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.

AM-GM inequality says that for any $ a_1, \dots , a_n > 0 $, we have $ \dfrac{a_1 + \cdots + a_n}{n} \geq  \sqrt[n]{a_1 \cdots a_n} $ with equality holding if and only if $ a_1 = \cdots = a_n $.
Here is Cauchy's Backward Induction proof, but with the backward part slightly modified (and hopefully more intuitive). We'll first prove the inequality, and then think about when equality holds. For brevity, let $ P_n $ mean "For any $a_1, \dots a_n > 0, \dfrac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n}$ ".
$P_1$ holds trivially.
$P_2$ is true because $ \dfrac{a_1 + a_2}{2} \geq \sqrt{a_1 a_2} $ can be written as $ a_1 + a_2 - 2\sqrt{a_1 a_2} \geq 0 $, which is just $ (\sqrt{a_1} - \sqrt{a_2})^2 \geq 0 $.
$ P_4 $ comes from $ P_2 $, as $ \dfrac{a_1 + a_2 + a_3 + a_4}{4} = \dfrac{\dfrac{a_1 + a_2}{2} + \dfrac{a_3 + a_4}{2}}{2} \geq \dfrac{\sqrt{a_1 a_2} + \sqrt{a_3 a_4}}{2} \geq \sqrt{\sqrt{a_1 a_2}\sqrt{a_3 a_4}} = \sqrt[4]{a_1 \cdots a_4} $.
Similarly $P_8$ comes from $P_4$, $P_{16}$ comes from $P_8$, etc.
In general, if we know a $ P_{2^k} $ holds then $ P_{2^{k+1}} $ must hold as
$ \dfrac{ a_1 + \cdots +a_{2^{k+1}} }{2^{k+1}} = \dfrac{ \dfrac{ a_1 + \cdots a_{2^k} }{2^k} + \dfrac{ a_{2^{k} + 1} + \cdots a_{2^{k+1}} }{2^k} }{2} \geq \dfrac{(a_1 \cdots a_{2^k})^{ \frac{1}{2^k} }  + (a_{2^k + 1} \cdots a_{2^{k+1}})^{ \frac{1}{2^k} } }{2} \geq \sqrt{ (a_1 \cdots a_{2^k})^{ \frac{1}{2^k} } (a_{2^k + 1} \cdots a_{2^{k+1}})^{ \frac{1}{2^k} } } = (a_1 \cdots a_{2^{k+1}})^{\frac{1}{2^{k+1}}} $.
So it is clear $P_k$ holds whenever $ k $ is a power of $2$.
It now suffices to show "If a $P_{k+1}$ holds, then $P_k$ must hold". [Informally, we showed we can jump to powers of $2$, but there are gaps so now we are trying to show we can also take single steps backwards]. So we'll try to prove this:
Fix any $ a_1, \dots, a_k > 0 $. Now we are to show $ \dfrac{a_1 + \cdots + a_k}{k} \geq (a_1 \cdots a_k)^{\frac{1}{k}} $ , but we have the information $ \dfrac{ a_1 + \cdots + a_k + t }{k+1} \geq (a_1 \cdots a_k t)^{\frac{1}{k+1}} $ for any $ t > 0 $. In order to match the info's RHS to that of the goal we can try picking a $ t $ such that $ (a_1 \cdots a_k t)^{\frac{1}{k+1}} = (a_1 \cdots a_k)^{\frac{1}{k}} $, i.e we can try setting $ t = (a_1 \cdots a_k)^{\frac{1}{k}} $, and in fact it works :
$ \dfrac{ a_1 + \cdots a_k + (a_1 \cdots a_k)^{\frac{1}{k}} }{k+1} \geq (a_1 \cdots a_k)^{\frac{1}{k}} $, giving the required inequality $ \dfrac{a_1 + \cdots + a_k}{k} \geq (a_1 \cdots a_k)^{\frac{1}{k}} $.
[We could've also tried to match info's LHS to that of the goal, i.e we could've tried setting $ t = \dfrac{a_1 + \cdots a_k}{k} $, and even this happens to work !].
So the inequality part is proved. Now we'll think about when the equality is true. The $P_2$-inequality becomes equality if and only if $ \sqrt{a_1} - \sqrt{a_2} = 0 $, i.e if and only it $ a_1 = a_2 $. The $P_4$-inequality becomes equality if and only if $ \sqrt{a_1 a_2} = \sqrt{a_3 a_4}, a_1 = a_2 $ and $ a_3 = a_4 $, i.e. if and only if $ a_1 = \cdots = a_4 $. Proceeding so, it is clear $P_{2^k}$-inequality becomes equality if and only if $ a_1 = \cdots = a_{2^k} $. Now looking at the backwards induction step above, we see that in general equality holds if and only if all the numbers taken are equal.
