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.

We use the method if infinite descent.
Let $P_n$ be the proposition that there is some set $S_n=\{a_1,a_2,\dots a_n\}$ with $a_i\geqslant0$ and $\sum a_i=n$ and for which $\prod a_i > 1$.
We would like to show that $P_n$ is always false. Suppose, for the sake of contradiction, that there is some $n$ for which $P_n$ is true and let $m$ be the smallest such $n$.
Clearly $P_1$ is false and so $m \geqslant 2$.
In order for $\sum a_i=n$ there must be at least one element bigger than $1$ and one less than $1$ (they cannot all equal $1$ because $\prod a_i > 1$) so we can say $a_1>1$ and $a_2<1$.
$$(a_1-1)(1-a_2)>0  \implies a_1+a_2-1 >a_1 a_2 \tag 1$$
We now replace $a_1$ with $a_1+a_2-1$ and $a_2$ with $1$. The sum ($\sum a_i$) doesn't change and, by (1), the product increases, so we still have $\prod a_i > 1$.
But now one of the elements of $S_m$ is equal to $1$ we can define $S_{m-1}=S_m \setminus  \{1\}$ and for this set we have $\sum a_i=m-1$ and $\prod a_i > 1$. This proves $P_{m-1}$ and so contradicts the definition of $m$ and so our assumption that such an $m$ exists must be false and so $P_n$ is false for all $n$.
With the hard work done, AM-GM follows immediately. Suppose
$$\frac{x_1+ \cdots + x_n}{n} < \sqrt[n]{x_1 \cdots x_n}$$
Then define
$$a_i = \frac{n \, x_i}{x_1+ \cdots + x_n}$$
This would prove $P_n$ which we have shown to be false and so no such inequality can be true.
