The Art and Craft of Problem Solving

Thus, for any real a, b, c, x, y, z, we have

5.5 INEQUALITIES 183

O:S (ay - bxf + ( az -cx)^2 + (bz -cy)^2.

This is equivalent to

2 (abxy+acxz + bcyz) :S a^2 i +b^2 � +a^2 i +c^2 � +b^2 z^2 +c^2 i. (8)

Adding

(ax)^2 + (by)^2 + (cz)^2

to both sides of (8) yields (7). This argument generalizes; it is true not just for n = 3.

5.5.21 Convince yourself that this method generalizes, by writing it out for the n = 4
case. Your starting point will be a sum of six squares.

Even though the Cauchy-Schwarz inequality is a fairly simple consequence of
AM-GM, it is a powerful tool, because it has so many "degrees of freedom." For
example, if we let a = b = c = 1 in (7), we get the appealing inequality

(x+y+z)^2 <

� 2 2

3

• +y +z.

We can derive another useful inequality from Cauchy-Schwarz if the variables are
positive. For example, if a, b, c, x, y, z > 0, then

(9)

is a simple consequence of (7)-just replace a by va, etc. This inequality (which of
course generalizes to any n) comes in quite handy sometimes, because it is a surprising
way to find the lower bound of the product of two often unrelated sums.
Here is a more interesting example that uses (9).

Example 5.5.22 (Titu Andreescu) Let P be a polynomial with positive coefficients.
Prove that if

p(�)^2 P�X)

holds for x = 1 then it holds for every x > O.

Solution: Write P(x) = Uo + U IX + U 2 x^2 + ... + unx". When x = 1, the inequality

is just P(l) 2 l/P(l), or

which reduces to

Uo + U I + U 2 + ... + Un 2 1

since the coefficients are positive. We wish to show that