SECTION 3.2 Classical Inequalities 149
since we have already shown that GM(x,y)≤ AM(x,y) forx, y≥ 0,
we now have
HM(x 1 ,x 2 ) =(
AM(
1
x 1+
1
x 2))− 1
≤(
GM(
1
x 1,
1
x 2))− 1
= GM(x 1 ,x 2 ).Finally, note that since 2x 1 x 2 ≤x^21 +x^22 (as proved in the above section),
(x 1 +x 2 )^2 =x^21 + 2(x 1 x 2 ) +x^22 ≤2(x^21 +x^22 ).Divide both sides of the above inequality by 4, take square roots and
infer that AM(x 1 ,x 2 )≤QM(x 1 ,x 2 ).
For a geometric argument showing HM≤GM≤AM, see Exercise 1,
below.
We turn next to proofs of the above inequalities in the general case.AM(x 1 ,...,xn)≤QM(x 1 ,...,xn): This is equivalent with saying that
(x 1 +···+xn)^2
n^2≤
x^21 +···+x^2 n
n,
which is equivalent with proving that
(x 1 +···+xn)^2 ≤n(x^21 +···+x^2 n).By induction, we may assume that
(x 1 +···+xn− 1 )^2 ≤(n−1)(x^21 +···+x^2 n− 1 ).Furthermore, note that for any real numbersx,y, we have 0≤(x−y)^2 =
x^2 +y^2 − 2 xy⇒ 2 xy≤x^2 +y^2. Armed with this, we proceed, as follows:
(x 1 +···+xn)^2 = (x 1 +···+xn− 1 )^2 + 2xn(x 1 +···+xn− 1 ) +x^2 n
≤ (n−1)(x^21 +···+x^2 n− 1 )
+(x^21 +x^2 n) +···+ (x^2 n− 1 +x^2 n) +x^2 n
= n(x^21 +···+x^2 n),which proves that AM≤QM. Notice that since, foranyx 1 , x 2 , ..., xn,
x 1 +x 2 +···+xn≤|x 1 |+|x 2 |+···+|xn|,