# Reverse Young's Inequality for Products

## Theorem

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

$\dfrac 1 p - \dfrac 1 q = 1$

Let $a \in \R_{\ge 0}$ be a positive real number.

Let $b \in \R_{> 0}$ be a strictly positive real number.

Then:

$a b \ge \dfrac {a^p} p - \dfrac {b^{-q} } q$

## Proof

Define:

 $\text {(1)}: \quad$ $\ds u$ $=$ $\ds \frac 1 p$ $\ds v$ $=$ $\ds \frac q p$ $\ds x$ $=$ $\ds \paren {a b}^p$ $\ds y$ $=$ $\ds b^{-p}$

Then:

 $\ds \frac 1 u + \frac 1 v$ $=$ $\ds p + \frac p q$ $\ds$ $=$ $\ds p \paren {1 + \frac 1 q}$ $\ds$ $=$ $\ds p \paren {\frac 1 p}$ because $\dfrac 1 p - \dfrac 1 q = 1$, by hypothesis $\ds$ $=$ $\ds 1$

Thus Young's Inequality for Products can be applied:

 $\ds x y$ $\le$ $\ds \frac {x^u} u + \frac {y^v} v$ $\ds \leadsto \ \$ $\ds \paren {a b}^p b^{-p}$ $\le$ $\ds \frac {\paren {\paren {a b}^p}^{1/p} } {1/p} + \frac {\paren {b^{-p} }^{q/p} } {q/p}$ substituting for $u, v, x, y$ from $(1)$ above $\ds \leadsto \ \$ $\ds a^p$ $\le$ $\ds p a b + p \frac {b^{-q} } q$ algebraic simplification $\ds \leadsto \ \$ $\ds \frac {a^p} p$ $\le$ $\ds a b + \frac {b^{-q} } q$ $\ds \leadsto \ \$ $\ds a b$ $\ge$ $\ds \frac {a^p} p - \frac {b^{-q} } q$

$\blacksquare$

## Source of Name

This entry was named for William Henry Young.