Reverse Young's Inequality for Products

Theorem
Let $p, q \in \R_{> 0}$ be strictly positive real numbers satisfying $\displaystyle \frac 1 p - \frac 1 q = 1$.

Then, for any $a \in \R_{\ge 0}$ and $b \in \R_{> 0}$:
 * $\displaystyle ab \ge \frac {a^p} {p} - \frac {b^{-q}}{q}$

Proof
Define:

Then $\displaystyle \frac 1 u + \frac 1 v = 1$.

By Young's Inequality for Products:
 * $\displaystyle xy \le \frac {x^u}{u} + \frac {y^v}{v}$

This gives the result.