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.
Pólya's Proof:

Let $f(x) = e^{x-1}-x$. The first derivative is $f'(x)=e^{x-1}-1$ and the second derivative is $f''(x) = e^{x-1}$. 

$f$ is convex everywhere because $f''(x) &gt; 0$, and has a minimum at $x=1$. Therefore $x \le e^{x-1}$ for all $x$, and the equation is only equal when $x=1$. 

Using this inequality we get 

$$\frac{x_1}{a} \frac{x_2}{a} \cdots \frac{x_n}{a} \le e^{\frac{x_1}{a}-1} e^{\frac{x_2}{a}-1} \cdots e^{\frac{x_n}{a}-1}$$

with $a$ being the arithmetic mean. The right side simplifies 

$$\exp \left(\frac{x_1}{a} -1 \ +\frac{x_1}{a} -1 \ + \cdots +  \frac{x_n}{a} -1  \right)$$

$$=\exp \left(\frac{x_1 + x_2 + \cdots + x_n}{a} - n \right) = \exp(n - n) = e^0 = 1$$

Going back to the first inequality 

$$\frac{x_1x_2\cdots x_n}{a^n} \le 1$$

So we end with 

$$\sqrt[n]{x_1x_2\cdots x_n} \le a$$
My idea is to use only the Cauchy Schwartz inequality by induction.


Applying the Cauchy_Schwartz inequality on the vectors $(\sqrt{a},\sqrt{a})^T$ and $(\sqrt{b},\sqrt{b})^T$ we get: 
$$\sqrt{ab} +\sqrt{ab} \le (a+b)^{1/2}(a+b)^{1/2} $$
that is $$\sqrt{ab} \le \frac{a+b}{2}~~~~~~\color{red}{\text{AM-GM for $n =2$}}$$



  Hypothesis of induction: Assume that for all $p\in\{1,2,\cdots, n-1\}$ we have, 
  \begin{align*}
\color{blue}{(a_1a_2\cdots a_p)^{1/p}\leq \frac{a_1+a_2+\cdots+a_p}{p}}
\end{align*}
  We want to prove that it is true for $n$. we will discuss two case: $n =2p$ and $n=2p-1$


If $n =2p$ Then we proceed as follows: 
Using that case: $n=2$  and the case $n=p$, we have, 

$$ \begin{align}\color{blue}{\frac{a_1+a_2+\cdots+a_n}{n}  }&amp;= \frac{1}{2}\left(\overbrace{\frac{a_1+a_2+\cdots+a_p}{p}}^a+\overbrace{\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}}^b  \right) \\
&amp;\ge\sqrt{\left(\frac{a_1+a_2+\cdots+a_p}{p}\right) \left(\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}\right)}\\
&amp;\ge\sqrt{(a_1a_2\cdots a_p)^{1/p}(a_{p+1}a_{p+2}\cdots a_n)^{1/p}}\\
&amp;=\sqrt{(a_1a_2\cdots a_n)^{1/p}}  = \color{blue}{(a_{p+1}a_{p+2}\cdots a_n)^{1/n} } ~~~~~~\color{red}{\text{AM-GM for $n =2p$}} \end{align}$$

If $n =2p-1$ (we use the case $n=2p$, as follows.we have,  $n+1 =2p $ and $p&lt;n$ : Then taking $$ a_{n+1} = \frac{a_1+a_2+\cdots+a_n}{n}$$
in the previous case we obtain, 

\begin{align}&amp;\frac{a_1+a_2+\cdots+a_{n+1}}{n+1}   \ge (a_1a_2\cdots a_{n+1})^{\frac{1}{n+1}}\\
&amp;\Longleftrightarrow\frac{1}{n+1}\left(a_1+a_2+\cdots+a_n+\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}}  \right) \ge \left(a_1a_2\cdots a_{n}\left(\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}}\right)  \right)^{\frac{1}{n+1}}\\
\\&amp;\Longleftrightarrow\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)  \right)^{\frac{1}{n+1}}\\
&amp;\Longleftrightarrow\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^{\frac{n}{n+1}} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n+1}}
\\&amp;\Longleftrightarrow \color{blue}{\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n}}}~~~~\color{red}{\text{AM-GM for $n =2p-1$}}\end{align}