Arithmetic and geometric means

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

  (a1+ ... + aN)/N

whereas their geometric mean is given by

  (a1· ... ·aN)1/N

The two quantities are always related in the following way

  (a1 + ... + aN)/N ≥ (a1· ... ·aN)1/N

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

  ai = (a1 + ... + aN)/N

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

  (a1 + ... + a2n+1)/2n+1 ≥ (a1· ... ·a2n+1)1/2n+1

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

  (a1 + ... + aN + (2n+1-N)(a1+ ... + aN)/N)/2n+1 ≥ (a1· ... ·aN)1/2n+1((a1 + ... + aN)/N)(2n+1-N)/2n+1

Adding similar terms on the left we get

  (N+2n+1-N)(a1 + ... + aN)/(N·2n+1)= (a1 + ... + aN)/N

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

  (a1 + ... + aN)/N ≥ (a1· ... ·aN)1/2n+1((a1 + ... + aN)/N)(2n+1-N)/2n+1

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

  ((a1 + ... + aN)/N)(1-(2n+1-N)/2n+1) ≥ (a1· ... ·aN)1/2n+1

or

  ((a1 + ... + aN)/N)N/2n+1 ≥ (a1· ... ·aN)1/2n+1

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

  (a1 + ... + aN)/N ≥ (a1· ... ·aN)1/N
Q.E.D.

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


Related material
Read more...

  • The Means
  • Averages, Arithmetic and Harmonic Means
  • Expectation
  • The Size of a Class: Two Viewpoints
  • Averages of divisors of a given integer
  • Family Statistics: an Interactive Gadget
  • Averages in a sequence
  • Geometric Meaning of the Geometric Mean
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    Copyright © 1996-2010 Alexander Bogomolny

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