1. Nguyen Duy Tung
567 Nice And Hard Inequalities
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3. IF
—A if a, b, c —re positive re—l num˜ersD then
a b c a2 + 1 b2 + 1 c2 + 1
+ + ≥ + + .
b c a b2 + 1 c2 + 1 a2 + 1
˜Avet a, b, c, d ˜e positive re—l num˜ersF€rove th—t
a2 − bd b2 − ca c2 − db d2 − ac
+ + + ≥ 0.
b + 2c + d c + 2d + a d + 2a + b a + 2b + c
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 a2 + b2
⇒ + =
b a 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 a2 a2 a2 b2 a2 + 1
= + ≥ + = .
b2 b2 + 1 b2 (b2 + 1) b2 + 1 b2 (b2 + 1) b2 + 1
„herefore
a 2 a b a2
1+ =1+2 + +
b b a b2
a2 + 1 b2 + 1 a2 + 1
≥1+2 + +
b2 + 1 a2 + 1 b2 + 1
2
a2 + 1
= 1+ .
b2 + 1
„herefore
a b c a2 + 1 b2 + 1 c2 + 1
+ + ≥ + +
b c a b2 + 1 c2 + 1 a2 + 1
—s requireF
˜Axoti™e th—t
2(a2 − bd) 2a2 + b2 + d2 + 2c(b + d)
+b+d=
b + 2c + d b + 2c + d
2 2
(a − b) + (a − d) + 2(a + c)(b + d)
= (1)
b + 2c + d
end simil—rlyD
2(c2 − db) (c − d)2 + (c − b)2 + 2(a + c)(b + d)
+b+d= (2)
d + 2a + b d + 2a + b
…sing g—u™hyEƒ™hw—rz9s inequ—lityDwe get
(a − d)2 (c − d)2 [(a − b)2 + (c − d)2 ]
+ ≥ (3)
b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b)
Q
4. (a − d)2 (c − b)2 [(a − d)2 + (c − b)2 ]2
+ ≥ (4)
b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b)
2(a + c)(b + d) 2(a + c)(b + d) 8(a + c)(b + d)
+ ≥ (5)
b + 2c + d d + 2a + b (b + 2c + d) + (d + 2a + b)
prom @IAD@PAD@QAD@RA —nd @SAD we get
a2 − bd c2 − db (a + c − b − d)2 + 4(a + c)(b + d)
2( + )+b+d≥ = a + b + c + d.
b + 2c + d d + 2a + b a+b+c+d
or
a2 − bd c2 − db a+c−b−d
+ ≥
b + 2c + d d + 2a + b 2
sn the s—me m—nnerDwe ™—n —lso show th—t
b2 − ca d2 − ac b+d−a−c
+ ≥
c + 2d + a a + 2b + 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 bc ca 3
+ + ≤
1 − c2 1 − a2 1 − b2 8
SolutionX prom the given ™ondition „he inequ—lity is equiv—lent to
4ab 3
≤
a2 + b2 + 2(ab + bc + ca) 2
˜ut from g—uhy ƒhw—rz inequ—lity
4ab
a2 + b2 + 2(ab + bc + ca)
ab ab
≤ + 2
a2 + ab + bc + ca b + ab + bc + ca
ab ab
= +
(a + b)(a + c) (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
1
a=b=c=
3
R
5. yr ‡e ™—n use the f—™t th—t
4ab 4ab
≤
a2 + b2 + 2(ab + bc + ca) (2ab + 2ac) + (2ab + 2bc)
ab ab
≤ +
2a(b + c) 2b(a + c)
1 b a
= +
2 b+c a+c
1 b c 3
= + =
2 b+c b+c 2
QD vet a, b, c ˜e the positive re—l num˜ersF €rove th—t
ab2 + bc2 + ca2 4. 3 (a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)
1+ ≥
(ab + bc + ca)(a + b + c) (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)(ab2 + bc2 + ca2 )
(a + b + c)2 +
ab + bc + ca
≥ 4. 3 (a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)
‡e h—ve
(a + b + c)(ab2 + bc2 + ca2 ) (a2 + ab + bc)(c + a)(c + b)
(a + b + c)2 + =
ab + bc + ca ab + bc + ca
fy using ewEqw inequ—lity ‡e get
(a2 + ab + bc)(c + a)(c + b) 3
(a2 + ab + bc)(b2 + bc + ca)(c2 + ca + ab)[(a + b)(b + c)(c + a)]2
≥ 3.
ab + bc + ca 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 1 1 1 1
1+ 1+ ··· 1 + ≤ 1+ 1+ ··· 1 +
a0 a1 − a0 an − a0 a0 a1 an
SolutionX ‡e will prove it ˜y indu™tionF
por n = 1 ‡e need to ™he™k th—t
1 1 1 1
1+ 1+ ≤ 1+ 1+
a0 a1 − a0 a0 a1
whi™h is equiv—lent to a0 (a1 − a0 − 1) ≥ 0, whi™h is true ˜y given ™onditionF
vet
1 1 1 1 1 1
1+ 1+ ··· 1 + ≤ 1+ 1+ ··· 1 +
a0 a1 − a0 ak − a0 a0 a1 ak
S
6. it rem—ins to prove th—tX
1 1 1
1+ 1+ ··· 1 + ≤
a0 a1 − a0 ak+1 − a0
1 1 1
≤ 1+ 1+ ··· 1 +
a0 a1 ak+1
fy our hypothesis
1 1 1
1+ 1+ ··· 1 + ≥
a0 a1 ak+1
1 1 1 1
≥ 1+ 1+ 1+ ··· 1 +
ak+1 a0 a1 − a0 ak − a0
id estD it rem—ins to prove th—tX
1 1 1 1
1+ 1+ 1+ ··· 1 + ≥
ak+1 a0 a1 − a0 ak − a0
1 1 1
≥1+ 1+ ··· 1 +
a0 a1 − a0 ak+1 − a0
fut
1 1 1 1
1+ 1+ 1+ ··· 1 + ≥
ak+1 a0 a1 − a0 ak − a0
1 1 1
≥1+ 1+ ··· 1 + ⇔
a0 a1 − a0 ak+1 − a0
1 1 1 1
⇔ + 1+ ··· 1 + ≥
ak+1 ak+1 a0 a1 − a0 ak − a0
1 1 1
≥ 1+ ··· 1 + ⇔
(ak+1 − a0 )a0 a1 − a0 ak − a0
1 1 1
⇔1≥ 1+ ··· 1 +
ak+1 − a0 a1 − a0 ak − a0
fut ˜y our ™onditions ‡e o˜t—inX
1 1 1
1+ ··· 1 + ≤
ak+1 − a0 a1 − a0 ak − a0
1 1 1
≤ 1+ ··· 1 + = 1.
k 1 k−1
„husD the inequ—lity is provenF
SD
qiven a, b, c > 0F €rove th—t
√
3
3 a2 + bc abc
≥ 9.
b2 + c2 (a + b + c)
Solution X „his ineq is equiv—lent toX
a2 + bc 9
≥ 3
3 2
abc(a2 + bc) (b2 + c2 ) (a + b + c)
fy ewEqw ineq D ‡e h—ve
a2 + bc
=
3 2
abc(a2 + bc) (b2 + c2 )
T
7. a2 + bc 3(a2 + bc)
= ≥
3
(a2 + bc)c(a2 + bc)b(b2 + c2 )a a2 b
sym
ƒimil—rlyD this ineq is true if ‡e prove th—tX
3(a2 + b2 + c2 + ab + bc + ca) 9
2b
≥ 3
a (a + b + c)
sym
a3 + b3 + c3 + 3abc ≥ a2 b
sym
‡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 1 1 2
+ + ≥ .
2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca
„he inequ—lity is equiv—lent to
ab + bc + ca
≥ 2, (1)
2a2 + bc
or
a(b + c) bc
+ ≥ 2.(2)
2a2 + bc bc + 2a2
…sing the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
2
bc ( bc)
≥ = 1.(3)
bc + 2a2 bc(bc + 2a2 )
„hereforeD it suffi™es to prove th—t
a(b + c)
≥ 1.(4)
2a2 + bc
ƒin™e
a(b + c) a(b + c)
2 + bc
≥
2a 2(a2 + bc)
it is enough to ™he™k th—t
a(b + c)
≥ 2, (5)
a2 + bc
whi™h is — known resultF
‚em—rkX
2ca + bc 2bc + ca 4c
2 + bc
+ 2 ≥ .
2a 2b + ca 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 1 1 1 12
+ 2 + 2 + ≥ .
2a2 + bc 2b + ca 2c + ab ab + bc + ca (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 1 4
+ ≥ .
2a2 + bc 2b2 + ca (a + b)(a + b + c)
U
8. in f—™tD this is used to repl—™ed for 4no two of whi™h —re zero4D so th—t the fr—™tions
1 1 1 1
, , ,
2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca
h—ve me—ningsF
fesidesD the i—ker —lso works for itX
1 1 1 2(ab + bc + ca)
+ 2 + 2 ≥
2a2 + bc 2b + ca 2c + ab a2 b2
+ 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 1 1 1 12
+ + + ≥ .
2a2 + bc 2b2 + ca 2c2 + ab ab + bc + ca (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) c2 (2b + a)
+
2a2 + bc 2b2 + ca
c4 (2a + b)(2b + a)
≥2
(2a2 + bc) (2b2 + ca)
2c2 (2a + b)(2b + a)
≥
(a2 + ab + b2 ) (a + b + c)
4c2 6abc c
= +
a+b+c a+b+c a2 + ab + b2
2c2 a + bc2 + 2ab2 + b2 c
2a2 + bc
c2 (2a + b) c2 (2b + a)
= +
2a2 + bc 2b2 + ca
4c2 6abc c
≥ +
a+b+c a+b+c a2 + ab + b2
4 a2 + b2 + c2 6abc c
= +
a+b+c a+b+c a2 + ab + b2
4 a2 + b2 + c2 6abc (a + b + c)2
≥ +
a+b+c a+b+c c (a2 + ab + b2 )
4 a2 + b2 + c2 6abc
= +
ab + c ab + bc + ca
2a2 b + 2ab2 + 2b2 c + 2bc2 + 2c2 a + 2ca2
⇒
2a2 + bc
V
9. 2c2 a + bc2 + 2ab2 + b2 c
= (b + c) +
2a2 + bc
4 a2 + b2 + c2 6abc
≥ (b + c) + +
a+b+c ab + bc + ca
8 a2 + b2 + c2 + ab + bc + ca 2 a2 b + ab2
= −
a+b+c ab + bc + ca
1 1
⇒ +
2a2 + bc ab + bc + ca
4 a2 + b2 + c2 + ab + bc + ca
≥
(a + b + c) ( (a2 b + ab2 ))
12
≥ .
(a + b + c)2
(a + b)(a + c) a2 + bc 12(ab + bc + ca)
<=> + −2≥
2a2 + bc 2a2 + bc (a + b + c)2
prom
2a2 + 2bc bc
2 + bc
−3= 2 ≥1
2a 2a + bc
‡e get
a2 + bc
−2≥0
2a2 + bc
xowD ‡e will prove the stronger
(a + b)(a + c) 12(ab + bc + ca)
2 + bc
≥
2a (a + b + c)2
prom ™—u™hyEs™h—rztD ‡e h—ve
(a + b)(a + c) 1 3(a + b)(b + c)(c + a)
= (a+b)(b+c)(c+a)( ≥
2a2 + bc (2a2 + bc)(b + c) ab(a + b) + bc(b + c) + ca(c + a)
pin—llyD ‡e only need to prove th—t
(a + b)(b + c)(c + a) 4(ab + bc + ca)
≥
ab(a + b) + bc(b + c) + ca(c + a) (a + b + c)2
(a + b + c)2 4[ab(a + b) + bc(b + c) + ca(c + a) 8abc
≥ =4−
ab + bc + ca (a + b)(b + c)(c + a) (a + b)(b + c)(c + a)
a2 + b2 + c2 8abc
+ ≥2
ab + bc + ca (a + b)(b + c)(c + a)
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 F €rove th—t
c
1 8
3 ≤ .
9
a+b+ 2(a + c)
SolutionX „his pro˜lem is r—ther e—syF …sing the ewEqw inequ—lityD ‡e h—veX
c+a c+a 3 (a + b)(c + a)
a+b+ 2(c + a) = a + b + + ≥3 .
2 2 2
W
10. ƒo th—tX
1 2
3 ≤ .
27(a + b)(c + a)
a+b+ 2(c + a)
„husD it9s enough to ™he™k th—tX
1
≤ 4 ⇐⇒ 6(a + b)(b + c)(c + a) ≥ a + b + c,
3(a + b)(c + a)
whi™h is true sin™e
9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca)
—nd
16(ab + bc + ca)2 3
16abc(a + b + c) ≥ ab + bc + ca ⇒ ≥ ab + bc + ca ⇐⇒ ab + bc + ca ≥ .
3 16
„he Solution is ™ompletedF iqu—lity holds if —nd only if a = b = c = 4 F
1
WD vet x, y, z ˜e positive re—l num˜ers su™h th—t xyz = 1F €rove th—t
x3 + 1 y3 + 1 z3 + 1 √
+ + ≥ 2 xy + yz + zx.
x4 + y + z y4 + z + x z4 + x + y
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 + y 2 z + yz 2 )(x2 y + x2 z + xyz)
≤ (x3 + y 2 z + yz 2 ) + (x2 y + x2 z + xyz)
(x + y + z)(x3 + 1)
= (x + y + z)(x2 + yz) = .
x
it follows th—t √
x3 + 1 2x xy + yz + zx
≥ .
x4 + y + z 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
√ 2 √ 2
5 5
(c + b) · −1 ·c · + 1 b · (c − b)
2 2
√ 5
5(b + c) √
≤ (c + b) ≤ 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) =
√ √
2 + 1; 2 − 1; 0 —nd —ll its ™y™li™ permut—tionsF 2
5 5
IH
11. IID vet a, b, c > 0 —nd a + b + c = 3F €rove th—t
1 1 1 3
2 + b2
+ 2 + c2
+ 2 + a2
≤
3+a 3+b 3+c 5
SolutionX ‡e h—veX
1 1 1 3
+ + ≤
3 + a2 + b2 3 + b2 + c2 3 + c2 + a2 5
3 3 3 9
<=> + + ≤
3 + a2 + b2 3 + b2 + c2 3 + c2 + a 2 5
a2 + b2 6
≥
3 + a2 + b2 5
…sing g—u™hyEƒ™hw—rz9s inequ—lityX
a2 + b2
( 3 + a2 + b2 ) ≥ ( a2 + b2 )2
3 + a2 + b2
„h—t me—ns ‡e h—ve to prove
6
( a2 + b2 )2 ≥ ( (3 + a2 + b2 ))
5
54 12
(a2 + b2 ) + 2 (a2 + b2 )(a2 + c2 ) ≥ + a2
5 5
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 1 1
+ + ≥1
4a2 − bc + 1 4b2 − ca + 1 4c2 − ab + 1
SolutionX in f—™tD the sh—rper inequ—lity holds
1 1 1 3
+ + ≥ .
4a2 − bc + 1 4b2 − ca + 1 4c2 − ab + 1 2
„he inequ—lity is equiv—lent to
1 1 1 3
+ + ≥ .
a(4a + b + c) b(4b + c + a) c(4c + a + b) 2
…sing the g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
2
1 4a + b + c 1 1
≥ = .
a(4a + b + c) a a a2 b2 c2
„hereforeD it suffi™es to prove th—t
2 4a + b + c 4b + c + a 4c + a + b
≥ + + .
3a2 b2 c2 a b c
ƒin™e
4a + b + c a+b+c (a + b + c)(ab + bc + ca)
= 3+ =9+
a a abc
a+b+c
=9+ ,
abc
II
12. this inequ—lity ™—n ˜e written —s
2
9a2 b2 c2 + abc(a + b + c) ≤ ,
3
whi™h is true ˜e™—use
3
ab + bc + ca 1
a2 b2 c2 ≤ = ,
3 27
—nd
(ab + bc + ca)2 1
abc(a + b + c) ≤ = .
3 3
IQD qiven a, b, c ≥ 0 su™h th—t ab + bc + ca = 1F €rove th—t
1 1 1
+ + ≥1
4a2 − bc + 2 4b2 − ca + 2 4c2 − ab + 2
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
[c(2a + b) + b(2a + c)]2
4a2 − bc + 2(ab + bc + ca) = (2a + b)(2a + c) ≤
4bc
(ab + bc + ca)2 1
= = .
bc bc
it follows th—t
1
≥ bc.
4a2 − bc + 2
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 1 1 1 1 1 9
( + + )( + + )≥ .
a b c 1+a 1+b 1+c 1 + abc
SolutionX „he origin—l inequ—lity is equiv—lent to
abc + 1 abc + 1 abc + 1 1 1 1
+ + + + ≥9
a b c a+1 b+1 c+1
or
1 + a2 c 1 1 1
+ + ≥9
cyc
a a+1 b+1 c+1
fy g—u™hy ƒ™hw—rz ineq —nd ewEqw ineqD
1 + a2 c c(1 + a)2
≥ ≥ 3 3 (1 + a)(1 + b)(1 + c)
cyc
a cyc
a(1 + c)
—nd
1 1 1 3
+ + ≥
a+1 b+1 c+1 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
3
a(b + 1) + b(c + 1) + c(a + 1) ≤ (a + 1)(b + 1)(c + 1)
2
IP
13. 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
9(a + 1)(b + 1)(c + 1)
( a(b + 1) + b(c + 1) + c(a + 1))2 ≤ 3(ab + bc + ca + a + b + c) ≤
4
„he inequ—lity is trueF g—sePF ifa + b + c + ab + bc + ca ≤ 3abc + 3.
9(a + 1)(b + 1)(c + 1)
<=> ≥ 2(a + b + c + ab + bc + ca) + 3abc + 3
4
fy ewEqw9s inequ—lity X
2 ab(b + 1)(c + 1) ≤ [ab(c + 1) + (b + 1)] = a + b + c + ab + bc + ca + 3abc + 3
9
=> ab + bc + ca + a + b + c + 2 ab(b + 1)(c + 1) ≤
4(a + 1)(b + 1)(c + 1)
3
=> ( a(b + 1) + b(c + 1) + c(a + 1))2 ≤ [ (a + 1)(b + 1)(c + 1)]2
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 1 1 10
+ 2 + 2 ≥
a2 + b2 b + c2 c + a2 (a + b + c)2
SolutionX essume c = min{a, b, c}F „hen
1 1 2
+ 2 ≥ ⇐⇒ (ab − c2 )(a − b)2 ≥ 0
a2 +c 2 b +c 2 ab + c2
end ˜y g—u™hyEs™hw—rz
1 2
((a2 + b2 ) + 8(ab + c2 )) + ≥ 25
a2 + b2 ab + c2
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 b2 (a + c)2 c2 (a + b)2
+ 2 + 2 ≤ a2 + b2 + c2
a2 + 3bc b + 3ac c + 3ab
Solution:
fy g—u™hyEƒ™hw—rz ineq D ‡e h—ve
2 3 3
a2 (b + c) a2 (b + c) a2 (b + c)
2 + bc
= 2 = 2 + c2 ) + c(a2 + b2 )
a (a + bc)(b + c) b(a
a2 (b + c) b2 c2 a2 (b + c) b c
≤ ( 2 2)
+ 2 + b2 )
)= ( 2 2
+ 2 )
4 b(a + c c(a 4 a +c a + b2
ƒimil—rlyD ‡e h—ve
b c c(a2 (b + c) + b2 (c + a))
LHS ≤ a2 (b + c)( 2 + c2
+ 2 )=
a a + b2 a2 + b2
IQ
14. abc(a + b) abc(a + b)
= a2 + b2 + c2 + ≤ a2 + b2 + c2 + ≤ a2 + b2 + c2 + ab + bc + ca
a2 + b2 a2 + b2
whi™h is true ˜y ewEqw ineq
„he origin—l inequ—lity ™—n ˜e written —s
(a + b)2 (a + c)2 8
≤ (a + b + c)2 .
a2 + bc 3
ƒin™e (a + b)(a + c) = (a2 + bc) + a(b + c) ‡e h—ve
(a + b)2 (a + c)2 (a2 + bc)2 + 2a(b + c)(a2 + bc) + a2 (b + c)2
=
a2 + bc a2 + bc
a2 (b + c)2
= a2 + bc + 2a(b + c) + 2 ,
a + bc
—nd thus the —˜ove inequ—lity is equiv—lent to
a2 (b + c)2 8
≤ (a + b + c)2 − a2 − 5 ab,
a2 + bc 3
or
a2 (b + c)2 5(a2 + b2 + c2 ) + ab + bc + ca
2 + bc
≤ .
a 3
ƒin™e
5(a2 + b2 + c2 ) + ab + bc + ca
≥ a2 + b2 + c2 + ab + bc + ca
3
it is enough show th—t
a2 (b + c)2
≤ a2 + b2 + c2 + ab + bc + ca.
a2 + bc
FiFh
IVD qiven
a1 ≥ a2 ≥ . . . ≥ an ≥ 0, b1 ≥ b2 ≥ . . . ≥ bn ≥ 0
n n
ai = 1 = bi
i=1 i=1
pind the m—xmium of
n
(ai − bi )2
i=1
‡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 − y 2 = −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
16. it is enough to show th—t
7 − 3ab 7 − 3bc 7 − 3ca
+ + ≥ 3,
7 − 3ac 7 − 3ba 7 − 3cb
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 1 1 3
+ + ≥
a + x2 (by + cz) a + y 2 (bz + cx) a + z 2 (bx + cy) a+b+c
SolutionX ƒetting
1 1
u= ,v =
x y
—nd
1
w=
z
—nd using the ™ondition
uvw = 1
the inequ—lity ™—n ˜e rewritten —s
u u2 3
= F
au + cv + bw au2 + cuv + bwu a+b+c
epplying g—u™hyD it suffi™es to prove
2
(u + v + w) 3
a u2 + (b + c) uv a+b+c
1
· (b + c − 2a) (x − y)2 0D
2
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
17. €rove th—t
1 1 1 3
+ + ≥
(1 + x2 )(1 + x7 ) (1 + y 2 )(1 + y7 ) (1 + z 2 )(1 + z7) 4
SolutionX pirst ‡e prove this ineq e—sy
1 3
≥ 9
(1 + x2 )(1 + x7 ) 4(x9 + x 2 + 1)
end this ineq ˜e™—meX
1 1 1
9 + 9 + 9 ≥1
x9 +x +1 2 y9 +y +1 2 z9 + z2 + 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 b c 3
√ +√ +√ ≤√ .
a+b b+c c+a 2
SolutionX vet
A = a2 + b2 + c2 , B = ab + bc + ca
3
2A + B = 2 a2 + ab ≤ 3 a2 + ab = 9.
4
fy g—u™hy ƒ™hw—rz inequ—lityD ‡e h—ve
a √ a
√ = a
a+b a+b
√ a
≤ a+b+c .
a+b
fy g—u™hy ƒ™hw—rz inequ—lity —g—inD ‡e h—ve
b b2
=
a+b b(a + b)
(a + b + c)2
≥
b(a + b)
A + 2B
=
A+B
a b A + 2B 2A + B
=3− ≤3− =
a+b a+b A+B A+B
hen™eD it suffi™es to prove th—t
2A + B 9
(a + b + c) · ≤
A+B 2
IU
18. gonsider
√
(a + b + c) 2A + B
= (A + 2B) (2A + B)
(A + 2B) + (2A + B)
≤
2
3
= (A + B)
2
2A + B 3√ 9
⇒ (a + b + c) · ≤ 2A + B ≤
A+B 2 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
√
2 a a+c a
LHS = ( ) ≤ (a2 + b2 + c2 + ab + bc + ca)( )
(a + b)(a + c) (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 2(ab + bc + ca) 9
= ≤
(a + b)(a + c) (a + b)(b + c)(c + a) 4(a + b + c)
end ‡e need to prove th—t
9(a2 + b2 + c2 + ab + bc + ca) 9 6 − (a2 + b2 + c2 )
≤ ⇔ ≤1
4(a + b + c) 2 24 − 5(a2 + b2 + c2 )
⇔ (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 8(a + b + c)2
≤
(a2 + bc)(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 b2 (c + a)2 c2 (a + b)2
+ 2 + 2 ≤ a2 + b2 + c2 + ab + bc + ca.
a2 + bc b + ca c + ab
a2 (b + c)2 b2 (c + a)2 c2 (a + b)2
+ 2 + 2 ≤ a2 + b2 + c2 + ab + bc + ca
a2 + bc b + ca c + ab
IV
19. PSF
qiven
a, b, c ≥ 0
su™h th—t
ab + bc + ca = 1
€rove th—t
1 1 1 9
8 2 + 8 2 + 8 2 ≥
5a + bc 5 b + ca 5 c + ab 4
essume ‡vyq
a≥b≥c
this ineq
1 5 1 5 1
8 2 − + 8 2 − + 8 2 −1≥0
5a + bc 8 5 b + ca
8 5 c + ab
8
8 − 8a2 − 5bc 8 − 8b2 − 5ca 1 − 5 c2 − ab
+ + ≥0
8a2 + 5bc 8b2 + 5ca c2 + 8 ab
5
8
8a(b + c − a) + 3bc 8b(a + c − b) + 5ac c(a + b − 5 c)
+ + ≥0
8a2 + 5bc 8b2 + 5ca c2 + 8 ab
5
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 8a
≥ 2
8b2 + 5ca 8a + 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 b c a+b+c abc(a + b + c)
+ 2 + 2 ≥ + 3
b2 +c2 a +c 2 a +b 2 ab + bc + ca (a + b3 + c3 )(ab + bc + ca)
a a2
=
b2 + c2 ab2 + c2 a
(a + b + c)2
≥ ,
(ab2 + c2 a)
it suffi™es to prove th—t
a+b+c 1 abc
≥ + ,
(ab2 + c2 a) ab + bc + ca (ab + bc + ca) (a3 + b3 + c3 )
IW
20. ˜e™—use
a+b+c 1
2 + c2 a)
−
(ab ab + bc + ca
3abc
= ,
(ab + bc + ca) (ab2 + ca2 )
it suffi™es to prove th—t
3 a3 + b3 + c3 ≥ ab2 + c2 a ,
whi™h is true ˜e™—use
2 a3 + b3 + c3 ≥ ab2 + c2 a .
‚em—rkX
a b c a+b+c 3abc(a + b + c)
+ 2 + 2 ≥ + .
b2 + c2 c + a2 a + b2 ab + bc + ca 2(a3 + b3 + c3 )(ab + bc + ca)
qive
a, b, c ≥ 0
€rove th—t
1 1 1 3 81a2 b2 c2
+ 2 + 2 ≥ +
a2 + bc b + ca c + ab ab + bc + ca 2(a2 + b2 + c2 )4
iqu—lity o™™ur if —nd if only
a = b = c, a = b, c = 0
or —ny ™y™li™ permutionF
it is true ˜e™—use
(1)
1 1 1 3 a2 + b2 + c2
+ 2 + 2 ≥ 3
a2 + bc b + ca c + ab a b + ab3 + b3 c + bc3 + c3 a + ca3
—nd
3 a2 + b2 + c2 3 81a2 b2 c2
(2) ≥ + .
a3 b + ab3 + b3 c + bc3 + c3 a + ca3 ab + bc + ca 2(a2 + b2 + c2 )4
fe™—use
a2 1
−
(a3 b + ab3 ) ab + bc + ca
abc(a + b + c)
= ,
(ab + bc + ca) ( (a3 b + ab3 ))
it suffi™es to prove th—t
4
2(a + b + c) a2 + b2 + c2 ≥ 27abc(ab + bc + ca) a3 b + ab3 ,
whi™h is true ˜e™—use
(a) (a + b + c) a2 + b2 + c2 ≥ 9abc,
(b) a2 + b2 + c2 ≥ ab + bc + ca,
2
(c) 2 a2 + b2 + c2 ≥3 a3 b + ab3 ,
whi™h
(c)
PH
21. 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 (c + a) c2 (a + b) 2(a2 + b2 + c2 )
+ 2 + 2 .
b2 + bc + c2 c + ca + a2 a + ab + b2 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)
≥ 2
(b2 + bc + c2 + ab + bc + ca)
4a2 (ab + bc + ca)
= ,
(b + c)(a + b + c)2
it suffi™es to prove
a2 (a + b + c) a2 + b2 + c2
≥ ,
b+c 2(ab + bc + ca)
or
a2 (a + b + c)3
+a ≥ ,
b+c 2(ab + bc + ca)
or
a (a + b + c)2
≥ ,
b+c 2(ab + bc + ca)
whi™h is true ˜y g—u™hyEƒ™hw—rz inequ—lity
a a2
=
b+c 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) a(b + c) a(b + c)(a2 + b2 + c2 + ab + bc + ca)
+ = ,
b2 + bc + c 2 a+b+c (b2 + bc + c2 )(a + b + c)
—nd
2(a2 + b2 + c2 ) a(b + c) 2(a2 + b2 + c2 + ab + bc + ca)
+ = .
a+b+c a+b+c a+b+c
„hereforeD the inequ—lity ™—n ˜e written in the form
a(b + c) b(c + a) c(a + b)
+ 2 + 2 ≥ 2,
b2 + bc + c 2 c + ca + a 2 a + ab + b2
xote th—t
a(b + c) 4a(b + c)(ab + bc + ca) 4a(ab + bc + ca)
= .
cyc
b2 + bc + c2 cyc
4(b2 + bc + c2 )(ab + bc + ca) cyc
(b + c)(a + b + c)2
PI
22. ƒo th—t ‡e h—ve to proveX
4a(ab + bc + ca)
2,
cyc
(b + c)(a + b + c)2
or
a (a + b + c)2
,
cyc
b+c 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
ab(a + b) + bc(b + c) + ca(c + a)
(a2 + ac + c2 )(b2 + bc + c2 ) ≤ .(1)
a+b
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
√ 2ab
(a2 + ac + c2 )(b2 + bc + c2 ) ≤ ab + c2 + c a + b − ab ≤ ab + c2 + c a + b −
a+b
ab(a + b) + bc(b + c) + ca(c + a)
= .
a+b
xowD from @IAD using the ewEqw inequ—lityD ‡e get
1 1 2
+ 2 ≥
a2 + ac + c2 b + bc + c2 (a2 + ac + c2 )(b2 + bc + c2 )
(2)
2(a + b)
≥ .
ab(a + b) + bc(b + c) + ca(c + a)
prom
(2)
‡e h—ve
a(b + c) 1 1
= ab + 2
b2 + bc + c2 a2
+ ac + c 2 b + bc + c2
2ab(a + b)
≥ = 2.
ab(a + b) + bc(b + c) + ca(c + a)
PWD if
a, b, c > 0
then the following inequ—lity holdsX
a2 (b + c) b2 (c + a) c2 (a + b) a3 + b3 + c3
+ 2 + 2 ≥2
b2 + bc + c2 c + ca + a2 a + ab + b2 a+b+c
„his inequ—lity is equiv—lent to
a2 (b + c)(a + b + c)
≥2 (a3 + b3 + c3 ) (a + b + c)
b2 + bc + c2
or
a2 (ab + bc + ca)
a2 + ≥2 (a3 + b3 + c3 ) (a + b + c),
b2 + bc + c2
PP
23. ˜e™—use
a3 + b3 + c3 (a + b + c)
2 (a3 + b3 + c3 ) (a + b + c) ≤ a2 + b2 + c2 + ,
a2 + b2 + c2
it suffi™es to prove th—t
a2 a3 + b3 + c3 (a + b + c)
≥ 2 ,
b2 + bc + c2 (a + b2 + c2 ) (ab + bc + ca)
˜y g—u™hyEƒ™hw—rz inequ—lityD ‡e h—ve
2 2
a2 a2 + b2 + c2 a2 + b2 + c2
2 + bc + c2
≥ 2 (b2 + bc + c2 )
= ,
b a 2 a2 b2 + a2 bc
it suffi™es to prove th—t
3
a2 + b2 + c2 (ab + bc + ca) ≥ a3 + b3 + c3 (a + b + c) 2 a2 b2 + a2 bc .
vet
1
A= a4 , B = a3 b + ab3 , C = a2 b2 , D = a2 bc,
2
‡e h—ve
2
a2 + b2 + c2 = 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.
„hereforeD it suffi™es to prove th—t
(A + 2C) (2B + D) ≥ (A + 2B) (2C + D) ,
or
2 (A − D) (B − C) ≥ 0,
whi™h is true ˜e™—use
A≥D
—nd
B≥C
QHD qiven
a, b, c ≥ 0
su™h th—t
a+b+c=1
€rove th—t
2 a2 b + b2 c + c2 a + ab + bc + ca ≤ 1
‚ewrite the inform inequ—lity —s
2 a2 b + b2 c + c2 a + ab + bc + ca ≤ (a + b + c)2
PQ
24. 2 (a2 b + b2 c + c2 a) (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
a2 b + b2 c + c2 a
2 (a2 b + b2 c + c2 a) (a + b + c) ≤ + b(a + b + c)
b
it is thus suffi™ient to prove the stronger inequ—lity
a2 b + b2 c + c2 a
a2 + b2 + c2 + ab + bc + ca ≥ + b(a + b + c)
b
„his inequ—lity is equiv—lent to
c(a − b)(b − c)
≥ 0,
b
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 + y 2 (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
a2 b2 b2 c2 c2 a2
A= + 3 2 + 3 2
c3 (a2 + b2 ) a (b + c2 ) b (c + a2 )
xo one like this pro˜lemc ƒetting
1 1 1
x= ,y = ,z =
a b c
PR
25. ‡e h—ve
x2 + y 2 + z 2 ≥ 1
‡e will prove th—t √
x3 y3 z3 3
+ 2 + 2 ≥
y2 + z2 x + z2 x + y2 2
…sing g—u™hyEƒ™hw—rzX
(x2 + y 2 + z 2 )2
LHS ≥
x(y 2 + z2) + y(x2 + z 2 ) + z(x2 + y 2 )
fy ewEqw ‡e h—veX
2 2 2 2 2
x(y 2 +z 2 )+y(x2 +z 2 )+z(x2 +y 2 ) ≤ (x +y +z )(x+y+z) ≤ √ (x2 +y 2 +z 2 ) x2 + y 2 + z 2
3 3
fe™—use
x2 + y 2 + z 2 ≥ 1
ƒo √
(x2 + y 2 + z 2 )2 3
2
≥
√ (x2 + y 2 + z 2 ) x2 + y 2 + z 2 2
3
‡e done3
QPF
vet xDyDz ˜e non neg—tive re—l num˜ers su™h th—t x2 + y 2 + z 2 = 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 − z 2 = g(x)(0 ≤ x ≤ 1 − z 2 ), then ‡e h—ve
x
g (x) = 1 − √ ,
1 − x2 − z 2
‡e get
1 − z2
g (x) > 0 ⇔ 0 ≤ x <
2
—nd
1 − z2
g (x) < 0 ⇔ <x≤ 1 − z2,
2
so ‡e h—ve
mmin = min{g(0), g( 1 − z 2 )} = 1 − z2
—nd
1 − z2
mmax = g = 2 − 2z 2 .
2
e™tu—llyD f —nd written —s
z z
f = f (m) = − m2 + m + 1 − z 2 + z,
2 2
e—sy to prove th—t the —xis of symmetry
1
m= > 2 − 2z 2
z
PS
26. so f (m) is in™re—sing in the interv—l of mD thusD ‡e h—ve
f (m) ≥ f ( 1 − z 2 ) = 1 − z2 + z
—nd
z3 z
f (m) ≤ f ( 2 − 2z 2 ) = + + 2 − 2z 2 .
2 2
ƒin™e
( 1 − z 2 + z)2 = 1 + 2z 1 − z2 ≥ 1
‡e get f (m) ≥ 1 —nd when two of xDyDz —re zero ‡e h—ve f = 1, soW egetfmin = 1.
vet
z3 z
h(z) = + + 2 − 2z 2 ,
2 2
e—sy to prove th—t
1 1
h (z) > 0 ⇔ 0 ≤ z < √ andh (z) < 0 ⇔ √ < z ≤ 1
3 3
then ‡e get √
1 8 3
f (m) ≤ h √ = ,
3 9
√ √
1 8 3 8 3
when x = y = z = √ W ehavef = D so ‡e getfmax = .
3 9 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 + y 2 + z 2 − xyz ≥
3 2
x2 + y 2 + z 2 ⇔ (x + y + z) x2 + y 2 + z 2 − xyz ≥
3
x2 + y 2 + z 2 ⇔ x2 y 2 z 2 + 2 x5 y + x3 y 3 + x3 y 2 z ≥ 0,
sym
1
the l—st inequ—lity is o˜vious trueD so ‡e got f ≥ 1; ‡henx = y = z = √ ‡e h—ve
√ 3
8 3
f= ,
9
—nd ‡e will prove th—t √
8 3
f≤
9
then ‡e ™—n get √
8 3
fmax =
9
e™tu—llyD ‡e h—ve
√ √
8 3 8 3
f≤ ⇔ x + y + z − xyz ≤ ⇔ (x + y + z) x2 + y 2 + z 2 − xyz ≤
9 9
√
8 3 3 2
x2 + y 2 + z 2 ⇔ 27 (x + y + z) x2 + y 2 + z 2 − xyz
9
3 1 2
≤ 64 x2 + y 2 + z 2 ⇔ S (x, y, z) (y − z) ≥ 0,
4 cyc
PT
27. where
S(x, y, z) = 17y 2 (2y−x)2 +17z 2 (2z−x)2 +56y 2 (z−x)2 ++56z 2 (y−x)2 +24x4 +6y 4 +6z 4 +57x2 (y 2 +z 2 )+104y 2 z 2
√
8 3
is o˜vious positiveD so the l—st inequ—lity is o˜vious trueD so ‡e gotfmax = .
9
QQD por positive re—l num˜ersD show th—t
a3 (b + c − a) b3 (c + a − b) c3 (a + b − c) ab + bc + ca
+ + ≤
a2 + bc b2 + ca c2 + ab 2
ineq
ab + bc + ca a3 (b + c − a)
a2 + b2 + c2 + ≥ + a2
2 a2 + bc
ab + bc + ca a2
a2 + b2 + c2 + ≥ (ab + bc + ca)( )
2 a2 + bc
bc 5
a2 + b2 + c2 + (ab + bc + ca)( ) ≥ (ab + bc + ca)
a2 + bc 2
a2 + b2 + c2 bc 5
+ ≥
ab + bc + ca a2 + bc 2
…se two ineq
bc ab ac 4abc
+ 2 + 2 ≥ + 1(1)
a2 + bc c + ab b + ac (a + b)(b + c)(c + a)
it is e—sy to proveF
a2 + b2 + c2 8abc
+ ≥ 2(2)
ab + bc + ca (a + b)(b + c)(c + a)
ƒo e—sy to see th—t
a2 + b2 + c2 bc a2 + b2 + c2 4abc
+ 2 + bc
≥ + +1
ab + bc + ca a ab + bc + ca (a + b)(b + c)(c + a)
a2 + b2 + c2 5
≥ +2≥
2(ab + bc + ca) 2
‡e h—ve done 3
a3 (b + c − a) b3 (c + a − b) c3 (a + b − c) 3abc(a + b + c)
2 + bc
+ 2 + ca
+ 2 + ab
≤
a b c 2(ab + bc + ca)
Solution
a2 + b2 + c2 bc 3abc(a + b + c)
+ + ≥3
ab + bc + ca a2 + bc 2(ab + bc + ca)2
end ‡e prove th—t
3abc(a + b + c) 4abc
2 ≥
2(ab + bc + ca) (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
8
3(a + b + c)(a + b)(b + c)(c + a) ≥ (a + b + c)2 (ab + bc + ca) ≥ 8(ab + bc + ca)2
3
PU
28. ƒo
a2 + b2 + c2 4abc 4abc
LHS ≥ + + +1≥3
ab + bc + ca (a + b)(b + c)(c + a) (a + b)(b + c)(c + a)
vet
a, b, c > 0
ƒhow th—t
a3 (b + c − a) b3 (c + a − b) c3 (a + b − c) 9abc
+ + ≤
a2 + bc b2 + ca c2 + ab 2(a + b + c)
pirstD‡e prove this lenm—X
a2 b2 c2 (a + b + c)2
+ 2 + 2 ≤
a2 + bc b + ca c + ab 2(ab + bc + ca)
bc ac ab a2 + b2 + c2
+ 2 + 2 + ≥2
a2 + bc b + ac c + ab 2(ab + bc + ca)
whi™h is true from
bc ac ab 4abc
+ + ≥1+
a2 + bc b2 + ac c2 + ab (a + b)(b + c)(c + a)
a2 + b2 + c2 4abc
+ ≥1
2(ab + bc + ca) (a + b)(b + c)(c + a)
equ—lity o™™ur if —nd if only
a=b=c
or
a = b, c = 0
or —ny ™y™li™ permutionF
‚eturn to your inequ—lityD‡e h—ve
a3 (b + c − a) 9abc
( + a2 ) ≤ a2 + b2 + c2 +
a2 + bc 2(a + b + c)
or
a2 9abc
(ab + bc + ca) ≤ a2 + b2 + c2 +
a2 + bc 2(a + b + c)
prom
a2 b2 c2 (a + b + c)2
+ 2 + 2 ≤
a2 + bc b + ca c + ab 2(ab + bc + ca)
‡e only need to prove th—t
(a + b + c)2 9abc
≤ a2 + b2 + c2 + or
2 2(a + b + c)
9abc
a2 + b2 + c2 + ≥ 2(ab + bc + ca)
a+b+c
‡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
or —ny ™y™li™ permutionF
QQD vet
a, b, c > 0
PV
29. su™h th—t
a+b+c=1
„hen
a3 + bc b3 + ca c3 + ab
+ + ≥2
a2 + bc b2 + ca c2 + ab
prom the ™ondition
a − 1 = −(b + c)
it follows th—t
a3 + bc a2 (b + c)
= − +1
a2 + bc a2 + bc
„hus it suffi™es to prove th—t
a2 (b + c)
a+b+c≥
a2 + bc
por
a, b, c
positive re—ls prove th—t
ab(a + b)
≥ a
c2 + ab
<=> a4 + b4 + c4 − b2 c2 − c2 a2 − a2 b2 ≥ 0
ab(a + b) c2 (a + b)
+ =2 a
c2 + ab c2 + ab
—nd our inequ—lity ˜e™omes
c2 (a + b)
≤ a
(c2 + ab)
˜ut
c2 (a + b) c2 (a + b)2 (ca + cb)2
= = ≤
(c2 + ab) (c2 + ab)(a + b) a(b2 + c2 ) + b(a2 + c2 )
c2 a2 c2 b2
+ = a
a(b2 + c2 ) b(a2 + c2 )
QR
vet
a, b, c ≥ 0
su™h th—t
a+b+c=1
„hen
a3 + bc b3 + ca c3 + ab
6(a2 + b2 + c2 ) ≥ + +
a2 + bc b2 + ca c2 + ab
Solution
a2 (b + c)
6(a2 + b2 + c2 ) + ≥3
a2 + bc
a2 (b + c)
6(a2 + b2 + c2 ) − 2(a + b + c)2 ≥ (a− )
a2 + bc
a(a − b)(a − c)
4 (a − b)(a − c) ≥
a2 + bc
a
(a − b)(a − c)(4 − 2 )≥0
a + bc
PW
30. essuming ‡vyq
a≥b≥c
then e—sy to see th—t
a
4− ≥0
a2 + bc
—nd
c
4− ≥0
c2 + ab
c a
(c − a)(c − b)(4 − ) ≥ 0and(a − b)(a − c)(4 − 2 )≥0
c2 + ab a + bc
‡e h—ve two ™—ses
g—se I
b
4− ≤0
b2 + ac
then
b
(b − c)(b − a)(4 − )≥0
b2 + ac
so this ineq is true
g—se P
b
4− ≤0
b2 + ac
e—sy to see th—t
c b
4− ≥4− 2
c2 + ab b + ac
ƒo
c b c
LHS ≥ (c − b)2 (4 − ) + (a − b)(b − c)( 2 − )≥0
c2 + ab b + ac c2 + ab
FiFh
QSF vet
x, y, z
˜e re—l num˜ers s—tisfyX
x2 y 2 + 2yx2 + 1 = 0
pind the m—ximum —nd minimum v—lues ofX
2 1 1
f (x, y) = + + y(y + + 2)
x2 x x
SolutionX €ut
1
t= ;k = y + 1
x
D ‡e h—veX
t2 + k 2 = 1
f (x, y) = t2 + tk
€ut
t = cos α; k = sin α
then
f (x, y) = cos α2 + cos α sinα =
1 1 π
sin 2α2 = + √ cos (2α − )
2 2 4
1 1
max f (x, y) = + √
2 2
QH
31. 1 1
min f (x, y) = −√
2 2
FiFh
F
QTF
ƒuppose —D˜D™Dd —re positive integers with ab + cd = 1.
„henD por W e = 1, 2, 3, 4,let (xi )2 + (yi )2 = 1, where xi —nd yi —re re—l num˜ersF
ƒhow th—t
a b c d
(ay1 + by2 + cy3 + dy4 )2 + (ax4 + bx3 + cx2 + dx1 )2 ≤ 2( + + + ).
b a d c
ƒolitionX
…se g—u™hyEƒ™hw—rtz D ‡e h—ve
(ay1 + by2 + cy3 + dy4 )2 ≤
(ay1 + by2 )2 (cy3 + dy4 )2
(ab + cd)( + )=
ab cd
(ay1 + by2 )2 (cy3 + dy4 )2
+
ab cd
ƒimil—rX
(ax4 + bx3 + cx2 + dx1 )2 ≤
(ax4 + bx3 )2 (cx2 + dx1 )2
(ab + cd)( + )
ab cd
(ax4 + bx3 )2 (cx2 + dx1 )2
= +
ab cd
futX
(ay1 + by2 )2 ≤ (ay1 + by2 )2 + (ax1 − bx2 )2 = a2 + b2 + 2ab(y1 y2 − x1 x2 )
ƒimil—rF
(cx2 + dx1 )2 ≤ c2 + d2 + 2cd(x1 x2 − y1 y2 )
D then ‡e getX
(ay1 + by2 )2 (cx2 + dx1 )2
+ ≤
ab cd
a b c d
+ + +
b a d c
@IA
„he s—me —rgument show th—tX
(cy3 + dy4 )2 (ax4 + bx3 )2
+ ≤
cd ab
a b c d
+ + +
b a 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
32. —nd —re— F
€rove th—t X √
3 3 2
F ≤ c
4
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
x+y−z+t
„hen = c —nd x ≥ y ≥ z ≥ t.
4
x+y−z+t 4
„hus ‡e h—veX xyzt ≤ 27( ) .
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
x+t 4
„hen ‡e just prove th—t xy 2 t ≤ 27( ) .
4
fe™—use xy t ≤ x tD ‡e just h—ve to prove
2 3
x+t 4
x3 t ≤ ( )
4
end then it follows th—t the —˜ove inequ—lity is —lso trueF
x x x
+ + +t≥
3 3 3
x x x
44 · · ·t
3 3 3
hen™e
x+y 4
27( ) ≥ x3 t
4
QVF
vet efg ˜e — tri—ngleF €rove th—tX
1 1 1 1 1 1
+ + ≤ + +
a b c a+b−c b+c−a c+a−b
SolutionX
IF
1 1 2b 2b 2b 2
+ = = 2 ≥ 2 =
a+b−c b+c−a (a + b − c)(b + c − a) b − (c − a)2 b b
ƒimil—rlyD ‡e h—ve
1 1 2
+ ≥
b+c−a c+a−b c
1 1 2
+ ≥
c+a−b a+b−c 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
1
f (x) =
x
QP
