The arithmetic-mean/geometric-mean inequality

1. Arithmetic and geometric means

A previous page showed from the monotone convergence theorem that in the reals, all positive x have an n th root, x 1 n , for each n . This page investigates these n th roots further by discussing an important and powerful inequality connecting arithmetic and geometric means in an ordered field.

Recall that given numbers x 1 , x 2 , , x n their mean is the number 1 n i = 1 n x i . This is often called the arithmetic mean to distinguish it from the geometric mean which is obtained by replacing addition with multiplication (and division by n by n th roots). I.e. the geometric mean of x 1 , x 2 , , x n is ( x 1 · x 2 · · x n ) 1 n . For non-negative numbers x 1 , x 2 , , x n , the arithmetic and geometric means are related by the inequality

( x 1 · x 2 · · x n ) 1 n 1 n · i = 1 n x i

We will prove this, and the proof will be by induction.

The following lemma will be the base case of the induction.

Lemma.

Given x 1 , x 2 and y 1 , y 2 with x 1 + x 2 = y 1 + y 2 , we have: x 1 · x 2 < y 1 · y 2 if and only if | x 1 - x 2 | > | y 1 - y 2 | . In particular

x 1 · x 2 x 1 + x 2 2 2

Proof.

Observe that

x 1 · x 2 = x 1 + x 2 2 2 - x 1 - x 2 2 2

with a similar equality for y 1 , y 2 . So x 1 + x 2 2 2 is the maximum possible value for x 1 · x 2 for a given value for x 1 + x 2 . Also, x 1 · x 2 < y 1 · y 2 holds if and only if x 1 - x 2 2 2 > y 1 - y 2 2 2 holds, which is if and only if | x 1 - x 2 | > | y 1 - y 2 | , completing the proof.

We can now prove the famous arithmetic-mean/geometric-mean inequality.

Arithmetic-mean/Geometric-mean Inequality.

Let n and x 1 , x 2 , , x n 0 . Then

x 1 · x 2 · · x n 1 n · i = 1 n x i n

Proof.

By induction on n , the induction hypothesis being the statement of the theorem. Note that the case n = 1 is easy and the case n = 2 is true by the previous lemma.

Assume n 3 and assume inductively that the theorem is true for n - 1 in place of n . Given x 1 , x 2 , , x n 0 let M = 1 n · i = 1 n x i . If x 1 = x 2 = = x n then each x i = M and it is easy to see that in fact equality holds in the above inequality, so we would be done. Otherwise we may assume (by re-ordering the x i if necessary) that x 1 < M < x n . So 2 x 1 < 2 M so x 1 + x n - 2 M < x n - x 1 . We set y 1 = M and y 2 = x 1 + x 2 - M . Then if y 1 > y 2 we have y 1 - y 2 = 2 M - x 1 - x n < x n - x 1 as 2 M - x n < x n ; also, if y 1 < y 2 we have y 2 - y 1 = x 1 + x 2 - 2 M < x n - x 1 as 2 M - x 1 > x 1 ; either way we have | y 2 - y 1 | < | x n - x 1 | and hence by the lemma we have

x 1 · x n < y 1 · y 2 = M ( x 1 + x n - M )

Thus

x 1 · x 2 · · x n < M · ( x 1 + x n - M ) · x 2 · · x n - 1

and by the induction hypothesis

( x 1 + x n - M ) · x 2 · · x n - 1 < 1 n - 1 ( ( x 1 + x n - M ) + i = 2 n - 1 x i ) n - 1 = n M - M n - 1 n - 1 = M n - 1

giving x 1 · x 2 · · x n < M n as required.

2. A useful corollary

We give as a corollary a useful inequality, reminiscent of Bernoulli's inequality which we saw earlier, but the inequality sign goes the other way round and the result is only valid for exponents at most one.

The Exponential Inequality.

Suppose r , s with 1 r s . Then for all x with -1 < x < 1 we have ( 1 + x ) r s 1 + r s x .

Proof.

With x > -1 and 1 r s integers apply the previous result to x 1 , , x r = 1 + x and x r + 1 , , x s = 1 . Then

( 1 + x ) r r · ( 1 + x ) + ( s - r ) · 1 s s = 1 + r s x s

so ( 1 + x ) r s 1 + r s x .