# Reverse Young's Inequality for Products

## Theorem

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

$\displaystyle \frac 1 p - \frac 1 q = 1$

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

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

Then:

$\displaystyle a b \ge \frac {a^p} p - \frac {b^{-q}} q$

## Proof

Define:

 $(1):\quad$ $\displaystyle u$ $=$ $\displaystyle \frac 1 p$ $\displaystyle v$ $=$ $\displaystyle \frac q p$ $\displaystyle x$ $=$ $\displaystyle \left({a b}\right)^p$ $\displaystyle y$ $=$ $\displaystyle b^{-p}$

Then:

 $\displaystyle \frac 1 u + \frac 1 v$ $=$ $\displaystyle p + \frac p q$ $\displaystyle$ $=$ $\displaystyle p \left({1 + \frac 1 q}\right)$ $\displaystyle$ $=$ $\displaystyle p \left({\frac 1 p}\right)$ because $\dfrac 1 p - \dfrac 1 q = 1$, by hypothesis $\displaystyle$ $=$ $\displaystyle 1$

Thus Young's Inequality for Products can be applied:

 $\displaystyle x y$ $\le$ $\displaystyle \frac {x^u} u + \frac {y^v} v$ $\displaystyle \implies \ \$ $\displaystyle \left({a b}\right)^p b^{-p}$ $\le$ $\displaystyle \frac {\left({\left({a b}\right)^p}\right)^{1/p} } {1/p} + \frac {\left({b^{-p} }\right)^{q/p} } {q/p}$ substituting for $u, v, x, y$ from $(1)$ above $\displaystyle \implies \ \$ $\displaystyle a^p$ $\le$ $\displaystyle pab + p \frac { b^{-q} } q$ algebraic simplification $\displaystyle \implies \ \$ $\displaystyle \frac {a^p} p$ $\le$ $\displaystyle ab + \frac { b^{-q} } q$ $\displaystyle \implies \ \$ $\displaystyle ab$ $\ge$ $\displaystyle \frac {a^p} p - \frac { b^{-q} } q$

$\blacksquare$

## Source of Name

This entry was named for William Henry Young.