# Contents

## Young’s inequality for products

###### Proposition

Given

• $a, \,b \,\in\, \mathbb{R}_{\gt 0}$, non-negative real numbers;

• $p, \, q \,\in\, \mathbb{R}_{\gt 1}$ such that

$\frac{1}{p} + \frac{1}{q} = 1$

then the following inequality holds:

$a b \;\leq\; \frac{a^p}{p} + \frac{b^q}{q} \,,$

which is an equality if and only if $a^p = b^q$.

One proof is by convexity of the exponential function: choosing $x, y, t$ such that $\exp(x) = a^p$, $\exp = b^q$ and $t = \frac1{p}$, Young’s inequality is identical to the convexity constraint

$\exp(tx + (1-t)y) \leq t\exp(x) + (1-t)\exp(y).$