12 Chapter 2
87 Problem
Shew that 1 + x1 + x
2
1 + x
4
1 + x
8
··· 1 + x
1024
= 1 − x
2048
1 − x .
88 Problem Shew that
a
2
+ b
2
+ c
2
− ab − bc − ca = 1
2 a − b
2
+ b − c
2
+ c − a
2
.
89 Problem
Prove that if r ≥ s ≥ t ≥ u ≥ v then
r
2
− s
2
+ t
2
− u
2
+ v
2
≥ r − s + t − u + v
2
2.7
90 Problem AIME 1987 Compute
10
4
+ 32422
4
+ 32434
4
+ 32446
4
+ 32458
4
+ 324 4
4
+ 32416
4
+ 32428
4
+ 32440
4
+ 32452
4
+ 324 .
91 Problem Write a
2
+ a + 1
2
as the sum of three squares.
2.2 Squares of Real Numbers
If x is a real number then x
2
≥ 0. Thus if a ≥ 0,b ≥ 0 then √
a − √
b
2
≥ 0 gives, upon expanding the square, a−2 √
ab + b ≥ 0,
or √
ab ≤
a + b 2
. Since
a + b 2
is the arithmetic mean of a , b and
√ ab is the geometric mean of a
, b the inequality √
ab ≤
a + b 2
2.8 is known as the Arithmetic-Mean-Geometric Mean AM-GM Inequality.
92 Example Let u
1
, u
2
, u
3
, u
4
be non-negative real numbers. By applying the preceding result twice, establish the AM-GM Inequality for four quantities:
u
1
u
2
u
3
u
4 1
4
≤ u
1
+ u
2
+ u
3
+ u
4
4 2.9
Solution: We have √
u
1
u
2
≤ u
1
+ u
2
2 and
√ u
3
u
4
≤ u
3
+ u
4
2 . Now, applying the AM-GM Inequality twice to
√ u
1
u
2
and √
u
3
u
4
we obtain
È
√ u
1
u
2
√ u
3
u
4
≤ √
u
1
u
2
+ √u
3
u
4
2 ≤
u
1
+u
2
2
+
u
3
+u
4
2
2 .
Simplification yields the desired result.
93 Example Let u
, v, w be non-negative real numbers. By using the preceding result on the four quantities u, v, w, and u + v + w
3 ,
establish the AM-GM Inequality for three quantities: uvw
1 3
≤ u + v + w
3 2.10
Solution: By the AM-GM Inequality for four values uvw
u + v + w 3
1 4
≤ u + v + w +
u+v+w 3
4 .
Some algebraic manipulation makes this equivalent to uvw
1 4
u + v + w 3
1 4
≤ u + v + w
4 +
u + v + w 12
or upon adding the fraction on the right uvw
1 4
u + v + w 3
1 4
≤ u + v + w
3 .
Squares of Real Numbers 13
Multiplying both sides by u + v + w
3
−1 4
we obtain uvw
1 4
≤ u + v + w
3
3 4
, from where the desired inequality follows.
94 Example
Let a 0, b 0. Prove the Harmonic-Mean-Geometric-Mean Inequality
2
1 a
+
1 b
≤ √
ab 2.11
Solution: By the AM-HM Inequality
r
1 a
· 1
b ≤
1 a
+
1 b
2 ,
from where the desired inequality follows.
95 Example Prove that if a
, b, c are non-negative real numbers then a + bb + cc + a
≥ 8abc. Solution: The result quickly follows upon multiplying the three inequalities a + b
≥ 2 √
ab, b + c ≥ 2
√ bc and c + a
≥ 2 √
ca.
96 Example
If a , b, c, d, are real numbers such that a
2
+ b
2
+ c
2
+ d
2
= ab + bc + cd + da , prove that a = b = c = d.
Solution: Transposing, a
2
− ab + b
2
− bc + c
2
− dc + d
2
− da = 0 ,
or a
2
2 − ab +
b
2
2 +
b
2
2 − bc +
c
2
2 +
c
2
2 − dc +
d
2
2 +
d
2
2 − da +
a
2
2 = 0
. Factoring,
1 2
a − b
2
+ 1
2 b − c
2
+ 1
2 c − d
2
+ 1
2 d − a
2
= 0 .
As the sum of non-negative quantities is zero only when the quantities themselves are zero, we obtain a = b , b = c, c = d, d = a,
which proves the assertion. We note in passing that from the identity
a
2
+ b
2
+ c
2
− ab − bc − ca = 1
2
a − b
2
+ b − c
2
+ c − a
2
2.12 it follows that
a
2
+ b
2
+ c
2
≥ ab + bc + ca 2.13
97 Example The values of a
, b, c, and d are 1, 2, 3 and 4 but not necessarily in that order. What is the largest possible value of ab + bc + cd + da?
Solution: ab + bc + cd + da
= a + cb + d
≤
a + c + b + d 2
2
=
1 + 2 + 3 + 4 2
2
= 25
, by AM-GM. Equality occurs when a + c = b + d. Thus one may choose, for example, a = 1
, c = 4, b = 2, d = 3.
14 Chapter 2
Practice
98 Problem If 0
a ≤ b, shew that 1
8 ·
b − a
2
b ≤
a + b 2
− √
ab ≤
1 8
· b − a
2
a
99 Problem Prove that if a
, b, c are non-negative real numbers then a
2
+ 1b
2
+ 1c
2
+ 1 ≥ 8abc
100 Problem The sum of two positive numbers is 100. Find their maximum possible
product.
101 Problem Prove that if a
, b, c are positive numbers then a
b +
b c
+ c
a ≥ 3.
102 Problem Prove that of all rectangles with a given perimeter, the square has the
largest area.
103 Problem
Prove that if 0 ≤ x ≤ 1 then x − x
2
≤ 1
4 .
104 Problem Let 0
≤ a,b,c,d ≤ 1. Prove that at least one of the products a1 − b
, b1 − c, c1 − d
, d1 − a is
≤ 1
4 .
105 Problem Use the AM-GM Inequality for four non-negative real numbers to prove
a version of the AM-GM for eight non-negative real numbers.
2.3 Identities with Cubes
