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Inequality Techniques
Inequality Techniques
  To prove x  y, it can be easier to prove x  y  0
Inequality Techniques
             To prove x  y, it can be easier to prove x  y  0
                           p2  q2
e.g. i 1995 Prove pq 
                               2
Inequality Techniques
               To prove x  y, it can be easier to prove x  y  0
                            p2  q2
e.g. i 1995 Prove pq 
                                 2
          p2  q2
                   pq
             2
Inequality Techniques
               To prove x  y, it can be easier to prove x  y  0
                              p2  q2
e.g. i 1995 Prove pq 
                                 2
          p2  q2          p 2  2 pq  q 2
                   pq 
             2                    2
Inequality Techniques
               To prove x  y, it can be easier to prove x  y  0
                              p2  q2
e.g. i 1995 Prove pq 
                                 2
          p2  q2          p 2  2 pq  q 2
                   pq 
             2                    2
                           p  q 2
                        
                               2
Inequality Techniques
               To prove x  y, it can be easier to prove x  y  0
                              p2  q2
e.g. i 1995 Prove pq 
                                 2
          p2  q2          p 2  2 pq  q 2
                   pq 
             2                    2
                           p  q 2
                        
                               2
                        0
Inequality Techniques
               To prove x  y, it can be easier to prove x  y  0
                              p2  q2
e.g. i 1995 Prove pq 
                                 2
          p2  q2          p 2  2 pq  q 2
                   pq 
             2                    2
                         
                            p  q2


                               2
                         0
                 p2  q2
                         pq
                    2
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                          b 2  c 2  2bc
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                         b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
                                             1
b) If a  b  c  1, prove ab  ac  bc 
                                             3
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
                                             1
b) If a  b  c  1, prove ab  ac  bc 
                                             3
              a  b  c  a  b  c   2ab  ac  bc 
               2     2     2               2
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
                                             1
b) If a  b  c  1, prove ab  ac  bc 
                                             3
              a  b  c  a  b  c   2ab  ac  bc 
               2     2     2               2


            a  b  c   2ab  ac  bc   ab  ac  bc
                          2
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
                                             1
b) If a  b  c  1, prove ab  ac  bc 
                                             3
              a  b  c  a  b  c   2ab  ac  bc 
               2     2     2               2


            a  b  c   2ab  ac  bc   ab  ac  bc
                          2


                               3ab  ac  bc   a  b  c 
                                                              2
ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
                        a  b 2  0
                  a 2  2ab  b 2  0
                        a 2  b 2  2ab
                          a 2  c 2  2ac
                           b 2  c 2  2bc
                2a 2  2b 2  2c 2  2ab  2ac  2bc
                     a 2  b 2  c 2  ab  ac  bc
                                             1
b) If a  b  c  1, prove ab  ac  bc 
                                             3
              a  b  c  a  b  c   2ab  ac  bc 
               2     2     2               2


            a  b  c   2ab  ac  bc   ab  ac  bc
                          2


                               3ab  ac  bc   a  b  c 
                                                              2


                             3ab  ac  bc   1
                                                1
                                ab  ac  bc 
                                                3
1
c) Prove     a  b  c   3 abc
           3
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
                            a 3  b 3  c 3  3abc  0
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
                            a 3  b 3  c 3  3abc  0
                                a  b  c   abc
                             1 3 3 3
                             3
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
                            a 3  b 3  c 3  3abc  0
                                a  b  c   abc
                             1 3 3 3
                             3
           1      1         1
let a  a , b  b , c  c
           3      3         3
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
                            a 3  b 3  c 3  3abc  0
                                a  b  c   abc
                             1 3 3 3
                             3
           1      1         1
let a  a , b  b , c  c
           3      3         3
                                                    1   1 1
                                1
                                  a  b  c   a 3 b 3 c 3
                                3
1
c) Prove     a  b  c   3 abc
           3
  a 2  b 2  c 2  ab  ac  bc
 a 2  b 2  c 2  ab  ac  bc  0
 a  b  c a 2  b 2  c 2  ab  ac  bc   0
 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c
                   a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
                            a 3  b 3  c 3  3abc  0
                                a  b  c   abc
                             1 3 3 3
                             3
           1      1         1
let a  a , b  b , c  c
           3      3         3
                                                    1   1 1
                                1
                                  a  b  c   a 3 b 3 c 3
                                3
                                1
                                  a  b  c   3 abc
                                3
Arithmetic Mean  Geometric Mean
 a1  a2    an n
                  a1a2  an
         n
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
                                     1
                                        x  y  z   3 xyz
                                     3
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
                                     1
                                        x  y  z   3 xyz
                                     3
                                         x  y  z  33 xyz
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
                                     1
                                        x  y  z   3 xyz
                                     3
                                         x  y  z  33 xyz

                                    xy  yz  xz  33  xy  yz  xz 
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
                                     1
                                        x  y  z   3 xyz
                                     3
                                         x  y  z  33 xyz

                                    xy  yz  xz  33  xy  yz  xz 

                                    xy  yz  xz  33 x 2 y 2 z 2
Arithmetic Mean  Geometric Mean
                  a1  a2    an n
                                   a1a2  an
                          n

d) Suppose 1  x 1  y 1  z   8, prove xyz  1

                             1  x 1  y 1  z   8
              1  x  y  xy  z  xz  yz  xyz  8
                                     1
                                        x  y  z   3 xyz
                                     3
                                         x  y  z  33 xyz

                                    xy  yz  xz  33  xy  yz  xz 

                                    xy  yz  xz  33 x 2 y 2 z 2
                                     xy  yz  xz  3 xyz 
                                                                2
                                                            3
1  x  y  z  xy  xz  yz  xyz  8
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                          2
          3         3
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                            2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2       3
     3         3            3
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                               2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2             3
     3         3               3


                        1    3   xyz   8
                                       3
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                               2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2             3
     3         3               3


                        1    3   xyz   8
                                       3



                            1  3 xyz  2
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                               2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2                3
     3         3               3


                        1    3   xyz   8
                                         3



                            1  3 xyz  2
                                   3   xyz  1
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                               2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2                3
     3         3               3


                        1    3   xyz   8
                                         3



                            1  3 xyz  2
                                   3   xyz  1
                                       xyz  1
1  x  y  z  xy  xz  yz  xyz  8
     1  3 xyz  3 xyz   xyz  8
                               2
          3         3


1  3 xyz  3 xyz    xyz   8
                        2                3
     3          3              3


                        1    3   xyz   8
                                         3



                            1  3 xyz  2
                                   3   xyz  1
                                       xyz  1

         Inequalities Sheet

              Exercise 10D

Note: Cambridge 8H (Book 1); 28

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X2 T08 04 inequality techniques (2011)

  • 2. Inequality Techniques To prove x  y, it can be easier to prove x  y  0
  • 3. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2
  • 4. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2 p2  q2  pq 2
  • 5. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2 p2  q2 p 2  2 pq  q 2  pq  2 2
  • 6. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2 p2  q2 p 2  2 pq  q 2  pq  2 2  p  q 2  2
  • 7. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2 p2  q2 p 2  2 pq  q 2  pq  2 2  p  q 2  2 0
  • 8. Inequality Techniques To prove x  y, it can be easier to prove x  y  0 p2  q2 e.g. i 1995 Prove pq  2 p2  q2 p 2  2 pq  q 2  pq  2 2   p  q2 2 0 p2  q2   pq 2
  • 9. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac
  • 10. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0
  • 11. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0
  • 12. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab
  • 13. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc
  • 14. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc
  • 15. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc
  • 16. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc 1 b) If a  b  c  1, prove ab  ac  bc  3
  • 17. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc 1 b) If a  b  c  1, prove ab  ac  bc  3 a  b  c  a  b  c   2ab  ac  bc  2 2 2 2
  • 18. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc 1 b) If a  b  c  1, prove ab  ac  bc  3 a  b  c  a  b  c   2ab  ac  bc  2 2 2 2  a  b  c   2ab  ac  bc   ab  ac  bc 2
  • 19. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc 1 b) If a  b  c  1, prove ab  ac  bc  3 a  b  c  a  b  c   2ab  ac  bc  2 2 2 2  a  b  c   2ab  ac  bc   ab  ac  bc 2 3ab  ac  bc   a  b  c  2
  • 20. ii 1994 a) Prove a 2  b 2  c 2  ab  bc  ac a  b 2  0 a 2  2ab  b 2  0  a 2  b 2  2ab a 2  c 2  2ac b 2  c 2  2bc 2a 2  2b 2  2c 2  2ab  2ac  2bc a 2  b 2  c 2  ab  ac  bc 1 b) If a  b  c  1, prove ab  ac  bc  3 a  b  c  a  b  c   2ab  ac  bc  2 2 2 2  a  b  c   2ab  ac  bc   ab  ac  bc 2 3ab  ac  bc   a  b  c  2 3ab  ac  bc   1 1 ab  ac  bc  3
  • 21. 1 c) Prove a  b  c   3 abc 3
  • 22. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc
  • 23. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0
  • 24. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0
  • 25. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0
  • 26. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0 a 3  b 3  c 3  3abc  0
  • 27. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0 a 3  b 3  c 3  3abc  0 a  b  c   abc 1 3 3 3 3
  • 28. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0 a 3  b 3  c 3  3abc  0 a  b  c   abc 1 3 3 3 3 1 1 1 let a  a , b  b , c  c 3 3 3
  • 29. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0 a 3  b 3  c 3  3abc  0 a  b  c   abc 1 3 3 3 3 1 1 1 let a  a , b  b , c  c 3 3 3 1 1 1 1 a  b  c   a 3 b 3 c 3 3
  • 30. 1 c) Prove a  b  c   3 abc 3 a 2  b 2  c 2  ab  ac  bc a 2  b 2  c 2  ab  ac  bc  0 a  b  c a 2  b 2  c 2  ab  ac  bc   0 a 3  ab 2  ac 2  a 2b  a 2 c  abc  a 2b  b 3  bc 2  ab 2  abc  b 2 c  a 2 c  b 2 c  c 3  abc  ac 2  bc 2  0 a 3  b 3  c 3  3abc  0 a  b  c   abc 1 3 3 3 3 1 1 1 let a  a , b  b , c  c 3 3 3 1 1 1 1 a  b  c   a 3 b 3 c 3 3 1 a  b  c   3 abc 3
  • 31. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n
  • 32. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1
  • 33. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8
  • 34. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8 1  x  y  z   3 xyz 3
  • 35. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8 1  x  y  z   3 xyz 3 x  y  z  33 xyz
  • 36. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8 1  x  y  z   3 xyz 3 x  y  z  33 xyz xy  yz  xz  33  xy  yz  xz 
  • 37. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8 1  x  y  z   3 xyz 3 x  y  z  33 xyz xy  yz  xz  33  xy  yz  xz  xy  yz  xz  33 x 2 y 2 z 2
  • 38. Arithmetic Mean  Geometric Mean a1  a2    an n  a1a2  an n d) Suppose 1  x 1  y 1  z   8, prove xyz  1 1  x 1  y 1  z   8 1  x  y  xy  z  xz  yz  xyz  8 1  x  y  z   3 xyz 3 x  y  z  33 xyz xy  yz  xz  33  xy  yz  xz  xy  yz  xz  33 x 2 y 2 z 2 xy  yz  xz  3 xyz  2 3
  • 39. 1  x  y  z  xy  xz  yz  xyz  8
  • 40. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3
  • 41. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3
  • 42. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3 1  3 xyz   8 3
  • 43. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3 1  3 xyz   8 3 1  3 xyz  2
  • 44. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3 1  3 xyz   8 3 1  3 xyz  2 3 xyz  1
  • 45. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3 1  3 xyz   8 3 1  3 xyz  2 3 xyz  1 xyz  1
  • 46. 1  x  y  z  xy  xz  yz  xyz  8 1  3 xyz  3 xyz   xyz  8 2 3 3 1  3 xyz  3 xyz    xyz   8 2 3 3 3 3 1  3 xyz   8 3 1  3 xyz  2 3 xyz  1 xyz  1 Inequalities Sheet Exercise 10D Note: Cambridge 8H (Book 1); 28