Difference between revisions of "Power Mean Inequality"
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The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> (where <math>M(t)</math> is the <math>t</math>th power mean) together with [[Jensen's Inequality]]. | The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> (where <math>M(t)</math> is the <math>t</math>th power mean) together with [[Jensen's Inequality]]. | ||
| + | == Proof == | ||
| + | <cmath> | ||
| + | \left(\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\right)^{\frac{1}{k_1}}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{1}{k_2}} | ||
| + | </cmath> | ||
| + | Raising both sides to the <math>k_1</math>th power, we get | ||
| + | <cmath> | ||
| + | \sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}} | ||
| + | </cmath> | ||
| + | We can see that the function <math>f(x)=x^{\frac{k_1}{k_2}}</math> is convex for all <math>x \> 0</math>, and so we can apply [[Jensen's Inequality]]. Therefore, | ||
| + | <cmath> | ||
| + | \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}= f\left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)\le \sum_{i=1}^n f\left(\frac{a_i}{n}\right)= \sum_{i=1}^n \frac{a_{i}^{k_1}}{n} | ||
| + | </cmath> | ||
{{stub}} | {{stub}} | ||
[[Category:Inequality]] | [[Category:Inequality]] | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 04:05, 13 April 2017
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.
Proof
![\[\left(\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\right)^{\frac{1}{k_1}}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{1}{k_2}}\]](http://latex.artofproblemsolving.com/0/5/9/059dcc989c9c5b8bdd5c2388a1aec7ba518b8948.png)
Raising both sides to the th power, we get
![\[\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}\]](http://latex.artofproblemsolving.com/1/2/2/122c41965c1491041edd9c9426eba00acf1922bd.png)
We can see that the function 
is convex for all 
, and so we can apply Jensen's Inequality. Therefore,
![\[\left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}= f\left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)\le \sum_{i=1}^n f\left(\frac{a_i}{n}\right)= \sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\]](http://latex.artofproblemsolving.com/8/6/a/86a1eb018c8dabb433cc9229b690d9dd960a609e.png)
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