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.

It remains to show that $ab=\frac{a^p}p+\frac{b^q}q$ if and only if $b=a^{p-1}$. One direction is nearly trivial: If $b=a^{p-1}$ then

The converse is less trivial, and we prove it by the contrapositive. So assume that $b\ne a^{p-1}$ and we will argue that $ab\ne \frac{a^p}p+\frac{b^q}q$. We separate this into cases, and begin with the case that $b< a^{p-1}$. Note that this is precisely what is diagrammed in the image above (identifying $a=\alpha,b=\beta$). Because the sum of the colored regions strictly exceeds the area of the rectangle, we have that $ab<\frac{a^p}p+\frac{b^q}q$. In the case that $b > a^{p-1}$ one draws a diagram similar to the one above, although integrated with respect to $u$ rather than $t$.

This shows that $b\ne a^{p-1}$ implies $ab\ne \frac{a^p}p+\frac{b^q}q$ which is the contrapositive of the claim that $ab=\frac{a^p}p+\frac{b^q}q$ implies $b=a^{p-1}$.