A fast way to verify the truth of the AM-GM

inequality is to use the fact that the

logarithm function is

concave (e.g. note that

log''x = -1/x

^{2} < 0 for all

positive x). Now, if some x

_{i} is 0, then their

arithmetic mean is non-negative (recall that all x

_{i}'s must be non-negative or their

geometric mean won't be defined), and the inequality holds. Otherwise, take logarithms and apply the

Jensen inequality (to the convex function -log x).

This method also works for suitably weighted versions.
Elementary proofs also exist, but are somewhat longer.

Equality is attained *iff* all elements are equal. In other words, the maximal product of a set of elements with constant sum is attained when they are all equal. This sometimes lets us solve multivariate optimization problems without using calculus!

It is customary to extend the AM-GM inequality to state that the geometric mean is itself no smaller than the harmonic mean. In fact, GM >= HM follows from AM >= GM.
Take y_{i}=1/x_{i}. Then AM >= GM for the y's means that

(1/x_{1} + ... + 1/x_{n})/n >= 1/(x_{1}...x_{n})^{1/n}

Taking inverses, we see that indeed

n / (1/x_{1} + ... + 1/x_{n}) <= (x_{1}...x_{n})^{1/n}

It is always fun to try to find new proof techniques for old theorems. Here's a (rigourous!) physical proof that the arithmetic mean is at least as large as the harmonic mean.