Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Power Mean Inequality"

(→‎Proof)
(→‎Inequality)
Line 1: Line 1:
 
The '''Power Mean Inequality''' is a generalized form of the multi-variable [[Arithmetic Mean-Geometric Mean]] Inequality.
 
The '''Power Mean Inequality''' is a generalized form of the multi-variable [[Arithmetic Mean-Geometric Mean]] Inequality.
  
−
== Inequality ==
+
== Description ==
−
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
+
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>, the power mean function <math>M:\mathbb{R}\rightarrow\mathbb{R}</math> is defined by
 
<cmath>
 
<cmath>
 
M(t)=
 
M(t)=
Line 11: Line 11:
 
</cmath>
 
</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
+
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=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}}
+
\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>  
 +
Considering the limiting behavior, we also have <math>\lim_{t\rightarrow +\infty} M(t)=\max\{x_i\}</math>, <math>\lim_{t\rightarrow -\infty} M(t)=\min\{x_i\}</math> and <math>\lim_{t\rightarrow 0} M(t)= M(0)</math>.
  
 
The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> together with [[Jensen's Inequality]].
 
The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> together with [[Jensen's Inequality]].

Revision as of 10:51, 30 July 2020

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

Description

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

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. $w_i=1/n$), if $k_1>k_2$, then \[\left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_1} \right)^{\frac{1}{k_1}} \ge \left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_2} \right)^{\frac{1}{k_2}}.\] Considering the limiting behavior, we also have $\lim_{t\rightarrow +\infty} M(t)=\max\{x_i\}$, $\lim_{t\rightarrow -\infty} M(t)=\min\{x_i\}$ and $\lim_{t\rightarrow 0} M(t)= M(0)$.

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 $0>k_1\ge k_2$, $f(x)$ becomes convex as $|k_1|\le |k_2|$, so the inequality sign when applying Jensen's Inequalitythe inequality sign is flipped. Thus, the inequality sign in $(1)$ is also flipped, but as $k_2<0$, $x^\frac{1}{k_2}$ is a decreasing function, so the inequality sign is flipped again, resulting in $M(k_1)\ge M(k_2)$ as desired.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Power_Mean_Inequality&oldid=129820"