The Art and Craft of Problem Solving

70 CHAPTER 3 TACTICS FOR SOLVING PROBLEMS

so (1) becomes u^2 - 2 + u + 1 = 0, or u^2 + u - 1 = 0, which has solutions

-1±V5

U=

2

Solving x +! = u, we get x^2 - ux + 1 = 0, or

x

u±Ju^2 -4

x=-----

2

Putting these together, we have

x=

-1±^0 � (_1±V5)

2

_

2

±

2

4

2

-1±V5±i\1t0±2V5

4

• 
  

The last few steps are mere "technical details." The two crux moves were to

increase the symmetry of the problem and then make the symmetrical substitution

u=x +x-^1 •

In the next example, we use symmetry to reduce the complexity of an inequality.

Example 3.1.11 Prove that

(a + b) (b + c) (c + a) 2 8abc

is true for all positive numbers a, band c, with equality only if a = b = c.

Solution: Observe that the alleged inequality is symmetric, in that it is unchanged

if we permute any of the variables. This suggests that we not multiply out the left side

(rarely a wise idea!) but instead look at the factored parts, for the sequence

a+b, b+c, c+a

can be derived by just looking at the term a + b and then performing the cyclic per­

mutation a t---+ b, b t---+ C, C t---+ a once and then twice.

The simple two-variable version of the Arithmetic-Geometric-Mean inequality

(see Section 5. 5 for more details) implies

a+b 2 2 vah.

Now, just perform the cyclic permutations

and

c +a 2 2vca·

The desired inequality follows by multiplying these three inequalities. •