Young's Inequality for Products/Geometric Proof

Proof

 * Holder's Ineq.jpg

In the above diagram, the $\color {blue} {\text {blue} }$ colored region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$ and the $\color {red} {\text {red} }$ colored region to $\ds \int_0^\beta u^{q - 1} \rd u$.

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:


 * $\ds a b \le \int_0^a t^{p - 1} \rd t + \int_0^b u^{q - 1} \rd u = \frac {a^p} p + \frac {b^q} q$

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.