Difference between revisions of "Power Mean Inequality"

Revision as of 11:16, 30 July 2020

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

Inequality

For $n$ positive real numbers $a_i$ and $n$ positive real weights $w_i$ with sum $\sum_{i=1}^n w_i=1$ , define the function $M:\mathbb{R}\rightarrow\mathbb{R}$ with $M(t)= \begin{cases} \prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\ \left(\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise} \end{cases}.$

The Power Mean Inequality states that for all real numbers $k_1$ and $k_2$ , $M(k_1)\ge M(k_2)$ if $k_1>k_2$ . In particular, for nonzero $k_1$ and $k_2$ , and equal weights (i.e. $w_i=\frac{1}{n}$ ), if $k_1>k_2$ , then $\left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_1} \right)^{\frac{1}{k_1}} \ge \left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_2} \right)^{\frac{1}{k_2}}$

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

Proof

We prove by cases:

1. $M(t)\ge M(0)\ge M(-t)$ for $t>0$

2. $M(k_1)\ge M(k_2)$ for $k_1 \ge k_2$ with $k_1k_2>0$

Case 1:

Note that $\begin{align*} && \left(\sum_{i=1}^n w_ia_i^{t} \right)^{\frac{1}{t}} &\ge \prod_{i=1}^n a_i^{w_i} \\ \Longleftarrow && \frac{1}{t} \ln\left( \sum_{i=1}^n w_i a_i^{t} \right) &\ge \sum_{i=1}^n w_i \ln{a_i} && \text{as } e^x \text{ is increasing} \\ \Longleftarrow && \ln\left( \sum_{i=1}^n w_i a_i^{t} \right) &\ge \sum_{i=1}^n w_i \ln{a_i^t} && \text{as } t>0 \end{align*}$ As $\ln(x)$ is concave, by Jensen's Inequality, the last inequality is true, proving $M(t)\ge M(0)$ . By replacing $t$ by $-t$ , the last inequality implies $M(0)\ge M(-t)$ as the inequality signs flip after multiplication by $-\frac{1}{t}$ .

Case 2:

For $k_1\ge k_2>0$ , $\begin{align} && \left(\sum_{i=1}^n w_ia_i^{k_1} \right)^{\frac{1}{k_1}} &\ge \left(\sum_{i=1}^n w_ia_i^{k_2} \right)^{\frac{1}{k_2}} \nonumber \\ \Longleftarrow && \left(\sum_{i=1}^n w_ia_i^{k_1} \right)^{\frac{k_2}{k_1}} &\ge \sum_{i=1}^n w_ia_i^{k_2} \label{eq} \end{align}$ As the function $f(x)=x^{\frac{k_2}{k_1}}$ is concave for all $x > 0$ , by Jensen's Inequality, $\left(\sum_{i=1}^n w_i a_i^{k_1} \right)^{\frac{k_2}{k_1}} = f\left(\sum_{i=1}^n w_i a_i^{k_1} \right) \geq \sum_{i=1}^n w_i f\left(a_i^{k_1}\right) =\sum_{i=1}^n w_i a_{i}^{k_2}$ For $0>k_1\ge k_2$ , the inequality sign in $(1)$ is flipped, but $f(x)$ becomes convex as $|k_1|\le |k_2|$ , and thus the inequality sign when applying Jensen's Inequality is also flipped.

@@ Line 21: / Line 21: @@
 We prove by cases:
-. <math>M(t)\ge M(-t)</math> for <math>t>0</math>
+. <math>M(t)\ge M(0)\ge M(-t)</math> for <math>t>0</math>
-. <math>M(t)\ge M(0)\ge M(-t)</math> for <math>t>0</math>
+. <math>M(k_1)\ge M(k_2)</math> for <math>k_1 \ge k_2</math> with <math>k_1k_2>0</math>
-. <math>M(k_1)\ge M(k_2)</math> for <math>k_1 \ge k_2</math> with <math>k_1k_2>0</math>
 Case 1:
-Note that
-<cmath>
-\begin{align*}
-&& \left(\sum_{i=1}^n w_ia_i^{t} \right)^{\frac{1}{t}} &\ge \left(\sum_{i=1}^n w_i a_i^{-t} \right)^{\frac{1}{-t}} \\
-\Longleftarrow && \sum_{i=1}^n w_i a_i^{t} &\ge \left( \sum_{i=1}^n w_ia_i^{-t} \right)^{-1} && \text{as } t>0\\
-\Longleftarrow && \left(\sum_{i=1}^n w_i a_i^{t}\right)\left(\sum_{i=1}^n w_i a_i^{-t}\right) &\ge 1 && \text{as }\sum_{i=1}^n w_ia_i^{-t} > 0
-\end{align*}
-</cmath>
-By [[Cauchy-Schwarz Inequality]],
-<cmath>
-\left(\sum_{i=1}^n w_i a_i^{t}\right)\left(\sum_{i=1}^n w_i a_i^{-t}\right) \ge \left( \sum_{i=1}^n\sqrt{w_ia_i^t}\sqrt{w_ia_i^{-t}} \right)^2
-= \left( \sum_{i=1}^n w_i \right)^2
-= 1
-</cmath>
-This concludes case 1.
-Case 2:
 Note that
@@ Line 58: / Line 38: @@
-Case 3:
+Case 2:
 For <math>k_1\ge k_2>0</math>,

Page

Toolbox

Search

Difference between revisions of "Power Mean Inequality"

Revision as of 11:16, 30 July 2020

Inequality

Proof