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Difference between revisions of "Minkowski Inequality"

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Minkowski Inequality states:
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The '''Minkowski Inequality''' states that if <math>r>s</math> are nonzero real numbers, then for any positive numbers <math>a_{ij}</math> the following holds:
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<math>\left(\sum_{j=1}^{m}\left(\sum_{i=1}^{n}a_{ij}^r\right)^{s/r}\right)^{1/s}\geq \left(\sum_{i=1}^{n}\left(\sum_{j=1}^{m}a_{ij}^s\right)^{r/s}\right)^{1/r}</math>
  
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Let <math>r>s</math> be a nonzero real number, then for any positive numbers <math>a_{ij}</math>, the following inequality holds:
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Notice that if either <math>r</math> or <math>s</math> is zero, the inequality is equivalent to [[Hölder's Inequality]].
  
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<math>(\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij}^r)^{s/r})^{1/s}\geq (\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij}^s)^{r/s})^{1/r}</math>
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== Equivalence with the standard form ==
  
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Notice that if one of <math>r,s</math> is zero, the inequality is equivelant to [[Holder's Inequality]].
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For <math>r>s>0</math>, putting <math>x_{ij}:=a_{ij}^s</math> and <math>p:=\frac rs>1</math>, the symmetrical form given above becomes
  
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<center><math> \sum_{j=1}^{m}\biggl(\sum_{i=1}^{n}x_{ij}^p\biggr)^{1/p}
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\geq\left(\sum_{i=1}^{n}\biggl(\sum_{j=1}^{m}x_{ij}\biggr)^p\right)^{1/p}</math>.</center>
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Putting <math>m=2</math> and <math>a_i:=x_{i1},b_i:=x_{i2}</math>, we get the form in which the Minkowski Inequality is given most often:
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<center><math>\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p}
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\geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}</math></center>
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As the latter can be iterated, there is no loss of generality by putting <math>m=2</math>.
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== Problems ==
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432791#432791 AIME 1991 Problem 15]
  
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[[Category:Algebra]]
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[[Category:Inequalities]]

Latest revision as of 15:49, 29 December 2021

The Minkowski Inequality states that if $r>s$ are nonzero real numbers, then for any positive numbers $a_{ij}$ the following holds: $\left(\sum_{j=1}^{m}\left(\sum_{i=1}^{n}a_{ij}^r\right)^{s/r}\right)^{1/s}\geq \left(\sum_{i=1}^{n}\left(\sum_{j=1}^{m}a_{ij}^s\right)^{r/s}\right)^{1/r}$

Notice that if either $r$ or $s$ is zero, the inequality is equivalent to Hölder's Inequality.

Equivalence with the standard form

For $r>s>0$, putting $x_{ij}:=a_{ij}^s$ and $p:=\frac rs>1$, the symmetrical form given above becomes

$\sum_{j=1}^{m}\biggl(\sum_{i=1}^{n}x_{ij}^p\biggr)^{1/p} \geq\left(\sum_{i=1}^{n}\biggl(\sum_{j=1}^{m}x_{ij}\biggr)^p\right)^{1/p}$.

Putting $m=2$ and $a_i:=x_{i1},b_i:=x_{i2}$, we get the form in which the Minkowski Inequality is given most often:

$\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p} \geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}$

As the latter can be iterated, there is no loss of generality by putting $m=2$.

Problems

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