Young's Inequality for Products/Proof by Convexity

Theorem
Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
 * $\dfrac 1 p + \dfrac 1 q = 1$

Then, for any $a, b \in \R_{\ge 0}$:
 * $a b \le \dfrac {a^p} p + \dfrac{b^q} q$

Proof
The result is obvious if $a=0$ or $b=0$, so assume WLOG that $a > 0$ and $b > 0$.

Then: