Young's Inequality

Young's inequality is a special case of the weighted AM-GM inequality. It is very useful in real analysis, including as a tool to prove Hölder's inequality. It is also a special case of a more general inequality known as Young's inequality for increasing functions.

Let $p,q$ be positive real numbers satisfying $\frac1{p} + \frac1{q} = 1.$ Then if $a,b$ are nonnegative real numbers, $$ab \le \frac{a^p}{p} + \frac{b^q}{q},$$ and equality holds if and only if $a^p=b^q.$

Proof 1 (using the weighted AM-GM inequality): Let $w_1 = \frac1p, w_2 = \frac1q.$ Then the weighted AM-GM inequality says that $$\frac{w_1 a^p + w_2 b^q}{w_1+w_2} \ge \big((a^p)^{w_1} (b^q)^{w_2}\big)^{1/(w_1+w_2)},$$ but $w_1+w_2=1,$ so the left side is $\frac{a^p}{p} + \frac{b^q}{q}$ and the right side is $(a^p)^{1/p} (b^q)^{1/q} = ab.$ The statement that equality holds if and only if $a^p=b^q$ also follows directly from the statement of the weighted AM-GM inequality.

Proof 2 (using logarithms): It suffices to prove the theorem for $a,b>0.$ The function $f(x) = \ln(x)$ is concave down, so for all positive $x,y$ and $t \in (0,1),$ $$\ln\big(tx+(1-t)y\big) \ge t\ln(x)+(1-t)\ln(y),$$ with equality holding if and only if $x=y.$

Now set $x=a^p, y = b^q, t=\frac1p$ $\big($so $1-t = \frac1q\big)$ and exponentiate both sides. $_\square$

The case $p=q=2$ is just the AM-GM inequality for $a^2,b^2$: $$ab \le \frac{a^2}2 + \frac{b^2}2.$$

Let $n$ be a positive integer. Find the minimum value of $f(x) = x+nx^{-1/n}$ for positive real numbers $x.$

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This could be done by AM-GM $\big($on $x$ and $n$ copies of $x^{-1/n}\big),$ but Young's inequality works as well: divide by $n+1$ to get $$\frac{f(x)}{n+1} = \frac{x}{n+1} + \frac{x^{-1/n}}{1+\frac1n},$$ and note that $p = n+1,$ $q = 1+\frac1n$ satisfy $\frac1{p} + \frac1{q} = 1,$ which suggests multiplying by $x^n$: $$\frac{x^n f(x)}{n+1} = \frac{x^{n+1}}{n+1} + \frac{x^{(n^2-1)/n}}{1+\frac1n}.$$ Now let $y = x^{n-1}.$ Then $$\frac{x^nf(x)}{n+1} = \frac{x^{n+1}}{n+1} + \frac{y^{1+1/n}}{1+\frac1n} \ge xy = x^n,$$ so $f(x) \ge n+1,$ and in fact equality holds when $x^{n+1} = x^{(n^2-1)/n},$ or $x=1.$ $_\square$

As mentioned in the introduction, Young's inequality is essential in the proof of Hölder's inequality; see the wiki for details.

Young's inequality for products is a special case of Young's inequality for increasing functions:

Let $f(x)$ be a continuous, increasing function defined for nonnegative real numbers $x,$ with $f(0)=0.$ Suppose $a,b$ are positive real numbers such that $a$ is in the domain of $f$ and $b$ is in the image of $f.$ Then $$ab \le \int_0^a f(x) \, dx + \int_0^b f^{-1}(x) \, dx,$$ with equality if and only if $f(a)=b.$

The proof is quite elegant: The shaded areas contain the rectangle. ^[1] This is the picture for $b < f(a);$ the picture for $b > f(a)$ is similar. Equality only holds if there is no extra area, which is when $b=f(a).$

Let $f(x) = x^{p-1}.$ Then $\int_0^a f(x) \, dx = \frac{a^p}{p}$ and $f^{-1}(x) = x^{1/(p-1)},$ so $$\int_0^b f^{-1}(x) \, dx = \frac{b^{p/(p-1)}}{\frac{p}{p-1}},$$ and $\frac{p}{p-1}$ is the $q$ of Young's inequality, so we recover Young's inequality: $$ab \le \frac{a^p}{p} + \frac{b^q}{q}.$$

1. Duke, N. Young Inequality. Retrieved July 28, 2011, from https://commons.wikimedia.org/wiki/File:Young_inequality.svg