Young's Inequality for Products/Geometric Proof

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


In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$.

Accordingly:
 * $u = t^{p-1} \iff t = u^{q-1}$

Let $a, b$ be any positive real numbers.

Since $a b$ is the area of the rectangle in the given figure, we have:


 * $\displaystyle a b \le \int_0^a t^{p-1} \ \mathrm d t + \int_0^b u^{q-1} \ \mathrm d u = \frac {a^p} p + \frac {b^q} q$

Note that even if the graph intersected the side of the rectangle corresponding to $t = a$, this inequality would hold.

Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially.