Difference between revisions of "Minkowski Inequality"

(Equivalence with the standard form)
(Equivalence with the standard form)
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<math>\biggl(\sum_{i=1}^{n}a_i^p\biggr)^{1/p}+ \biggl(\sum_{i=1}^{n}b_i^p\biggr)^{1/p}
 
<math>\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}\biggl(a_i+b_i\biggr)^p\right)^{1/p}</math>
+
\geq\left(\sum_{i=1}^{n}\Bigl(a_i+b_i\Bigr)^p\right)^{1/p}</math>
  
 
As the latter can be iterated, there is no loss of generality by putting <math>m=2</math> .
 
As the latter can be iterated, there is no loss of generality by putting <math>m=2</math> .

Revision as of 13:59, 12 November 2010

The Minkowski Inequality states that if $r>s$ is a nonzero real number, 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 Holder's Inequality.

Equivalence with the standard form

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

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

Put $m=2, a_i:=x_{i1},b_i:=x_{i2}$ and 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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