Young's Inequality for Products

Theorem
Let $p$ and $q$ be positive numbers such that $\dfrac 1 p + \dfrac 1 q = 1$.

Then, for any nonnegative numbers $a$ and $b$:
 * $\displaystyle ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}$

Proof
The result is obvious if $a=0$ or $b=0$, so assume without loss of generality that $a>0$ and $b>0$.

Then: