Power Mean Inequality

Revision as of 05:05, 13 April 2017 by Original user name (talk | contribs) (Added proof)

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 \[\sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n}}\]

which can be written more concisely as \[\sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n}}\]

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

Proof

\[\left(\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\right)^{\frac{1}{k_1}}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{1}{k_2}}\] Raising both sides to the $k_1$th power, we get \[\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}\]

We can see that the function $f(x)=x^{\frac{k_1}{k_2}}$ is convex for all $x \> 0$, and so we can apply Jensen's Inequality. Therefore, \[\left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}= f\left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)\le \sum_{i=1}^n f\left(\frac{a_i}{n}\right)= \sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\] This article is a stub. Help us out by expanding it.