Difference between revisions of "Power Mean Inequality"
(→Proof) |
Durianaops (talk | contribs) |
||
Line 2: | Line 2: | ||
== Inequality == | == Inequality == | ||
− | For | + | For <math>n</math> positive real numbers <math>a_i</math> and <math>n</math> positive real weights <math>w_i</math> with sum <math>\sum_{i=1}^n w_i=1</math>, define the function <math>M:\mathbb{R}\rightarrow\mathbb{R}</math> with |
+ | <cmath> | ||
+ | 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}. | ||
+ | </cmath> | ||
− | + | The Power Mean Inequality states that for all real numbers <math>k_1</math> and <math>k_2</math>, <math>M(k_1)\ge M(k_2)</math> if <math>k_1>k_2</math>. In particular, for nonzero <math>k_1</math> and <math>k_2</math>, and equal weights (i.e. <math>w_i=\frac{1}{n}</math>), if <math>k_1>k_2</math>, then | |
<cmath> | <cmath> | ||
− | \ | + | \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}} |
</cmath> | </cmath> | ||
− | + | The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> together with [[Jensen's Inequality]]. | |
− | < | ||
− | |||
− | </ | ||
− | |||
== Proof == | == Proof == | ||
+ | We prove by cases: | ||
+ | |||
+ | 1. <math>M(t)\ge M(-t)</math> for <math>t>0</math> | ||
+ | |||
+ | 2. <math>M(t)\ge M(0)\ge M(-t)</math> for <math>t>0</math> | ||
+ | |||
+ | 3. <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> | <cmath> | ||
− | \left(\sum_{i=1}^n \frac{ | + | \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> | </cmath> | ||
− | + | By [[Cauchy-Schwarz Inequality]], | |
<cmath> | <cmath> | ||
− | \sum_{i=1}^n \ | + | \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> | </cmath> | ||
+ | This concludes case 1. | ||
+ | |||
+ | Case 2: | ||
− | + | Note that | |
<cmath> | <cmath> | ||
− | \left(\sum_{i=1}^n | + | \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*} | ||
</cmath> | </cmath> | ||
− | {{ | + | As <math>\ln(x)</math> is concave, by [[Jensen's Inequality]], the last inequality is true, proving <math>M(t)\ge M(0)</math>. By replacing <math>t</math> by <math>-t</math>, the last inequality implies <math>M(0)\ge M(-t)</math> as the inequality signs flip after multiplication by <math>-\frac{1}{t}</math>. |
+ | |||
+ | |||
+ | Case 3: | ||
+ | |||
+ | For <math>k_1\ge k_2>0</math>, | ||
+ | <cmath> | ||
+ | \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} | ||
+ | </cmath> | ||
+ | As the function <math>f(x)=x^{\frac{k_2}{k_1}}</math> is concave for all <math>x > 0</math>, by [[Jensen's Inequality]], | ||
+ | <cmath> | ||
+ | \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} | ||
+ | </cmath> | ||
+ | For <math>0>k_1\ge k_2</math>, the inequality sign in <math>(1)</math> is flipped, but <math>f(x)</math> becomes convex as <math>|k_1|\le |k_2|</math>, and thus the inequality sign when applying Jensen's Inequality is also flipped. | ||
+ | |||
[[Category:Inequality]] | [[Category:Inequality]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Revision as of 10:10, 30 July 2020
The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality.
Inequality
For positive real numbers and positive real weights with sum , define the function with
The Power Mean Inequality states that for all real numbers and , if . In particular, for nonzero and , and equal weights (i.e. ), if , then
The Power Mean Inequality follows from the fact that together with Jensen's Inequality.
Proof
We prove by cases:
1. for
2. for
3. for with
Case 1:
Note that By Cauchy-Schwarz Inequality, This concludes case 1.
Case 2:
Note that As is concave, by Jensen's Inequality, the last inequality is true, proving . By replacing by , the last inequality implies as the inequality signs flip after multiplication by .
Case 3:
For , As the function is concave for all , by Jensen's Inequality, For , the inequality sign in is flipped, but becomes convex as , and thus the inequality sign when applying Jensen's Inequality is also flipped.