
bc
,
√
bc,
√
bc) ≥ 5.
ab FiFh
PFFFFFFFFF
essume th—t a ≥ b, cF ‡rite x =
√
a, y = b
c F „hen x ≥ 1 —nd the inequ—lity
(ab + bc + ca)(a + b + c) + 6 ≥ 5(a + b + c)
˜e™omes
x3
(y + y−1
) − 5x2
+ (y2
+ 9 + y−2
) − 5x−1
(y + y−1
) + x−3
(y + y−1
) ≥ 0
„his ™—n ˜e seper—ted —s
2x3
− 5x2
+ 11 − 10x−1
+ 2x−3
≥ 0
—nd
x3
(y + y−1
− 2) + (y2
+ y−2
− 2) − 5x−1
(y + y−1
− 2) + x−3
(y + y−1
− 2) ≥ 0
„he first one is e—syF eõut the se™ond oneD
xote th—t x3
+ x−3
≥ 2 ≥ 2x−1
—nd (y2
+ y−2
− 2) ≥ 3(y1
+ y−1
− 2) ≥ 3x−1
(y1
+ y−1
− 2)
sin™e
y2
− 3y + 4 − 3y−1
+ y−2
= (y − 1)2
(y − 1 + y−1
)
QFFFFFFFFF
vem— of †—ile girto—je
(a + b) (b + c) (c + a) + 7 ≥ 5 (a + b + c)
@g—n e—sy prove ˜y w†A
fut
(a + b) (b + c) (c + a) = a2
b + a2
c + b2
c + b2
a + c2
a + c2
b + 2abc
= a2
b + a2
c + b2
c + b2
a + c2
a + c2
b + 3abc − abc
UR

= (a + b + c) (bc + ca + ab) − abc = (a + b + c) (bc + ca + ab) − 1
where ‡e h—ve used th—t —˜™ a I in the l—st step of our ™—l™ul—tionF „husD ‡e h—ve
((a + b + c) (bc + ca + ab) − 1) + 7 ≥ 5 (a + b + c)
Y in other wordsD
(a + b + c) (bc + ca + ab) + 6 ≥ 5 (a + b + c)
…pon division ˜y — C ˜ C ™D this ˜e™omes
(bc + ca + ab) +
6
a + b + c
≥ 5
pin—llyD sin™e abc = 1D
‡e h—ve bc = 1
a D ca = 1
b —nd ab = 1
c D —nd thus ‡e get
1
a
+
1
b
+
1
c
+
6
a + b + c
≥ 5
WVF
vet a, b, c > 0 —nd with —ll k ≥ −3/2 F €rove the inequ—lityX
a3
+ (k + 1)abc
b2 + kbc + c2
≥ a + b + c
SolutionX
yur inequ—lity is equiv—lent to
a(a2
+ bc − b2
− c2
)
b2 + kbc + c2
+
b(b2
+ ca − c2
− a2
)
c2 + kca + a2
+
c(c2
+ ab − a2
− b2
)
a2 + kab + b2
≥ 0,
(a2
−b2
)
a
b2 + kbc + c2
−
b
a2 + kac + c2
+c
a(b − c)
b2 + kbc + c2
+
b(a − c)
a2 + kac + c2
+
c2
+ ab − a2
− b2
a2 + kab + b2
≥ 0.
prom nowD ‡e see th—t
a
b2 + kbc + c2
−
b
a2 + kac + c2
=
(a − b)(a2
+ b2
+ c2
+ ab + kac + kbc)
(b2 + kbc + c2)(a2 + kac + c2)
,
c2
+ ab − a2
− b2
a2 + kab + b2
=
(c − a)(c − b) − a(b − c) − b(a − c) − (a − b)2
a2 + kab + b2
,
a(b − c)
b2 + kbc + c2
−
a(b − c)
a2 + kab + b2
=
a(a − c)(b − c)(a + c + kb)
(a2 + kab + b2)(b2 + kbc + c2)
,
b(a − c)
a2 + kac + c2
−
b(a − c)
a2 + kab + b2
=
a(a − c)(b − c)(b + c + ka)
(a2 + kab + b2)(a2 + kac + c2)
.
„hereforeD the inequ—lity ™—n ˜e rewritten —s
A(a − b)2
+
c(a − c)(b − c)
a2 + kab + b2
B ≥ 0,
where
A =
(a + b)(a2
+ b2
+ c2
+ ab + kac + kbc)
(a2 + kac + c2)(b2 + kbc + c2)
−
c
a2 + kab + b2
,
—nd
B =
a(a + c + kb)
b2 + kbc + c2
+
b(b + c + ka)
a2 + kac + c2
+ 1.
US

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