The Art and Craft of Problem Solving

184 CHAPTER 5 ALGEBRA

for all positive x. Since x and the Uj are positive, we can define the real sequences

ao = yfuO, al = JuI/x,a 2 = VU 2 /x^2 , ... , an = Junlxn

and

bo = yfuO,bl = yUIX,b 2 = JU 2 x^2 , ... ,bn = Junxn.

Notice that

for each i. Hence when we apply the Cauchy-Schwarz inequality to the aj and bj

sequences, we get

UO+UI +U 2 +···+Un �p(�)P(X).

But Uo + UI + U 2 + ... + Un 2 I, so we conclude that

p(�)P(x)2l. •

Here is another example, a solution to the IMO problem started with Exam­

ple 5.2.23 on page ISS.

Example 5.5.23 (lMO 19 95) Let a , b, c be positive real numbers such that abc = l.

Prove that

1 1 1 3

---=---+ + >-.

a^3 ( b + c) b^3 ( e + a) e^3 ( a + b) - 2

Solution: Recall that the substitutions x = lla , y = lib, z = lie transform the

original problem into showing that

x^2 y^2 z^2 3

-+-+-> - (10)

y+z z+x x+y -2'

where xyz = l.

Denote the left-hand side of (10) by S. Notice that

S= (

Jy

x

+z

)

2

+ (J:+

x

)

2

+ (

Jx

z

+y

)

2

,

and thus Cauchy-Schwarz implies that

S(u^2 +v^2 +w^2 ) 2 ( �+ �+ �)

2

,

yy+z yZ+x yx+y

(11)

for any choice of u, v, w. Is there a helpful choice?

Certainly U = Jy + z, v = J z + x, w = J x + y is a natural choice to try, since this

immediately simplifies the right-hand side of (1 1) to just (x + y + z)^2. But better yet,

with this choice we also get