# Hölder's Inequality

**Hölder's inequality** is a statement about sequences that generalizes the Cauchy-Schwarz inequality to multiple sequences and different exponents.

## Definition

Hölder's inequality states that, for sequences \(\{a_i\}, \{b_i\}, \ldots, \{z_i\}\), the inequality

\[(a_1+a_2+\cdots+a_n)^{\lambda_a}\cdots(z_1+z_2+\cdots+z_n)^{\lambda_z} \geq a_1^{\lambda_a}b_1^{\lambda_b}\cdots z_1^{\lambda_z}+\cdots+a_n^{\lambda_a}b_n^{\lambda_b}\cdots z_n^{\lambda_z}\]

holds for all \(\lambda_a+\lambda_b+\cdots+\lambda_z=1\). For instance, in the case of \(\lambda_a=\lambda_b=\frac{1}{2}\), Hölder's inequality reduces to

\[(a_1+a_2+\cdots+a_n)^{\frac{1}{2}}(b_1+b_2+\cdots+b_n)^{\frac{1}{2}} \geq (a_1b_1)^{\frac{1}{2}}+(a_2b_2)^{\frac{1}{2}}+\cdots+(a_nb_n)^{\frac{1}{2}},\]

which is the Cauchy-Schwarz inequality.

## Proof

Hölder's inequality is a consequence of Young's inequality, which states that if \(p,q\) are positive reals satisfying \(\frac{1}{p}+\frac{1}{q}=1\), then

\[ab \leq \frac{a^p}{p}+\frac{b^q}{q}\]

for all nonnegative reals \(a,b\).

Young's inequality follows from the concavity of the logarithm function; in particular,

\[\log\left(\frac{1}{p}a^{p}+\frac{1}{q}b^q\right) \geq \frac{1}{p}\log a^p+\frac{1}{q}\log b^q=\log a+\log b=\log ab\]

and exponentiating gives Young's inequality.

## Strategies and Applications

Hölder's inequality is often used to deal with square (or higher power) roots of expressions in inequalities, since those can be eliminated through successive multiplication. For instance,

## Let \(a,b,c\) be positive reals satisfying \(a+b+c=3\). What is the minimum possible value of

\[\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}?\]

By Hölder's inequality,

\[\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)(a+b+c) \geq (1+1+1)^3.\]

Since \(a+b+c=3\), the minimum possible value of \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\) is \(3\). \(_\square\)

(2001 IMO, Problem 2)

Let \(a,b,c\) be positive real numbers. Prove that \[\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1.\]

By Hölder's inequality,

\[\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}.\]

Thus it suffices to show that \[(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc,\] which follows immediately from the famous inequality (incidentally, easily provable with Hölder's) \[(a+b)(b+c)(c+a)\ge 8abc.\ _\square\]

A cycling strategy can also be used, as in the demonstration that Hölder's inequality implies AM-GM:

By Hölder's inequality,

\[(a_1+a_2+\ldots+a_n)^{\frac{1}{n}}(a_2+a_3+\cdots+a_1)^{\frac{1}{n}}\cdots(a_n+a_1+\cdots+a_{n-1})^{\frac{1}{n}} \geq n\sqrt[n]{a_1a_2\cdots a_n},\]

which rearranges to AM-GM immediately. \(_\square\)

Another useful strategy is to insert constants (especially 1) as members of a sequence, especially to "reduce" powers. For instance,

Let \(a,b\) be positive real numbers. Show that

\[4\left(a^3+b^3\right) \geq (a+b)^3.\]

By Hölder's inequality,

\[\left(a^3+b^3\right)^{\frac{1}{3}}(1+1)^{\frac{1}{3}}(1+1)^{\frac{1}{3}} \geq \left(a^3\right)^{\frac{1}{3}}1^{\frac{1}{3}}1^{\frac{1}{3}}+\left(b^3\right)^{\frac{1}{3}}1^{\frac{1}{3}}1^{\frac{1}{3}},\]

which rearranges to the desired inequality. \(_\square\)

## Minkowski's inequality

Minkowski's inequality easily follows from Holder's inequality.

Minkowski's inequalitystates that\[\left(\sum _{ n=1 }^{ k } ({ x }_{ n }+{ y }_{ n })^p\right)^{ \frac { 1 }{ p } }\le \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right) ^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }\]

for \(p>1\) and \({x}_{n},{y}_{n}\ge0.\) \(_\square\)

We know

\[\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p }=\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }({ x }_{ n }+{ y }_{ n })^{ p-1 } } } +\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }({ x }_{ n }+{ y }_{ n } } )^{ p-1 }.\]

Let's define \(a\) as \(a=\frac { p }{ p-1 } ,\) then by Holder's inequality we have

\[\sum_{n \mathop = 1}^k x_n \left({x_n + y_n}\right)^{p-1} + \sum_{n \mathop = 1}^k y_n \left({x_n + y_n}\right)^{p-1} \le \left[ \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }\right] \left[ \sum _{ n=1 }^{ k }{ ({ x }_{ k }+{ y }_{ k })^{ p } } \right]^{ \frac { 1 }{ a } }.\]

Dividing by \(\left[\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } \right]^{ \frac { 1 }{ a }}, \) we get

\[\left(\sum _{ n=1 }^{ k } ({ x }_{ n }+{ y }_{ n })^p\right)^{ \frac { 1 }{ p } }\le \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right) ^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }.\] Hence proved. \(_\square\)

## See Also

**Cite as:**Hölder's Inequality.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/holders-inequality/