Young's Inequality for Products

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

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

Equality occurs :
 * $b = a^{p - 1}$

Also presented as
Statements of Young's inequality will commonly insist that $p, q > 1$.

This is no different to the statement presented here, since if:


 * $\ds \frac 1 p + \frac 1 q = 1$

for positive $p$ and $q$ we must have:


 * $\ds \frac 1 p \le 1$

and:


 * $\ds \frac 1 q \le 1$

So $p, q \ge 1$.

But we cannot have $p = 1$ or $q = 1$, otherwise $\dfrac 1 q = 0$ or $\dfrac 1 p = 0$ respectively.

This is understood to be solved by $q = \infty$ and $p = \infty$ respectively, but there is no real meaning to the in this case, so we arrive at the usual hypotheses of $p, q > 1$.