In the following I'll consider sets of positive real numbers a₁, ..., a_N, N a positive integer. Arithmetic mean of the given numbers is defined as

whereas their geometric mean is given by

The two quantities are always related in the following way

Here I am not going to prove the well known inequality but just emphasize a fact that was used by Cauchy in his proof. Namely, if the inequality holds for all N = 2ⁿ then it holds for all N ≥ 1. This would afford another example of a general proposition implied by its special case.

Thus, assume the inequality holds for all N = 2ⁿ and let N = 2ⁿ + m, where 0 < m < 2ⁿ and n > 0. For i = N+1, ..., 2ⁿ⁺¹, define the "missing" a's as

Since the inequality holds for N = 2ⁿ⁺¹ we have

Substituting a_i = (a₁ + ... + a_N)/N for i = N+1, ..., 2ⁿ⁺¹ results in

Adding similar terms on the left we get

which actually says that the arithmetic mean has not been changed by addition of new terms.

Dividing by the rightmost term and with one more step to go

or

Now raising both sides to the power of 2ⁿ⁺¹/N we finally get

There is a way to derive a complete proof of the inequality from the Pythagorean Theorem.

Copyright © 1996-2008 Alexander Bogomolny