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This very useful theorem (an oxymoron?) states that the arithmetic mean of a sequence of numbers is always greater than their geometric mean.
For any sequence of numbers x1, x2, ..., xn

(x1 + x2+ ... + xn)/n ≥ (x1x2...xn)1/n

It applies to weighted means as well.

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/x2 < 0 for all positive x). Now, if some xi is 0, then their arithmetic mean is non-negative (recall that all xi'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 yi=1/xi. Then AM >= GM for the y's means that

(1/x1 + ... + 1/xn)/n >= 1/(x1...xn)1/n
Taking inverses, we see that indeed
n / (1/x1 + ... + 1/xn) <= (x1...xn)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.

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