# Young's Inequality for Products

## 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:

 $$\displaystyle ab$$ $$=$$ $$\displaystyle \exp \left({\ln\left({ab}\right)}\right)$$ Exponential of Natural Logarithm $$\displaystyle$$ $$=$$ $$\displaystyle \exp \left({ \ln a + \ln b }\right)$$ Sum of Logarithms $$\displaystyle$$ $$=$$ $$\displaystyle \exp \left({ \frac 1 p p \ln a + \frac 1 q q \ln b }\right)$$ Definitions of Multiplicative Identity and Multiplicative Inverse $$\displaystyle$$ $$=$$ $$\displaystyle \exp \left({ \frac 1 p \ln \left( {a^p} \right) + \frac 1 q \ln \left( {b^q} \right) }\right)$$ Logarithms of Powers $$\displaystyle$$ $$\le$$ $$\displaystyle \frac 1 p \exp \left( {\ln \left( {a^p} \right)} \right) + \frac 1 q \exp \left( {\ln \left( {b^q} \right)} \right)$$ Exponential is Strictly Increasing and Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{a^p} p + \frac{b^q} q$$ Exponential of Natural Logarithm

$\blacksquare$

## Proof 2

The blue colored region corresponds to $\displaystyle \int_0^\alpha t^{p-1} \mathrm d t$ and the red colored region to $\displaystyle \int_0^\beta u^{q-1} \mathrm d u$.

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

 $$\displaystyle \frac 1 p + \frac 1 q$$ $$=$$ $$\displaystyle 1$$ $$\displaystyle \implies$$ $$\displaystyle p + q$$ $$=$$ $$\displaystyle p q$$ multiplying both sides by $p q$ $$\displaystyle \implies$$ $$\displaystyle p + q - p - q + 1$$ $$=$$ $$\displaystyle p q - p - q + 1$$ adding $1 - p - q$ to both sides $$\displaystyle \implies$$ $$\displaystyle 1$$ $$=$$ $$\displaystyle \left({p - 1}\right) \left({q - 1}\right)$$ elementary algebra $$\displaystyle \implies$$ $$\displaystyle \frac 1 {p - 1}$$ $$=$$ $$\displaystyle q - 1$$

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:

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

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.