# Reverse Young's Inequality for Products

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## 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:

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

Then:

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

$\blacksquare$

## Source of Name

This entry was named for William Henry Young.