In class (Sats 1.1 and exercise 1.1) we proved the case n = 2 of the following well-known theorem

The AMGM inequality Let a_{1}, a_{2}, ..., a_{n} be positive real numbers. Then

\frac{1}{n} (a_{1} + ⋯ + a_{n}) ≥ ^{n} \sqrt{a_{1} ⋯ a_{n}},  (1)

with equality if and only if a_{1} = a_{2} = ... = a_{n}.

There are many different proofs out there of this result and various generalisations of it, and every so often a new one appears in some popular mathematics journal. Here is one fairly short proof, taken from the book ‘A course of pure mathematics’, by G.H. Hardy.

Proof : Denote the geometric mean (rhs of (1)) by G. Let a_{r} and a_{s} be the greatest and least of the a’s respectively (if there are several of either, choose any one arbitrarily). If we replace a_{r}, a_{s} by

a′_{r} := G and a′_{s} := \frac{a_{r}a_{s}}{G} ,

then the rhs of (1) is unchanged, whereas since

(a′_{r} + a′_{s}) − (a_{r} + a_{s}) = \frac{1}{G}(a_{r} − G)(a_{s} − G) ≤ 0,

the lhs of (1) is not increased, and is in fact decreased unless a_{r} = a_{s} = G, which is the case if and only if all a_{i} = G.

Now repeat this procedure until each of a_{1}, ..., a_{n} has been replaced by G - at most n steps are needed. At the end, the arithmetic mean (lhs of (1)) has become equal to G, so at the beginning it must have been bigger, unless all a_{i} = G at the outest. This completes the proof !