# Power Mean Inequality

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