# Young's Inequality for Products

From ProofWiki

## Contents

## Theorem

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

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

Then, for any $a, b \in \R_{\ge 0}$:

- $a b \le \dfrac {a^p} p + \dfrac{b^q} q$

Equality occurs if and only if $b=a^{p-1}$.

## Proof 1

The result is obvious if $a=0$ or $b=0$, so assume WLOG that $a > 0$ and $b > 0$.

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle ab\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \exp \left({\ln\left({ab}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | Exponential of Natural Logarithm | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \exp \left({ \ln a + \ln b }\right)\) | \(\displaystyle \) | \(\displaystyle \) | Sum of Logarithms | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \exp \left({ \frac 1 p p \ln a + \frac 1 q q \ln b }\right)\) | \(\displaystyle \) | \(\displaystyle \) | Definitions of Multiplicative Identity and Multiplicative Inverse | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \exp \left({ \frac 1 p \ln \left( {a^p} \right) + \frac 1 q \ln \left( {b^q} \right) }\right)\) | \(\displaystyle \) | \(\displaystyle \) | Logarithms of Powers | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 p \exp \left( {\ln \left( {a^p} \right)} \right) + \frac 1 q \exp \left( {\ln \left( {b^q} \right)} \right)\) | \(\displaystyle \) | \(\displaystyle \) | Exponential is Strictly Increasing and Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac{a^p} p + \frac{b^q} q\) | \(\displaystyle \) | \(\displaystyle \) | Exponential of Natural Logarithm |

$\blacksquare$

## Proof 2

In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$.

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac 1 p + \frac 1 q\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle p + q\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle p q\) | \(\displaystyle \) | \(\displaystyle \) | multiplying both sides by $p q$ | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle p + q - p - q + 1\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle p q - p - q + 1\) | \(\displaystyle \) | \(\displaystyle \) | adding $1 - p - q$ to both sides | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle 1\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({p - 1}\right) \left({q - 1}\right)\) | \(\displaystyle \) | \(\displaystyle \) | elementary algebra | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \frac 1 {p - 1}\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle q - 1\) | \(\displaystyle \) | \(\displaystyle \) |

Accordingly:

- $u = t^{p-1} \iff t = u^{q-1}$

Let $a, b$ be any positive real numbers.

Since $a b$ is the area of the rectangle in the given figure, we have:

- $\displaystyle a b \le \int_0^a t^{p-1} \ \mathrm d t + \int_0^b u^{q-1} \ \mathrm d u = \frac {a^p} p + \frac {b^q} q$

Note that even if the graph intersected the side of the rectangle corresponding to $t = a$, this inequality would hold.

Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially.

$\blacksquare$

## Source of Name

This entry was named for William Henry Young.

## Sources

- Walter Rudin:
*Principles of Mathematical Analysis*(1953): Exercise $6.10 a$