# Arithmetic and geometric means

In the following I'll consider sets of positive real numbers $a_{1}, \ldots , a_{N},\;$ $N\;$ a positive integer. Arithmetic mean of the given numbers is defined as

$\displaystyle A(a)=A(a_1,\ldots,a_n)=\frac{a_{1}+ \ldots + a_{N}}{N},$

whereas their geometric mean is given by

$G(a)=G(a_1,\ldots,a_n)=(a_{1}\cdot \ldots \cdot a_{N})^{\displaystyle\frac1{N}}.$

The two quantities always relate in the following manner known as the *Arithmetic Mean - Geometric Mean Inequality* (*AM-GM*, for short), $A(a)\ge G(a),\;$ or more explicitly

$\displaystyle \frac{a_{1}+ \ldots + a_{N}}{N}\ge (a_{1}\cdot \ldots \cdot a_{N})^{\displaystyle\frac1{N}}.$

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 $N = 2^{n}\;$ then it holds for all $N \ge 1.\;$ This would afford another example of a general proposition implied by its special case.

Thus, assume the inequality holds for all $N = 2^{n}\;$ and let $N = 2^{n} + m,\;$ where $0 \lt m \lt 2^{n}\;$ and $n \gt 0.\;$ For $i = N+1, \ldots , 2^{n+1},\;$ define the "missing" a's as

$\displaystyle a_{i} = \frac{a_{1} + \ldots + a_{N}}{N}.$

Since the inequality holds for $N = 2^{n+1},$ we have

$\displaystyle\frac{a_{1} + \ldots + a_{2^{n+1}}}{2^{n+1}} \ge (a_{1}\cdot \ldots \cdot a_{2^{n+1}})^{\displaystyle\frac{1}{2^{n+1}}}.$

Substituting $\displaystyle a_{i} = \frac{a_{1} + \ldots + a_{N}}{N}\;$ for $i = N+1, \ldots , 2^{n+1}\;$ results in

$\displaystyle\frac{a_{1} + \ldots + a_{N} + \frac{\displaystyle (2^{n+1}-N)(a_{1}+ \ldots + a_{N})}{N}}{2^{n+1}} \ge (a_{1}\cdot \ldots \cdot a_{N})^{1/2^{n+1}}\left(\frac{a_{1} + \ldots + a_{N}}{N}\right)^{(2^{n+1}-N)/2^{n+1}}.$

Adding similar terms on the left we get

$\displaystyle\frac{(N+2^{n+1}-N)(a_{1} + \ldots + a_{N})}{N\cdot 2^{n+1}}= \frac{a_{1} + \ldots + a_{N}}{N}$

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

$\displaystyle\frac{a_{1} + \ldots + a_{N}}{N} \ge (a_{1}\cdot \ldots \cdot a_{N})^{1/2^{n+1}} \cdot\left(\frac{a_{1} + \ldots + a_{N}}{N}\right)^{(2^{n+1}-N)/2^{n+1}}.$

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

$\displaystyle \left(\frac{a_{1} + \ldots + a_{N}}{N}\right)^{1-(2^{n+1}-N)/2^{n+1}} \ge (a_{1}\cdot \ldots \cdot a_{N})^{1/2^{n+1}}$

or

$\displaystyle \left(\frac{a_{1} + \ldots + a_{N}}{N}\right)^{N/2^{n+1}} \ge (a_{1}\cdot \ldots \cdot a_{N})^{1/2^{n+1}}.$

Now raising both sides to the power of $\displaystyle\frac{2^{n+1}}{N}\;$ we finally get

$\displaystyle\frac{a_{1} + \ldots + a_{N}}{N} \ge (a_{1}\cdot \ldots \cdot a_{N})^{1/N}.$

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

|Contact| |Front page| |Contents| |Generalizations| |Store|

Copyright © 1996-2017 Alexander Bogomolny

62105953 |