Difference between revisions of "Power Mean Inequality"

m (Inequality)
(Inequality)
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>
\left( \frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n} \right) ^ {\frac{1}{k_1}}\ge \left( \frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n} \right) ^ {\frac{1}{k_2}}
+
\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} \right)}
 +
</cmath>
 +
 
 +
which can be written more concisely as
 +
<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} \right)}
 
</cmath>  
 
</cmath>  
  

Revision as of 21:12, 11 January 2014

The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality.

Inequality

For real numbers $k_1,k_2$ and positive real numbers $a_1, a_2, \ldots, a_n$, $k_1\ge k_2$ implies the $k_1$th power mean is greater than or equal to the $k_2$th.

Algebraically, $k_1\ge k_2$ implies that

\[\sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n} \right)}\] (Error compiling LaTeX. Unknown error_msg)

which can be written more concisely as

\[\sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n} \right)}\] (Error compiling LaTeX. Unknown error_msg)

The Power Mean Inequality follows from the fact that $\frac{\partial M(t)}{\partial t}\geq 0$ (where $M(x)$ is the $t$th power mean) together with Jensen's Inequality.

This article is a stub. Help us out by expanding it.