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4unit_inequalities1

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									First published by
John Kinny-Lewis in 2011
c   John Kinny-Lewis 2011

National Library of Australia

Cataloguing-in-publication data

ISBN: 978-0-9872782-2-7

This book is copyright. Apart from any fair dealing for the purposes
of private study, research, criticism or review as permitted under the
Copyright Act 1968, no part may be reproduced, stored in a retrieval
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mission. Enquiries to be made to John Kinny-Lewis.

Copying for educational purposes.

Where copies of part or the whole of the book are made under Sec-
tion 53B or Section 53D of the Copyright Act 1968, the law requires
that records of such copying be kept. In such cases the copyright owner
is entitled to claim payment.




Typeset by Christopher Hines


Edited by Christopher Hines
                     Preface


This revision of Harder Extension 1 Topics has been

written to follow the syllabus for the NSW Higher School

Certificate course Extension 2 Mathematics.



It is assumed that the student is familiar with the con-

tent of the corresponding Mathematics and Extension

1 Mathematics syllabuses.




           Other Publications

HSC Extension 1 Mathematics Harder Questions by
Topic

HSC Mathematics Harder Questions by Topic
2                                                         INEQUALITIES

Example 1

             a+b √                                  a     b
Prove that      ≥ ab and hence show that              +     ≥ 2, for a > 0 and b > 0.
              2                                     b     a
Solution

Beginning with the fact that (x − y)2 ≥ 0 for any real x and y,

                                 (x − y)2 ≥ 0

                           x2 − 2xy + y 2 ≥ 0

                                   x2 + y 2 ≥ 2xy




              Letting a = x2 , b = y 2
                         √         √
                     x = a, y = b         (a and b are positive)
                                  √ √
                     ∴a+b≥2 a b
                        a+b √
                    ∴      ≥ ab
                         2


a+b                                       √
    is the arithmetic mean of a and b, and ab is the geometric mean.
 2

                                   a+b √
                             Now      ≥ ab
                                    2
                                   a+b
                                   √ ≥2
                                    ab
                             a   b
                            √ +√ ≥2
                              ab ab
                             √   √
                            a a b b
                             √ +√ ≥2
                            a b  ab

                                a        b
                                  +        ≥ 2.
                                b        a
INEQUALITIES                                                                   3

Example 2

For a > 0, b > 0 and given that (a − b)4 ≥ 0, prove that a4 + b4 ≥ a3 b + ab3 .
Solution
       (a − b)4 = a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 ≥ 0

                                             a4 + b4 ≥ 4a3 b + 4ab3 − 6a2 b2

                                     a4 + b4 + 6a2 b2
                                 ∴                    ≥ a3 b + ab3
                                            4
                                                    a4 + b4 + 6a2 b2
Now using (a2 − b2 )2 we prove that a4 + b4 ≥                        .
                                                           4
                         (a2 − b2 )2 ≥ 0

                            a4 + b4 ≥ 2a2 b2

                          3a4 + 3b4 ≥ 6a2 b2

                        ∴ 4a4 + 4b4 ≥ a4 + b4 + 6a2 b2

                                           a4 + b4 + 6a2 b2
                          ∴ a4 + b 4 ≥
                                                  4
Then finally

                                 a4 + b4 + 6a2 b2
                    a4 + b 4 ≥                    ≥ a3 b + ab3
                                        4
                                         ∴ a4 + b4 ≥ a3 b + ab3
4                                                        INEQUALITIES

Example 3
Here are a series of proofs which show that for a > 0, b > 0 and c > 0,
                            a   b   c   3
                              +   +    ≥ .
                           b+c c+a a+b  2
(i) Prove that a2 + b2 + c2 ≥ ab + bc + ca.
(ii) Show that

        a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ac)

and hence show that
                                  a+b+c √
                                        3
                                       ≥ abc
                                    3
as well as
                             1 1 1     3
                               + + ≥ √ .
                                     3
                             a b c     abc
(iii) Using part (ii) show that,
                                       1 1 1
                           (a + b + c)( + + ) ≥ 9.
                                       a b c
(iv) Hence show that

                            a   b   c   3
                              +   +    ≥ .
                           b+c c+a a+b  2
Solution

(i) From Example 1 we have that:

                                 a2 + b2 ≥ 2ab    (1)

                                 b2 + c2 ≥ 2bc    (2)

                                 a2 + c2 ≥ 2ac    (3)




             (1) + (2) + (3) :       2a2 + 2b2 + 2c2 ≥ 2(ab + bc + ac)

                                      ∴ a2 + b2 + c2 ≥ ab + bc + ac
INEQUALITIES                                                                     5

(ii)


 (a + b + c)(a2 + b2 + c2 − ab − bc − ac) = a3 + ab2 + ac2 − a2 b − abc − a2 c

                                          + b3 + a2 b + bc2 − ab2 − b2 c − abc

                                          + c3 + a2 c + b2 c − abc − bc2 − ac2

                                          = a3 + b3 + c3 − 3abc (∗)

From (i) we have: x2 + y 2 + z 2 ≥ xy + yz + xz.


           ∴ x2 + y 2 + z 2 − xy − yz − xz ≥ 0

(x + y + z)(x2 + y 2 + z 2 − xy − yz − xz) ≥ 0         (x, y and z are all positive)

                   ∴ x3 + y 3 + z 3 − 3xyz ≥ 0         (from (∗))

                               x3 + y 3 + z 3 ≥ 3xyz
                                           √        √      √
Let a = x3 , b = y 3 and c = z 3 , ∴ x = 3 a, y = 3 b, z = 3 c.
                                            √ √ √
                                                3
                           ∴a+b+c≥33a b3c
                             a+b+c √
                                   3
                         ∴        ≥ abc               (4)
                               3
          1      1       1
Let a =     , b = and c = ,
          α      β       γ

                                      1  1 1    3
                                  ∴     + + ≥ √
                                      α β γ   3
                                                αβγ
                                      1 1 1    3
             so replacing symbols,     + + ≥ √ .
                                             3
                                                                      (5)
                                      a b c    abc



(iii) Both sides of (4) and (5) are positive given that a, b and c are positive.
Hence (4) × (5):

                  a+b+c         1 1 1             √
                                                  3             3
                        ×        + +          ≥       abc × √
                                                            3
                    3           a b c                           abc
                                1 1 1
                ∴ (a + b + c)    + +          ≥9
                                a b c
6                                                            INEQUALITIES

(iv)

              a   b   c
Let Exp =       +   +
             b+c a+c a+b
             a+b+c a+b+c a+b+c                                          a+b+c    a
         =        +     +      −3                               since         =     +1
              b+c   a+c   a+b                                            b+c    b+c
                           1   1   1
         = (a + b + c)       +   +                      −3
                          b+c a+c a+b

Now let x = b + c, y = a + c and z = a + b,

                     x + y + z = 2a + 2b + 2c
                                 x+y+z
                 ∴a+b+c=
                                   2
                                 x+y+z      1 1 1
                       ∴ Exp =               + +                −3
                                   2        x y z


                                 1 1 1
                  (x + y + z)     + +           ≥9           (from (iii))
                                 x y z
                     x+y+z       1 1 1              9
                 ∴                + +           ≥
                       2         x y z              2
                 x+y+z       1 1 1                  9
             ∴                + +        −3≥          −3
                   2         x y z                  2
                                                    3
                                         Exp ≥
                                                    2
The result is then proven.
INEQUALITIES                                                               7

Example 4

                     Prove for x ≥ 1, x > tan−1 (1 − x)
Solution

Consider y = x − tan−1 (1 − x),

                          dy          −1
                             =1−
                          dx     1 + (1 − x)2
                                        1
                            =1+
                                   1 + (1 − x)2
                                1 + (1 − x)2 + 1
                            =
                                  1 + (1 − x)2

                            >0       since (1 − x)2 > 0
                          dy
                      ∴      >0
                          dx
When x = 1, y = 1 − tan−1 (0) = 1. Since the curve increasing for all x, then
y > 0 for all x ≥ 1.
                           ∴ x > tan−1 (1 − x)
8                                                       INEQUALITIES

                                      2
               a2 + b 2       a+b
1) Prove that:          >
                  2            2
2) a, b and c are positive numbers, show that
           a b
     i)     + >2
           b a
                                           1 1 1
     ii) Hence show that: (a + b + c)       + +         >9
                                           a b c
3) Prove that x2 + x + 1 ≥ 0 for all real x.

     Hence, or otherwise, prove that:
     i) a2 + ab + b2 ≥ 0
     ii) a4 + b4 ≥ a3 b + ab3


4) Given that x2 + y 2 + z 2 ≥ xy + xz + yz and x > 0, y > 0 and z > 0
      Prove that (x + y + z)2 ≥ 3(xy + yz + zx)


5) Let x > 0, y > 0, z > 0, you are given:

           x3 + y 3 + z 3 − 3xyz = (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx)

      Show that
     i) x3 + y 3 + z 3 ≥ 3xyz
         x+y+z   √
     ii)       ≥ 3 xyz
           3
         x y z
     iii) + + ≥ 3
         y z x

6) For x > 0, y > 0 and z > 0 show that
     i) (x + y)(y + z)(z + x) ≥ 8xyz
     ii) Now given x + y + z = 1, show that

                                (1 − x)(1 − y)(1 − z) ≥ 8xyz
INEQUALITIES                                                                9

7) Given the Arithmetic-Geometric Mean Inequality:

                     a1 + a2 + a3 + · · · + an   √
                                               ≥ n a1 a2 a3 . . . an
                                n
      For a > 0, b > 0, c > 0 and d > 0 show that:
       (a + b + c)(a + b + d)(a + c + d)(b + c + d) ≥ 81abcd


8) Given that x2 + y 2 + z 2 ≥ xy + xz + yz and x > 0, y > 0, z > 0, show that

                       x2 y 2 + y 2 z 2 + z 2 x2 ≥ xyz (x + y + z)

9) If x2 + y 2 = a2 + b2 = 1, prove that |ax + by| ≤ 1
10) Three positive numbers a, b, c satisfy the conditions that a ≥ b ≥ c and
     a + b + c ≤ 1. By considering (a + b + c)2 or otherwise, prove that

                                 a2 + 3b2 + 5c2 ≤ 1
11) Given that x + y = s prove that for x > 0, y > 0

           1 1  4
     i)     + ≥
           x y  s

           1     1   8
     ii)     2
               + 2 ≥ 2
           x    y   s

12) For x > 0, y > 0 and z > 0 show that
                   1 1 1
       x+y+z+       + + ≥6
                   x y z
                                                                 1 1 1
13) For x > 0, y > 0 and z > 0, given that (x + y + z)            + +     ≥9
                                                                 x y z
     and x + y + z = 1 Show that:

           1 1 1
     i)     + + ≥9
           x y z
           1   1   1
     ii)     + 2 + 2 ≥ 27
           x2 y   z
14) Suppose a, b, c are the sides of a triangle.
     i) Explain why (b − c)2 < a2
     ii) Deduce that (a + b + c)2 < 4(ab + bc + ca)
10                                                       INEQUALITIES

                                                                a+b  √
15) For a > 0, b > 0, c > 0 and d > 0 and given that                ≥ ab, show
                                                                 2
     that
                            a+b+c+d √      4
                                        ≥ abcd
                                  4
                                                         a+b+c+d √
                                                                 4
16) For a > 0, b > 0, c > 0 and d > 0 and given that            ≥ abcd,
                                                            4
     show that
                                a b c d
                                  + + + ≥4
                                b c d a
                                                         a+b+c+d √
                                                                 4
17) For a > 0, b > 0, c > 0 and d > 0 and given that            ≥ abcd,
                                                            4
     show that
                                 a+b+c √     3
                                          ≥ abc
                                    3
18) If x > 0, y > 0, a > 0 and b > 0, show that

                              (ab + xy)(ax + by) ≥ 4abxy

19) The Schwartz inequality for a > 0, b > 0, c > 0 and d > 0 is

                            ab + cd ≤    (a2 + c2 )(b2 + d2 )

     Prove this inequality.
20) Heron’s formula for the area of a triangle is:

                            A=      s(s − a)(s − b)(s − c)

     Where a, b and c are the lengths of the sides of the triangle and
        a+b+c
     s=          . Show that
            2
                                        a2 + b 2 + c 2
                                   A≤
                                             4
21) Prove for x > 0
                                    x ≥ ln(1 + x)
                        π
22) Show that for x >
                        2
                                          2 cos x
                                   x> √
                                          2 − sin x
INEQUALITIES                                                             11

23) i) Show that ex > 1 + x for x > 0
     ii) Let f (x) = ex − 1 − x + x2 . Show that the graph of y = f (x) is
          concave up for all x.
     iii) By considering the first two derivatives of f (x), show for x > 0:

                                        e x > 1 + x − x2
                      √
24) Suppose 0 ≤ t ≤       2
                            2t2
     i) Show that 0 ≤            ≤ t2
                          4 − t2
     ii) Hence show that
                                      2   2
                                0≤      +    − 2 ≤ t2
                                     2+t 2−t

                                                           in
     iii) By integrating the expressions in the inequality √ part (ii) with
          respect to t from t = 0 to t = x (where 0 ≤ x ≤ 2), show that

                                           2+x                x3
                               0 ≤ 2 ln                − 2x ≤
                                           2−x                3
                                            √
     iv) Hence show that for 0 ≤ x ≤         2
                                                 2
                                        2+x                   1 3
                                1≤                   e−2x ≤ e 3 x
                                        2−x




END OF CHAPTER
12                                    INEQUALITIES - SOLUTIONS

  1) Prove that:
                                                             2
                                 a2 + b2         a+b
                                         >
                                    2             2


                                      2
                a2 + b2         a+b           a2 + b2 a2 + 2ab + b2
                        −                 =          −
                   2             2               2          4

                                              2a2 + 2b2 − a2 − 2ab − b2
                                          =
                                                          4
                                              a2 − 2ab + b2
                                          =
                                                    4
                                              (a − b)2
                                          =
                                                 4
                                          >0
                                                         2
                                a2 + b2        a+b
                            ∴           >
                                   2            2

2) a, b and c are positive numbers, show that
   a b
i) + > 2
   b a

                                          (a − b)2 > 0

                                 a2 − 2ab + b2 > 0

                                          a2 + b2 > 2ab

                                       a2   b2
                                          +    >2
                                       ab ab
                                      a b
                                  ∴    + >2
                                      b a

                                      1 1 1
ii) Hence show that: (a + b + c)       + +           >9
                                      a b  c

                      1 1 1                   a a    b b   c c
        (a + b + c)    + +            =1+       + +1+ + +1+ +
                      a b  c                  b  c   a c   a b

                                                a b              b c       a c
                                      =3+        +           +    +    +    +
                                                b a              c b       c a


                      a b                     b c                c a
                Now    +         >2⇒           +         > 2,     +  >2
                      b a                     c b                a c
INEQUALITIES - SOLUTIONS                                                     13

                                       1 1 1
                      ∴ (a + b + c)     + +            >3+2+2+2
                                       a b  c

                                       1 1 1
                      ∴ (a + b + c)     + +            >9
                                       a b  c



3) Prove that x2 + x + 1 ≥ 0 for all real x.

Let f (x) = x2 + x + 1
∆ = 12 − 4 ∗ 1 ∗ 1 = 1 − 4 = −3

∆ < 0 and the coefficient of x2 > 0
∴ f (x) has no real roots and is concave up, hence is positive definite
                                           ∴ f (x) > 0

                                      ∴ x2 + x + 1 > 0
Hence, or otherwise, prove that:

i) a2 + ab + b2 ≥ 0

                                          a2 ab b2
                  a2 + ab + b2 = b2          + 2 + 2
                                          b2  b   b

                                           a   2       a
                                 = b2              +     +1
                                           b           b
                                                                      a
                                 = b2 (x2 + x + 1)          let x =
                                                                      b
                                 ≥0        (proven above)

                                 ∴ a2 + ab + b2 ≥ 0

ii) a4 + b4 ≥ a3 b + ab3


                 a4 + b4 − a3 b − ab3 = a3 (a − b) + b3 (b − a)

                                         = a3 (a − b) − b3 (a − b)

                                         = (a − b)(a3 − b3 )

                                         = (a − b)(a − b)(a2 + ab + b2 )

                                         = (a − b)2 (a2 + ab + b2 )

                                         ≥0        (using the result in i)

                            ∴ a4 + b4 ≥ a3 b + ab3
14                                    INEQUALITIES - SOLUTIONS

4) Given that x2 + y 2 + z 2 ≥ xy + xz + yz and x > 0, y > 0 and z > 0

Prove that (x + y + z)2 ≥ 3(xy + yz + zx)


               Since (x + y + z)2 = x2 + y 2 + z 2 + 2(xy + yz + zx)

                 and given x2 + y 2 + z 2 ≥ xy + yz + zx

         x2 + y 2 + z 2 + 2(xy + yz + zx) ≥ xy + yz + zx + 2(xy + yz + zx)

                            ∴ (x + y + z)2 ≥ 3(xy + yz + zx)

5) Let x > 0, y > 0, z > 0, you are given:

          x3 + y 3 + z 3 − 3xyz = (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx)

 Show that
i) x3 + y 3 + z 3 ≥ 3xyz

                             (x − y)2 ≥ 0 ⇒ x2 + y 2 ≥ 2xy

                             (x − z)2 ≥ 0 ⇒ x2 + z 2 ≥ 2xz

                             (z − y)2 ≥ 0 ⇒ z 2 + y 2 ≥ 2zy


                        Adding these 2x2 + 2y 2 + 2z 2 ≥ 2xy + 2xz + 2yz

                                           ∴ x2 + y 2 + z 2 ≥ xy + xz + yz

                           x2 + y 2 + z 2 − xy − xz − yz ≥ 0

           (x + y + z)(x2 + y 2 + z 2 − xy − xz − yz) ≥ 0

                                 ∴ x3 + y 3 + z 3 − 3xyz ≥ 0

                                           ∴ x3 + y 3 + z 3 ≥ 3xyz



      x+y+z   √
ii)         ≥ 3 xyz
        3
Since x3 + y 3 + z 3 ≥ 3xyz,
                                      √
                                      3            3
                                                                √
                                                                3
                            Let x =       x ,y =       y ,z =       z
                   √
                   3
                                       √
                                       3
                ∴ ( x )3 + ( 3 y )3 + ( z )3 ≥ 3 3 x y z

                        x +y +z           3             x+y+z   √
                    ∴           ≥             xyz ⇒           ≥ 3 xyz
                           3                              3
INEQUALITIES - SOLUTIONS                                             15

       x y  z
iii)     + + ≥3
       y  z x
                                            √
                               x + y + z ≥ 3 3 xyz
                                     a      b        c
                           Let x =     , y = and z =
                                     b      c        a

                              a b c   3 a  b c
                               + + ≥3     × ×
                              b c a     b c a
                           a b c
                          ∴  + + ≥3
                           b c a
                           x y  z
                          ∴ + + ≥3
                           y  z x

6) For x > 0, y > 0 and z > 0 show that
i) (x + y)(y + z)(z + x) ≥ 8xyz

                                √   √
                               ( x − y)2 ≥ 0
                                             √
                                    x + y ≥ 2 xy
                                       x+y √
                                   ∴      ≥ xy
                                        2

                                          y+z √
                              Similarly      ≥ yz
                                           2
                                          z+x √
                                             ≥ zx
                                           2

                          (x + y)(y + z)(z + x)
                      ∴                         ≥    x2 y 2 z 2
                                    8
                      ∴ (x + y)(y + z)(z + x) ≥ 8xyz

OR

                                  (x + y)(y + z)(z + x) ≥ 8xyz

                          ⇒ (x + y)(zy + xy + z 2 + zx) ≥ 8xyz

                 2xyz + x2 y + xz 2 + zx2 + zy 2 xy 2 z 2 y ≥ 8xyz
                               x z  x y y z
                                 + + + + + ≥6
                               z  x y x z y
Solution Continues...
16                                 INEQUALITIES - SOLUTIONS

This can be proven using the previous result in question 2.

                                  x z  x y y z
                        LHS =       + + + + +
                                  z  x y x z y

                               >2+2+2

                               >6

                           x z  x y y z
                       ∴     + + + + + ≥6
                           z  x y x z y

                           ∴ (x + y)(y + z)(z + x) ≥ 8xyz


ii) Now given x + y + z = 1, show that (1 − x)(1 − y)(1 − z) ≥ 8xyz


                               If x + y + z = 1

                                       x+y =1−z

                                       y+z =1−x

                                       z+x=1−y


                                (x + y)(y + z)(z + x) ≥ 8xyz

                     ∴ (1 − x)(1 − y)(1 − z) ≥ 8xyz

7) Given the Arithmetic-Geometric Mean Inequality:
                    a1 + a2 + a3 + · · · + an   √
                                              ≥ n a1 a2 a3 . . . an
                               n
For a > 0, b > 0, c > 0 and d > 0 show that

              (a + b + c)(a + b + d)(a + c + d)(b + c + d) ≥ 81abcd
Choosing n = 3:

                                     a+b+c √      3
                                               ≥ abc
                                         3
                                                   √3
                                     a + b + c ≥ 3 abc
                                                   √3
                           Similarly a + b + d ≥ 3 abd
                                                   √3
                                     a + c + d ≥ 3 acd
                                                   √3
                                     b + c + d ≥ 3 bcd
INEQUALITIES - SOLUTIONS                                                         17

                                                              √
                                                              3
           ∴ (a + b + c)(a + b + d)(a + c + d)(b + c + d) ≥ 34 a3 b3 c3 d3

           ∴ (a + b + c)(a + b + d)(a + c + d)(b + c + d) ≥ 81abcd



8) Given that x2 + y 2 + z 2 ≥ xy + xz + yz and x > 0, y > 0, z > 0, show that

                       x2 y 2 + y 2 z 2 + z 2 x2 ≥ xyz (x + y + z)

 If (x )2 + (y )2 + (z )2 ≥ x y + x z + y z
Let x = xy, y = yz and z = zx


                   ∴ x2 y 2 + y 2 z 2 + z 2 x2 ≥ x2 yz + xy 2 z + xyz 2

                   ∴ x2 y 2 + y 2 z 2 + z 2 x2 ≥ xyz(x + y + z)

9) If x2 + y 2 = a2 + b2 = 1, prove that |ax + by| ≤ 1


                                    (x2 + y 2 )(a2 + b2 ) = 1

                           a2 x2 + b2 y 2 + a2 y 2 + b2 x2 = 1                   (*)


                                 Now (ay − bx)2 ≥ 0

                           a2 y 2 − 2abxy + b2 x2 ≥ 0

                                   ∴ a2 y 2 + b2 x2 ≥ 2abxy


                     From (*) a2 y 2 + b2 x2 = 1 − a2 x2 − b2 y 2

                         ∴ 1 − a2 x2 − b2 y 2 ≥ 2abxy

                      a2 x2 + 2abxy + b2 y 2 ≤ 1

                                  (ax + by)2 ≤ 1

                                  (ax + by)2 ≤ 1

                                 ∴ |ax + by| ≤ 1
18                               INEQUALITIES - SOLUTIONS

10) Three positive numbers a, b, c satisfy the conditions that a ≥ b ≥ c and
a + b + c ≤ 1. By considering (a + b + c)2 or otherwise, prove that

                               a2 + 3b2 + 5c2 ≤ 1


                                              a+b+c≤1

                                           (a + b + c)2 ≤ 1

                      ∴ a2 + b2 + c2 + 2ab + 2bc + 2ac ≤ 1


                            Now a ≥ b ⇒ 2ab ≥ 2b2

                                 b ≥ c ⇒ 2bc ≥ 2c2

                                 a ≥ c ⇒ 2ac ≥ 2c2


                        a2 + b2 + c2 + 2ab + 2bc + 2ac ≤ 1

                      ∴ a2 + b2 + c2 + 2b2 + 2c2 + 2c2 ≤ 1

                                      ∴ a2 + 3b2 + 5c2 ≤ 1


11) Given that x + y = s prove that for x > 0, y > 0
     1 1  4
i)    + ≥
     x y  s
                         1 1 4 1 1 4
                          + ≥ ⇒ + − ≥0
                         x y s x y s


                  1 1  4    y(x + y) + x(x + y) − 4xy
                   + −    =
                  x y x+y           xy(x + y)

                                     xy + y 2 + x2 + xy − 4xy
                                 =
                                             xy(x + y)

                                     x2 − 2xy + y 2
                                 =
                                       xy(x + y)

                                      (x − y)2
                                 =
                                     xy(x + y)

                                 ≥0

                            1 1  4
                        ∴    + ≥
                            x y  s
INEQUALITIES - SOLUTIONS                                                                19

      1     1   8
ii)     2
          + 2 ≥ 2
      x    y   s
                         1     1   8   1   1   8
                           2
                             + 2 ≥ 2 ⇒ 2 + 2 − 2 ≥0
                         x    y   s   x   y   s




           1   1  8   y 2 (x + y)2 + x2 (x + y)2 − 8x2 y 2
             + 2− 2 =
           x2 y  s               x2 y 2 (x + y)2

                             y 2 x2 + 2xy 3 + y 4 + x4 + 2x3 y + x2 y 2 − 8x2 y 2
                         =
                                                x2 y 2 (x + y)2

                             y 4 − 2x2 y 2 + x4 + 2xy 3 − 4x2 y 2 + 2x3 y
                         =
                                             x2 y 2 (x + y)2

                             (y 2 − x2 )2 + 2xy(y 2 − 2xy + x2 )
                         =
                                         x2 y 2 (x + y)2

                             (y 2 − x2 )2 + 2xy(y − x)2
                         =
                                    x2 y 2 (x + y)2

                         ≥0

                  1   1    8
              ∴     +    ≥ 2
                  x2 y 2  s
                                                             1 1 1
12) For x > 0, y > 0 and z > 0 show that x + y + z +          + + ≥6
                                                             x y z

                                  Now (x − 1)2 ≥ 0

                                  ∴ x2 − 2x + 1 ≥ 0

                                          x2 + 1 ≥ 2x
                   1              1               1
                     ≥ 2 and ∴ y + ≥ 2 and ∴ z + ≥ 2
                   ∴x+
                   x              y               z
                      1 1 1
             ∴x+y+z+ + + ≥6          (on adding these terms)
                      x y z
                                                                   1 1 1
13) For x > 0, y > 0 and z > 0, given that (x + y + z)              + +             ≥ 9 and
                                                                   x y z
x + y + z = 1 Show that:
      1 1 1
i)     + + ≥9
      x y z
                                                          1 1 1
        Substituting x + y + z = 1 into (x + y + z)        + +          ≥ 9 yeilds:
                                                          x y z
Solution Continues...
20                                  INEQUALITIES - SOLUTIONS


                                    1 1 1
                                     + + ≥9
                                    x y z
      1     1   1
ii)     2
          + 2 + 2 ≥ 27
      x    y   z
                                                  1 1 1
                      Squaring both sides of       + + ≥9
                                                  x y z

                                                        2
                                             1 1 1
                                              + +           ≥ 81
                                             x y z

                     1    1   1             1   1   1
                      2
                        + 2 + 2 +2            +   +         ≥ 81
                     x   y   z             xy yz xz

                          1     1   1          x+y+z
                            2
                              + 2 + 2 +2                    ≥ 81
                          x    y   z            xyz
                                       1     1  1   2
                                         2
                                           + 2+ 2+     ≥ 81
                                       x    y  z   xyz



                                                            x+y+z   √
        From the Arithmetic-Geometric Mean Inequality:            ≥ 3 xyz
                                                              3

                                         1 √
                                           ≥ 3 xyz
                                         3
                                         1
                                    ∴      ≥ xyz
                                        27
                                        1
                                   ∴       ≥ 27
                                       xyz
                                        2
                                           ≥ 54
                                       xyz

                                 1     1  1   2
                    so finally,     2
                                     + 2+ 2+     ≥ 81
                                 x    y  z   xyz
                                        1     1   1         2
                                          2
                                            + 2 + 2 ≥ 81 −
                                        x    y   z         xyz
                                        1   1   1
                                    ∴     +   +    ≥ 27
                                        x2 y 2 z 2
INEQUALITIES - SOLUTIONS                                                        21

14) Suppose a, b, c are the sides of a triangle.

i) Explain why (b − c)2 < a2

                         For a triangle to exist a + c > b

                                               (b − c) < a

                                            ∴ (b − c)2 < a2

ii) Deduce that (a + b + c)2 < 4(ab + bc + ca)

                      So (b − c)2 < a2 ⇒ b2 + c2 − 2bc < a2

                          (a − c)2 < b2 ⇒ a2 + c2 − 2ac < b2

                          (a − b)2 < c2 ⇒ a2 + b2 − 2ab < c2


     ∴ 2(a2 + b2 + c2 ) − 2(ab + bc + ac) < a2 + b2 + c2

                           ∴ a2 + b2 + c2 < 2(ab + bc + ac)

           a2 + b2 + c2 + 2(ab + bc + ac) < 2(ab + bc + ac) + 2(ab + bc + ac)

                           ∴ (a + b + c)2 < 4(ab + bc + ca)

                                                       a+b √
15) For a > 0, b > 0, c > 0 and d > 0 and given that      ≥ ab, show that
                                                        2
                               a+b+c+d √
                                       4
                                      ≥ abcd
                                  4
          a+b         c+d
Let x =       and y =
           2           2
                                   x+y   a+b+c+d
                               ∴       =
                                    2       4

                                         x+y √
                                   Now      ≥ xy
                                          2
                               a+b+c+d             √ √
                           ∴           ≥            ab cd
                                  4
                               a+b+c+d √
                                       4
                                      ≥ abcd
                                  4



Solution Continues...
22                                       INEQUALITIES - SOLUTIONS

                                                                  a+b+c+d √
                                                                          4
16) For a > 0, b > 0, c > 0 and d > 0 and given that                     ≥ abcd, show
                                                                     4
that
                                a b c d
                                  + + + ≥4
                                b c d a
    a +b +c +d         √4
 If                ≥ abcd
          4
              a       b      c      d
Then let a = b = c = d =
              b       c      d      a
                       1    a b c d                 4   a b c d
                   ∴         + + +             ≥         × × ×
                       4    b c d a                     b c d a

                            a b c d
                       ∴     + + +             ≥4
                            b c d a
                                                     a+b+c+d √4
17) For a > 0, b > 0, c > 0 and d > 0 and given that         ≥ abcd, show
                                                        4
that
                                 a+b+c √     3
                                          ≥ abc
                                    3
         a+b+c         a+b+c a+b+c             1          a+b+c
                   =            +           =     a+b+c+
             3             4          12       4            3
          a+b+c
Let d =
            3
                                   a+b+c+d  √
                                            4
                           Given           ≥ abcd
                                      4

              1          a+b+c                          4         a+b+c
                × a+b+c+                           ≥        abc
              4            3                                        3
                                                                           1
                                       a+b+c         1             a+b+c   4
                                     ∴       ≥ (abc) 4
                                         3                           3
                                               3
                                    a+b+c      4              1
                                                   ≥ (abc) 4
                                      3
                                         a+b+c √
                                               3
                                     ∴        ≥ abc
                                           3
18) If x > 0, y > 0, a > 0 and b > 0, show that

                              (ab + xy)(ax + by) ≥ 4abxy

                                     (ab − xy)2 ≥ 0

                                   a2 b2 + x2 y 2 ≥ 2abxy

                       a2 b2 + x2 y 2 + 2abxy ≥ 2abxy + 2abxy

                                   ∴ (ab + xy)2 ≥ 4abxy
INEQUALITIES - SOLUTIONS                                                         23

Likewise (ax + by)2 ≥ 4abxy


                          (ab + xy)2 (ax + by)2 ≥ (4abxy)2

                         ∴ (ab + xy)(ax + by) ≥ 4abxy

19) The Schwartz inequality for a > 0, b > 0, c > 0 and d > 0 is

                           ab + cd ≤     (a2 + c2 )(b2 + d2 )

Prove this inequality.



                             ab + cd ≤      (a2 + c2 )(b2 + d2 )

                          (ab + cd)2 ≤ (a2 + c2 )(b2 + d2 )

                   a2 b2 + 2abcdc2 d2 ≤ a2 b2 + a2 d2 + c2 b2 + c2 d2

                            ⇒ 2abcd ≤ a2 d2 + c2 b2

If the above is true, then the Schwartz inequality must be true. Consider:


                     (ad − cd)2 ≥ 0

                   a2 d2 + c2 d2 ≥ 2abcd ⇒ 2abcd ≤ a2 d2 + c2 b2

                      ∴ ab + cd ≤      (a2 + c2 )(b2 + d2 )

20) Heron’s formula for the area of a triangle is

                            A=      s(s − a)(s − b)(s − c)

                                                                        a+b+c
where a,b,c are the lengths of the sides of the triangle and s =              . Show
                                                                          2
that
                                      a2 + b2 + c2
                                A≤
                                           4
        a+b+c     b+c−a
s−a=          −a=
          2         2

                                               a+c−b
                               Also s − b =
                                                 2
                                               a+b−c
                                      s−c=
                                                 2
Solution Continues...
24                                   INEQUALITIES - SOLUTIONS



                    a+b+c          b+c−a         a+c−b           a+b−c
         A=
                      2              2             2               2

                1
            =       (2bc − a2 + b2 + c2 )(2bc + a2 − c2 − b2 )
                4
                1
        A2 =       (2a2 b2 + 2a2 c2 + 2b2 c2 − a4 − b4 − c4 )
                16
                1
            =      (2(a2 b2 + a2 c2 + b2 c2 ) + a4 + b4 + c4 − (2a4 + 2b4 + 2c4 ))
                16
                1 2                  1
            =      (a + b2 + c2 )2 − (2a4 + 2b4 + 2c4 )
                16                  16
                               2
              a2 + b2 + c2
            ≤
                   16
              a2 + b2 + c2
       ∴A≤
                   4
21) Prove for x > 0, x ≥ ln(1 + x)

                         x ≥ ln (1 + x) ⇒ x − ln (1 + x) ≥ 0

                               Let f (x) = x − ln (1 + x)

                                                  1
                                   f (x) = 1 −
                                                 1+x


We seek the values of x such that f (x) > 0

                                            1
                                      1−       >0
                                           1+x
                                            1
                                               <1
                                           1+x
                                       (1 + x)2
                                                < (1 + x)2
                                        1+x
                                         (1 + x) < (1 + x)2

                             (1 + x)2 − (1 + x) > 0

                                       x(1 + x) > 0
                                   ∴ x > 0 or x < −1
f (0) = 0 and since the curve is increasing for x > 0, then f (x) > 0.

                                    ∴ x ≥ ln (1 + x)
INEQUALITIES - SOLUTIONS                                                   25

                          π
22) Show that for x >
                          2
                                               2 cos x
                                      x> √
                                               2 − sin x

                                     2 cos x
               Let f (x) = x − √
                                     2 − sin x
                                                √
                                     (−2 sin x)( 2 − sin x) + (2 cos2 x)
                   ∴ f (x) = 1 −                 √
                                                ( 2 − sin x)2
                                 √
                                2 2 sin x − 2
                          =1+ √
                                ( 2 − sin x)2
                             √                √
                            ( 2 − sin x)2 + 2 2 sin x − 2
                          =         √
                                   ( 2 − sin x)2
                                 √                    √
                            2 − 2 2 sin x + sin2 x + 2 2 sin x − 2
                          =              √
                                        ( 2 − sin x)2
                                               2
                                    sin x
                          =    √
                                   2 − sin x

                          >0
    π       π
f       =     −0
    2       2
                                         π          π
                           ∴ for x >       , f (x) > > 0
                                         2          2
                                       2 cos x
                              ∴x− √              >0
                                       2 − sin x
                                                        2 cos x
                                               ∴x> √
                                                        2 − sin x
23)
i) Show that ex > 1 + x for x > 0



                                   Let f (x) = ex − 1 − x

                                       f (x) = ex − 1

                                   Now ex > 1 (for x > 0)

                               ∴ ex − 1 > 0

                                   ∴ f (x) > 0

Solution Continues...
26                                 INEQUALITIES - SOLUTIONS

 ∴ the curve is increasing for x > 0.
now f (0) = 0, hence f (x) > 0 for x > 0.

                               ∴ ex − 1 − x > 0

                                        ∴ ex > 1 + x

ii) Let f (x) = ex − 1 − x + x2 . Show that the graph of y = f (x) is concave up for
all x.

                              f (x) = ex − 1 − x + x2

                              f (x) = ex − 1 + 2x

                             f (x) = ex + 2

                                    >0       (for all x)

∴ f (x) is concave up for all x.

iii) By considering the first two derivatives of f (x), show for x > 0:

                                   ex > 1 + x − x2


                               Now ex > 1 + x         (from i)

                            ex − 1 − x > 0

                       ex − 1 − x + 3x > 0 + 3x > 0

                         ∴ ex − 1 + 2x > 0

 ∴ if g(x) = ex − 1 − x + x2 then g (x) > 0.
g(0) = 0 and since the curve is increasing for x > 0, then g(x) > 0


                           ex − 1 − x + x2 > 0

                                       ∴ e x > 1 + x − x2
INEQUALITIES - SOLUTIONS                                             27

                      √
24) Suppose 0 ≤ t ≤       2

                    2t2
i) Show that 0 ≤          ≤ t2
                   4 − t2
                                                √
                                     0≤t≤            2

                                     0 ≤ t2 ≤ 2

                                     0 ≥ −t2 ≥ −2

                                     4 ≥ 4 − t2 ≥ 2

                                     1    1      1
                                       ≤     2
                                               ≤
                                     4   4−t     2
                                    t2    2t2
                                       ≤        ≤ t2
                                    2    4 − t2
                                             2t2
                                    ∴0≤            ≤ t2
                                            4 − t2
ii) Hence show that
                                     2   2
                              0≤       +    − 2 ≤ t2
                                    2+t 2−t



                2   2      2(2 − t) + 2(2 + t) − 2(4 − t2 )
                  +    −2=
               2+t 2−t                 (4 − t2 )

                                         4 − 2t + 4 + 2t − 8 + 2t2
                                     =
                                                 (4 − t2 )

                                           2t2
                                     =
                                         (4 − t2 )


                                      2t2
                               0≤          ≤ t2          (from i)
                                    4 − t2
                                     2   2
                              ∴0≤      +    − 2 ≤ t2
                                    2+t 2−t
28                                          INEQUALITIES - SOLUTIONS

iii) By integrating the expressions in the inequality in part (ii) with respect to t
                                     √
from t = 0 to t = x (where 0 ≤ x ≤ 2), show that

                                              2+x                      x3
                                   0 ≤ 2 ln                 − 2x ≤
                                              2−x                      3



                      x                 x                                        x
                                             2   2
                          0 dt ≤               +    −2                dt ≤           t2 dt
                  0                 0       2+t 2−t                          0

                                                                             x3
                            0 ≤ 2 ln(2 + x) − 2 ln(2 − x) − 2x ≤
                                                                             3
                               2+x                              x3
                            0 ≤ 2 ln              − 2x ≤
                               2−x                              3
                                √
iv) Hence show that for 0 ≤ x ≤ 2
                                                    2
                                            2+x                      1 3
                                   1≤                   e−2x ≤ e 3 x
                                            2−x

Raise all the expressions in (iii) as the powers of e.
                                               2+x               x3
                                    e0 ≤ e2 ln( 2−x )−2x ≤ e      3


                                                        2
                                              2+x                      x3
                                   ∴1≤                      e−2x ≤ e    3
                                              2−x

								
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