Young's Inequality for Products/Proof by Calculus

Theorem

Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:

$\dfrac 1 p + \dfrac 1 q = 1$

Then:

$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$

Equality occurs if and only if:

$b = a^{p - 1}$

Proof

Without loss of generality 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:

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

if and only if:

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

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.

$\blacksquare$