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