Young's Inequality

The AM-GM inequality is a very special cases of Young's inequality. We derive it in a general form, using the coupling functions introduced by Moreau. Suppose $f$ is a real-valued function on $\mathcal{X}$ and $g$ is a real-valued function on $\mathcal{X}\times\mathcal{Y}$, called the coupling function. Here $\mathcal{X}$ and $\mathcal{Y}$ are arbitrary. Define the $g$-conjugate of $f$ by $$f^\circ_g(y)\mathop{=}\limits^{\Delta}\sup_{x\in\mathcal{X}}\ \{g(x,y)-f(x)\}$$ Then $g(x,y)-f(x)\leq f^\circ_g(y)$ and thus $$g(x,y)\leq f(x)+f^\circ_g(y)$$ , which is the generalized Young's inequality. We can also write this in the form that directly suggests minorization $$f(x)\geq f^\circ_g(y)+g(x,y).$$

The classical coupling function is $g(x,y)=xy$ with both $x$ and $y$ in the positive reals. If we take $f(x)=\frac{1}{p}x^p$, with $p>1$, then $$f^\circ_g(y)=\sup_{x}\ \{xy-\frac{1}{p}x^p\}.$$ The $\sup$ is attained for $x=y^{\frac{1}{p-1}}$, from which we find $f^\circ_g(y)=\frac{1}{q}y^q,$ with $q\mathop{=}\limits^{\Delta}\frac{p}{p-1}$.

Thus if $p,q>1$ such that $$\frac{1}{p}+\frac{1}{q}=1.$$ Then for all $x,y>0$ we have $$xy\leq\frac{x^p}{p}+\frac{y^q}{q},$$ with equality if and only if $y=x^{p-1}$.