Young's Inequality for Products/Geometric Proof

Proof

 * Holder's Ineq.jpg

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

From Positive Real Numbers whose Reciprocals Sum to 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 still hold.

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

It remains to prove the equality condition:
 * $a b = \dfrac {a^p} p + \dfrac {b^q} q$


 * $b = a^{p - 1}$
 * $b = a^{p - 1}$

Necessary Condition
Let $b = a^{p - 1}$.

Then:

That is:
 * $b = a^{p - 1} \implies a b = \dfrac {a^p} p + \dfrac {b^q} q$

Sufficient Condition
Let:
 * $a b = \dfrac {a^p} p + \dfrac {b^q} q$

$b \ne a^{p - 1}$.

Case $1$: $b < a^{p - 1}$
Referring to the diagram above, let us identify:
 * $a = \alpha$
 * $b = \beta$

Thus $b < a^{p - 1}$ is precisely what is diagrammed in the image above.

We note that the sum of the $\color {blue} {\text {blue} }$ and $\color {red} {\text {red} }$ regions strictly exceeds the area of the rectangle contained by $\alpha$ and $\beta$.

Hence we have that:
 * $a b < \dfrac {a^p} p + \dfrac {b^q} q$

Case $2$: $b > a^{p - 1}$
We draw a diagram similar to the one above, although integrated $u$ rather than $t$.

This shows that:
 * $b \ne a^{p - 1} \implies a b \ne \dfrac {a^p} p + \dfrac {b^q} q$

These two cases together contradict our assertion that $a b = \dfrac {a^p} p + \dfrac {b^q} q$.

Hence by Proof by Contradiction:
 * $a b = \dfrac {a^p} p + \dfrac {b^q} q \implies b = a^{p - 1}$