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Difference between revisions of "Hlder's inequality"

(Created page with 'The '''Hölder's Inequality,''' a generalization of the '''Cauchy-Schwarz inequality''', states that, For all <math>a_i, b_i > 0 , p,q > 0</math> such that <math>\frac {1}{p} \pl…')
 
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The '''Hölder's Inequality,''' a generalization of the '''Cauchy-Schwarz inequality''', states that,
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''See also: [[Hölder's Inequality]]''
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For all <math>a_i, b_i > 0 , p,q > 0</math> such that <math>\frac {1}{p} \plus{} \frac {1}{q} \equal{} 1,</math> we have:<br>
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<math>\sum_{i \equal{} 1}^n a_ib_i\leq \left(\sum_{i \equal{} 1}^n a_i^p\right)^{\frac {1}{p}}\left(\sum _{i \equal{} 1}^n b_i^q\right)^{\frac {1}{q}}.</math>
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'''Hölder's Inequality,''' a generalization of the '''Cauchy-Schwarz inequality''', states that,
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For all <math>a_i, b_i > 0 , p,q > 0</math> such that <math>\frac {1}{p}+ \frac {1}{q} =1,</math> we have:<br>
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<math>\sum_{i =1}^n a_ib_i\leq \left(\sum_{i=1}^n a_i^p\right)^{\frac {1}{p}}\left(\sum _{i =1}^n b_i^q\right)^{\frac {1}{q}}.</math>
 
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Letting <math>p=q=2</math> in this inequality leads to the Cauchy-Schwarz Inequality.
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Letting <math>p=q=2</math> in this inequality leads to the [[Cauchy-Schwarz]] Inequality.
 
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<br>
 
This can also be generalized further to <math> n</math> sets of variables with a similar form.
 
This can also be generalized further to <math> n</math> sets of variables with a similar form.

Latest revision as of 11:12, 29 October 2016

See also: Hölder's Inequality

Hölder's Inequality, a generalization of the Cauchy-Schwarz inequality, states that, For all $a_i, b_i > 0 , p,q > 0$ such that $\frac {1}{p}+ \frac {1}{q} =1,$ we have:
$\sum_{i =1}^n a_ib_i\leq \left(\sum_{i=1}^n a_i^p\right)^{\frac {1}{p}}\left(\sum _{i =1}^n b_i^q\right)^{\frac {1}{q}}.$
Letting $p=q=2$ in this inequality leads to the Cauchy-Schwarz Inequality.
This can also be generalized further to $n$ sets of variables with a similar form.

Applications

1. Given $a_i, b_i>0; \ \ i=1,2,\cdots, n$ we have,
$\frac{a_1^k}{b_1}+\frac{a_2^k}{b_2}+\cdots+\frac{a_n^k}{b_n}\geq \frac{\left(a_1+\cdots+a_n\right)^k}{n^{k-2}\cdot\left(b_1+\cdots+b_n\right)}.$

2. Power-mean inequality: For $a_1, a_2,\cdots,a_n>0$ and $k\geq l,$ we have
$\sqrt[k]{\frac{a_1^k+\cdots+a_n^k}{n}}\geq \sqrt[l]{\frac{a_1^l+\cdots+a_n^l}{n}}.$

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