# 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:

$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$

Equality occurs if and only if:

$b = a^{p - 1}$

## Proof by Convexity

The result follows directly if $a = 0$ or $b = 0$.

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

Consider:

$x := \map \ln {a^p}$
$y := \map \ln {b^q}$
$\alpha := p^{-1}$
$\beta := q^{-1}$

Note that by hypothesis:

$\alpha + \beta = 1$

Thus by definition of strictly convex real function:

$(1): \quad x \ne y \implies \map \exp {\alpha x + \beta y} < \alpha \map \exp x + \beta \map \exp y$

On the other hand:

$(2): \quad x = y \implies \map \exp {\alpha x + \beta y} = \alpha \map \exp x + \beta \map \exp y$

since:

$\map \exp {\alpha x + \beta y} = \map \exp x$

and:

$\alpha \map \exp x + \beta \map \exp y = \alpha \map \exp x + \beta \map \exp x = \map \exp x$

Therefore:

 $\ds a b$ $=$ $\ds \map \exp {\map \ln {a b} }$ 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 $\ds$ $\le$ $\ds \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {\map \ln {b^q} }$ by $(1)$ and $(2)$ $\ds$ $=$ $\ds \frac {a^p} p + \frac {b^q} q$ Exponential of Natural Logarithm

By $(1)$ and $(2)$, the equality:

$a b = \dfrac {a^p} p + \dfrac {b^q} q$

occurs if and only if:

$\map \ln {a^p} = \map \ln {b^q}$

That is, if and only if:

$b = a^{p - 1}$

$\blacksquare$

## Geometric Proof In the above diagram:

the $\color {blue} {\text {blue} }$ region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
the $\color {red} {\text {red} }$ region corresponds to $\ds \int_0^\beta u^{q - 1} \rd u$.

From Positive Real Numbers whose Reciprocals Sum to 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 \ \$ $\ds 1$ $=$ $\ds \paren {p - 1} \paren {q - 1}$ simplifying $\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 still hold.

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

$\Box$

It remains to prove the equality condition:

$a b = \dfrac {a^p} p + \dfrac {b^q} q$
$b = a^{p - 1}$

### Necessary Condition

Let $b = a^{p - 1}$.

Then:

 $\ds \frac {a^p} p + \frac {b^q} q$ $=$ $\ds \frac {a^p} p + \frac {\paren {a^{p - 1} }^q} q$ by hypothesis $\ds$ $=$ $\ds \frac {a^p} p + \frac {\paren {a^{p/q} }^q} q$ by hypothesis: $\dfrac 1 p + \dfrac 1 q = 1$ $\ds$ $=$ $\ds a^p \paren {\frac 1 p + \frac 1 q}$ simplifying $\ds$ $=$ $\ds a \cdot a^{p - 1}$ by hypothesis: $\dfrac 1 p + \dfrac 1 q = 1$ $\ds$ $=$ $\ds a b$ by hypothesis

That is:

$b = a^{p - 1} \implies a b = \dfrac {a^p} p + \dfrac {b^q} q$

$\Box$

### Sufficient Condition

Let:

$a b = \dfrac {a^p} p + \dfrac {b^q} q$

Aiming for a contradiction, suppose $b \ne a^{p - 1}$.

#### Case $1$: $b < a^{p - 1}$

Referring to the diagram above, let us identify:

$a = \alpha$
$b = \beta$

Thus $b < a^{p - 1}$ is precisely what is diagrammed in the image above.

We note that the sum of the $\color {blue} {\text {blue} }$ and $\color {red} {\text {red} }$ regions strictly exceeds the area of the rectangle contained by $\alpha$ and $\beta$.

Hence we have that:

$a b < \dfrac {a^p} p + \dfrac {b^q} q$

#### Case $2$: $b > a^{p - 1}$

We draw a diagram similar to the one above, although integrated with respect to $u$ rather than $t$.

This shows that:

$b \ne a^{p - 1} \implies a b \ne \dfrac {a^p} p + \dfrac {b^q} q$

These two cases together contradict our assertion that $a b = \dfrac {a^p} p + \dfrac {b^q} q$.

$a b = \dfrac {a^p} p + \dfrac {b^q} q \implies b = a^{p - 1}$

$\blacksquare$

## Proof by Calculus

Without loss of generality assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$.

Define $f : \hointr 0 \infty \to \R$ by:

$\ds \map f t = \frac {t^p} p + \frac 1 q - t$

We have, from Derivative of Power and Sum Rule for Derivatives:

$\ds \map {f'} t = t^{p - 1} - 1$

So for $t \ge 1$ we have:

$\ds \map {f'} t \ge 0$

So, from Real Function with Positive Derivative is Increasing, we have:

$f$ is increasing on $\hointr 1 \infty$.

That is:

$\map f t \ge \map f 1$

for all $t \ge 1$.

Since $a^p \ge b^q$, we have:

$a^p b^{-q} \ge 1$

so:

$a b^{-q/p} \ge 1$

So:

$\map f {a b^{-q/p} } \ge \map f 1$

That is:

$\ds \frac {a^p b^{-q} } p + \frac 1 q - a b^{-q/p} \ge \frac 1 p + \frac 1 q - 1$

By hypothesis, we have:

$\ds \frac 1 p + \frac 1 q = 1$

So it follows that:

$\ds \frac {a^p b^{-q} } p + \frac 1 q \ge a b^{-q/p}$

So:

$\ds \frac {a^p} p + \frac {b^q} q \ge a b^{q \paren {1 - 1/p} }$

Finally, we have:

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

so:

$\ds q \paren {1 - \frac 1 p} = 1$

giving:

$\ds \frac {a^p} p + \frac {b^q} q \ge a b$

For the equality case, note that:

$\map {f'} t > 0$

for $t > 1$.

So, from Real Function with Strictly Positive Derivative is Strictly Increasing, we have:

$f$ is strictly increasing for $t > 1$.

So:

$\map f {a b^{-q/p} } = \map f 1$

that is:

$\ds a b = \frac {a^p} p + \frac {b^q} q$
$a b^{-q/p} = 1$

That is:

$b = a^{p/q}$

We have:

$\ds \frac 1 p + \frac 1 q = 1$

and so:

$\ds 1 + \frac p q = p$

giving:

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

So we have:

$b = a^{p - 1}$

as required.

$\blacksquare$

## Parameter Inequalities

Statements of Young's Inequality for Products will commonly insist that $p, q > 1$.

However, from Positive Real Numbers whose Reciprocals Sum to 1 we have that if:

$p, q > 0$

and:

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

it follows directly that $p, q > 1$.

## Source of Name

This entry was named for William Henry Young.