Young's Inequality for Products/Proof by Calculus

Proof
assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$.

Define $f : \hointr 0 \infty \to \R$ by:


 * $\ds \map f t = \frac {t^p} p + \frac 1 q - t$

We have, from Derivative of Power and Sum Rule for Derivatives:


 * $\ds \map {f'} t = t^{p - 1} - 1$

So for $t \ge 1$ we have:


 * $\ds \map {f'} t \ge 0$

So, from Real Function with Positive Derivative is Increasing, we have:


 * $f$ is increasing on $\hointr 1 \infty$.

That is:


 * $\map f t \ge \map f 1$

for all $t \ge 1$.

Since $a^p \ge b^q$, we have:


 * $a^p b^{-q} \ge 1$

so:


 * $a b^{-q/p} \ge 1$

So:


 * $\map f {a b^{-q/p} } \ge \map f 1$

That is:


 * $\ds \frac {a^p b^{-q} } p + \frac 1 q - a b^{-q/p} \ge \frac 1 p + \frac 1 q - 1$

By hypothesis, we have:


 * $\ds \frac 1 p + \frac 1 q = 1$

So it follows that:


 * $\ds \frac {a^p b^{-q} } p + \frac 1 q \ge a b^{-q/p}$

So:


 * $\ds \frac {a^p} p + \frac {b^q} q \ge a b^{q \paren {1 - 1/p} }$

Finally, we have:


 * $\ds 1 - \frac 1 p = \frac 1 q$

so:


 * $\ds q \paren {1 - \frac 1 p} = 1$

giving:


 * $\ds \frac {a^p} p + \frac {b^q} q \ge a b$

For the equality case, note that:


 * $\map {f'} t > 0$

for $t > 1$.

So, from Real Function with Strictly Positive Derivative is Strictly Increasing, we have:


 * $f$ is strictly increasing for $t > 1$.

So:


 * $\map f {a b^{-q/p} } = \map f 1$

that is:


 * $\ds a b = \frac {a^p} p + \frac {b^q} q$




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

That is:


 * $b = a^{p/q}$

We have:


 * $\ds \frac 1 p + \frac 1 q = 1$

and so:


 * $\ds 1 + \frac p q = p$

giving:


 * $\ds \frac p q = p - 1$

So we have:


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

as required.