Young's Inequality for Products/Proof by Convexity

Proof
The result follows directly if $a = 0$ or $b = 0$.

, assume that $a > 0$ and $b > 0$.

Recall Exponential is Strictly Convex.

Consider:
 * $x := \map \ln {a^p}$
 * $y := \map \ln {b^q}$
 * $\alpha := p^{-1}$
 * $\beta := q^{-1}$

Note that :
 * $\alpha + \beta = 1$

Thus by definition of strictly convex real function:
 * $(1): \quad x \ne y \implies \map \exp {\alpha x + \beta y} < \alpha \map \exp x + \beta \map \exp y$

On the other hand:
 * $(2): \quad x = y \implies \map \exp {\alpha x + \beta y} = \alpha \map \exp x + \beta \map \exp y$

since:
 * $\map \exp {\alpha x + \beta y} = \map \exp x$

and:
 * $\alpha \map \exp x + \beta \map \exp y = \alpha \map \exp x + \beta \map \exp x = \map \exp x$

Therefore:

By $(1)$ and $(2)$, the equality:
 * $a b = \dfrac {a^p} p + \dfrac {b^q} q$

occurs :
 * $\map \ln {a^p} = \map \ln {b^q}$

That is, :
 * $b = a^{p - 1}$