# Young's Inequality for Products

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## 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 follows directly if $a = 0$ or $b = 0$.

Without loss of generality, assume that $a > 0$ and $b > 0$.

Then:

 $\ds a b$ $=$ $\ds \map \exp {\map \ln {a b} }$ Exponential of Natural Logarithm $\ds$ $=$ $\ds \map \exp {\ln a + \ln b}$ Sum of Logarithms $\ds$ $=$ $\ds \map \exp {\frac 1 p p \ln a + \frac 1 q q \ln b}$ Definition of Multiplicative Identity and Definition of Multiplicative Inverse $\ds$ $=$ $\ds \map \exp {\frac 1 p \map \ln {a^p} + \frac 1 q \map \ln {b^q} }$ Logarithms of Powers $\ds$ $\le$ $\ds \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {\map \ln {b^q} }$ Exponential is Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$ $\ds$ $=$ $\ds \frac {a^p} p + \frac {b^q} q$ Exponential of Natural Logarithm

$\blacksquare$

## Proof 2 The blue colored region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$ and the red colored region to $\ds \int_0^\beta u^{q - 1} \rd u$.

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

 $\ds \frac 1 p + \frac 1 q$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds p + q$ $=$ $\ds p q$ multiplying both sides by $p q$ $\ds \leadsto \ \$ $\ds p + q - p - q + 1$ $=$ $\ds p q - p - q + 1$ adding $1 - p - q$ to both sides $\ds \leadsto \ \$ $\ds 1$ $=$ $\ds \paren {p - 1} \paren {q - 1}$ elementary algebra $\ds \leadsto \ \$ $\ds \frac 1 {p - 1}$ $=$ $\ds q - 1$

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:

$\ds a b \le \int_0^a t^{p - 1} \rd t + \int_0^b u^{q - 1} \rd u = \frac {a^p} p + \frac {b^q} q$

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.