Difference between revisions of "Power Mean Inequality"

Revision as of 14:50, 28 August 2018

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.

@@ Line 24: / Line 24: @@
 </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,
+We can see that the function <math>f(x)=x^{\frac{k_2}{k_1}}</math> is concave 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)\leq \sum_{i=1}^n \frac{1}{n}f\left(a_i^{k_2}\right)= \sum_{i=1}^n \frac{a_{i}^{k_1}}{n}
+\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}
 </cmath>
 {{stub}}
 [[Category:Inequality]]
 [[Category:Theorems]]

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Difference between revisions of "Power Mean Inequality"

Revision as of 14:50, 28 August 2018

Inequality

Proof