Difference between revisions of "Power Mean Inequality"
(fixed latex) |
AlcumusGuy (talk | contribs) m (Fixed LaTeX.) |
||
| Line 6: | Line 6: | ||
Algebraically, <math>k_1\ge k_2</math> implies that | Algebraically, <math>k_1\ge k_2</math> implies that | ||
<cmath> | <cmath> | ||
| − | \sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n | + | \sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n}} |
</cmath> | </cmath> | ||
which can be written more concisely as | which can be written more concisely as | ||
<cmath> | <cmath> | ||
| − | \sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n | + | \sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n}} |
</cmath> | </cmath> | ||
Revision as of 22:40, 18 April 2015
The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality.
Inequality
For real numbers 
and positive real numbers 
, 
implies the 
th power mean is greater than or equal to the 
th.
Algebraically, implies that
![\[\sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n}}\]](http://latex.artofproblemsolving.com/5/1/c/51ca907a73ebeaf6f520c77f7281c0c0b5f662ec.png)
which can be written more concisely as
![\[\sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n}}\]](http://latex.artofproblemsolving.com/6/f/b/6fb4cbf03b94f3e494719f49c0cf2504c93add08.png)
The Power Mean Inequality follows from the fact that 
(where 
is the 
th power mean) together with Jensen's Inequality.
This article is a stub. Help us out by expanding it.
