Identities and inequalities in analysis and probability

Many inequalities are proved by identities representing the thing which must be proved to be non-negative as integral (or sum, or expectation) of squares. For example: CBS inequality $$ \int_X f^2 \cdot \int_X g^2 -\left(\int_X fg\right)^2=\frac12\int_{X\times X} \left(f(x)g(y)-f(y)g(x)\right)^2\geqslant 0. $$ Or more involved: Hardy inequality $$4\int_0^\infty f^2(x)dx-\int_0^\infty \left(\frac1x \int_0^x f(t)dt\right)^2dx=\int_0^\infty\int_0^\infty (f(x)\sqrt{x}-f(y)\sqrt{y})^2\frac{dxdy}{2\sqrt{xy}\max(x,y)}$$

Probably the Fenchel inequality counts? $\newcommand{\RR}{\mathbb{R}}$

For a function $f:X\to\RR\cup\{\infty\}$ on a vector space $X$ and it's convex conjugate $f^*(x^*) = \sup_{x\in X}\langle x^*,x\rangle - f(x)$ there always holds that $$\langle x^*,x\rangle\leq f(x) + f^*(x^*).$$

This include Cauchy-Schwarz and Young's inequality but can sometimes be used to get more specific inequalities.

I am not sure if this one fits to this category. In this case it is a PDE (or PDI -partial differential inequality) ruling the inequality. And (I think) you can extract some identity after ``integrating'' this PDE (or PDI).

Let $\Omega \subset \mathbb{R}^{2}$ be a rectangular subset, and let $H(x,y) : \Omega \to \mathbb{R}$ be a bounded smooth function such that $H_{x}, H_{y} \neq 0$ on $\Omega$. Fix some $\lambda \in (0,1)$. Then inequality $$ \int_{\mathbb{R}^{k}}\sup_{\lambda x+(1-\lambda)y=z}H(f(x),g(y)) d\gamma_{k}(z)\geq H\left(\int_{\mathbb{R}^{k}}f\; d\gamma_{k},\int_{\mathbb{R}^{k}}g\; d\gamma_{k} \right) $$ holds for all $(f(x),g(y)):\mathbb{R}^{k}\times \mathbb{R}^{k} \to \Omega$ where $d\gamma_{k}=e^{-x^{2}/2}\frac{1}{(2\pi)^{k/2}}dx$ is $k$ dimensional gaussian measure if and only if $$ (1-\lambda^{2}-(1-\lambda)^{2})\frac{H_{xy}}{H_{x}H_{y}}+\lambda^{2}\frac{H_{xx}}{H_{x}^{2}}+(1-\lambda)^{2}\frac{H_{yy}}{H_{y}^{2}}\geq 0 \quad (1). $$ Here is the reference to the formal statement

Applications:

You can try to play with some functions (or you can actually try to describe all solutions of PDE (1) -- when you have equality instead of inequality). For example, if you try $H(x,y)=\Psi(\lambda \Psi^{-1}(x)+(1-\lambda)\Psi^{-1}(y))$ then (1) takes the form $$ \frac{1}{\Psi'(\lambda P+(1-\lambda)Q)}\left(\frac{\Psi''(\lambda P+(1-\lambda)Q)}{\Psi'(\lambda P+(1-\lambda)Q)}-\lambda \frac{\Psi''(P)}{\Psi'(P)}-(1-\lambda)\frac{\Psi''(Q)}{\Psi'(Q)}\right), $$ where $P=\Psi^{-1}(x), Q=\Psi^{-1}(y)$, which is nonnegative if and only if $\Psi'>0$ and $\frac{\Psi''}{\Psi'}$ is concave (or $\Psi'<0$ and $\frac{\Psi''}{\Psi'}$ is convex).

1) (Prekopa--Leindler inequality) Indeed, take $\Psi(x)=e^{x}$ in the previous example. Then you will get $H(x,y)=x^{\lambda}y^{1-\lambda}$.

2) (Ehrhard inequality). Take $\Psi(t)=\int_{-\infty}^{t}\frac{e^{-x^{2}/2}dx}{\sqrt{2\pi}}$. This recovers Ehrhard inequality.