Proofs of AM-GM inequality The arithmetic - geometric mean inequality states that
$$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$
I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?
Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.

Another answer. We can prove this alternative result:
If $a_1, \ldots, a_n$ are positive reals such that $a_1 +  \dots + a_n = n$, then $a_1 \dots a_n \leq 1$.
We can suppose wlog that $a_1 \leq \dots \leq a_n$.
Notice that we can suppose $a_n \not= 1$, or else we could just solve the same problem for $n-1$.
By "pigeonhole principle", we know that $a_1 \leq 1 \leq a_n$. If we change them by their arithmetical mean, the sum stays the same, but the product increases. Indeed, the product "increases" by $(\frac{a_1+a_n}{2})^2-a_1a_n = (\frac{a_1-a_n}{2}) \geq 0$, with equality if and only if $a_1=a_n$.
That way, we can always increase the sum whenever there are at least two different terms. The only way it can't be done is if all terms are equal; and this is the supremum value.
(Another possible solution would be change $a_n$ by $1$ and $a_1$ by $(a_1+a_n)-1$; the sum stays the same but the product "increases" by $a_1+a_n-1-a_1a_n = (1-a_1)(a_n-1) \geq 0$; that way the maximum occurs if all terms are $1$).
