The AM-GM inequality is implied by an identity.
For the function $f=f(x_1,x_2,\dots,x_n)$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Thus $$Px_1^n=(n-1)! (x_1^n+x_2^n+\dots+x_n^n),$$ $$Px_1x_2\dots x_n=n!\ x_1x_2\dots x_n.$$
Consider the functions $\phi_k$, $k=1,2,\dots,n-1$, obtained in the following manner $$\phi_1=P[(x_1^{n-1}-x_2^{n-1})(x_1-x_2)],$$ $$\phi_2=P[(x_1^{n-2}-x_2^{n-2})(x_1-x_2)x_3],$$ $$\phi_2=P[(x_1^{n-3}-x_2^{n-3})(x_1-x_2)x_3x_4],$$ $$...$$ $$\phi_{n-1}=P[(x_1-x_2)(x_1-x_2)x_3x_4\dots x_n].$$
We see that $$\phi_1=Px_1^n+Px_2^n-Px_1^{n-1}x_2-Px_2^{n-1}x_1=2Px_1^n-2Px_1^{n-1}x_2.$$ Similarly, $$\phi_2=2Px_1^{n-1}x_2-2Px_1^{n-2}x_2x_3,$$ $$\phi_3=2Px_1^{n-2}x_2x_3-2Px_1^{n-3}x_2x_3x_4,$$ $$...$$ $$\phi_{n-1}=2Px_1^2x_2x_3\dots x_{n-1}-2Px_1x_2\dots x_n.$$ Adding these identities, we have $$\phi_1+\phi_2+\dots\phi_{n-1}=2Px_1^n-2Px_1x_2\dots x_n,$$ or, $$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+ \dots \phi_n).$$ It is easy to see that each of the functions $\phi_k(x)$ is nonnnegative for $x_i\geq 0$, since $$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}].$$
The proof isTwo examples due to Hurwitz (1891). I have reproduced it from Inequalities by Beckenbach and Bellman.
The AM-GM inequality. For the function $f=f(x_1,x_2,\dots,x_n)$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Then $$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+ \dots \phi_n),$$ where $$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}]\geq0.$$ The proof can be found in Inequalities by Beckenbach and Bellman.
The isoperimetric inequality. Let the boundary of $\Omega\subset \mathbb R^2$ be a rectifiable Jordan curve $\partial \Omega=\{((x(s),y(s))|\ s\in[0,2\pi))\}$. Then $$L^2-4\pi A=2\pi^2\sum\limits_{n=1}^{\infty}\left[(na_n-d_n)^2+(nb_n+c_n)^2+ (n^2-1)(c_n^2+d_n^2)\right],$$ where $$x(s)=\sum\limits_{n=0}^{\infty}(a_n\cos ns+b_n\sin ns),\quad y(s)=\sum\limits_{n=0}^{\infty}(c_n\cos ns+d_n\sin ns).$$