Advanced High-School Mathematics

SECTION 3.2 Classical Inequalities 151

Proof. We define a quadratic function ofx:

Q(x) = (xx 1 −y 1 )^2 +···(xxn−yn)^2

= (x^21 +···+x^2 n)x^2 −2(x 1 y 1 +···xnyn)x+ (y 12 +···+yn^2 ).

SinceQ(x)≥0, we see that the discriminant must be≤0:

4(x 1 y 1 +···+xnyn)^2 −4(x^21 +···+x^2 n)(y 12 +···+yn^2 )≤ 0 ,

so we’re done! Pretty slick, eh?^3 See Exercise 4, below.

Exercises.

1. The diagram to the right depicts
   a semicircle with center O and di-
   ameter XZ. If we write XZ =
   a+b, identify AM(a,b), GM(a,b),
   and HM(a,b) as lengths of segments
   in the diagram.

O

Q

P

X Y Z

2. Suppose that a 1 , a 2 , a 3 , ... is a sequence of non-negative terms
   such that for each indexi >1, ai = QM(ai− 1 ,ai+1). Show that
   a^21 , a^22 , a^23 , ...is an arithmetic sequence.


3. Show that ifa, b >0 then

a+b

2

≤

2

3

Ñ

a^2 +ab+b^2

a+b

é

.

4. Show how AM ≤ QM is a direct consequence of the Cauchy-
   Schwarz inequality.


5. State a necessary and sufficient condition for AM(x 1 ,...,xn) =
   QM(x 1 ,...,xn).

(^3) The Cauchy-Schwarz inequality can be generalized to complex numbers where it reads:
|x 1 y 1 +x 2 y 2 +···+xnyn|^2 ≤(|x 1 |^2 +|x 2 |^2 +···+|xn|^2 )(|y 1 |^2 +|y 2 |^2 +···+|yn|^2 ).