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"

(→‎Inequality)
m (→‎Inequality)
 
(5 intermediate revisions by 4 users not shown)
Line 10: Line 10:
 
\end{cases}.
 
\end{cases}.
 
</cmath>
 
</cmath>
 +
 +
(<math>M(0)</math> is the [[AM-GM_Inequality#Weighted_AM-GM_Inequality|weighted geometric mean]].)
  
 
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
 
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
Line 56: Line 58:
 
=\sum_{i=1}^n w_i a_{i}^{k_2}
 
=\sum_{i=1}^n w_i a_{i}^{k_2}
 
</cmath>
 
</cmath>
−
For <math>0>k_1\ge k_2</math>, <math>f(x)</math> becomes convex as <math>|k_1|\le |k_2|</math>, so the inequality sign when applying Jensen's Inequalitythe inequality is flipped. Thus, the inequality sign in <math>(1)</math> is flipped, but as <math>k_2<0</math>, <math>x^\frac{1}{k_2}</math> is a decreasing function, so the inequality sign is flipped again after applying <math>x^{\frac{1}{k_2}}</math>, resulting in <math>M(k_1)\ge M(k_2)</math> as desired.
+
For <math>0>k_1\ge k_2</math>, <math>f(x)</math> becomes convex as <math>|k_1|\le |k_2|</math>, so the inequality sign when applying Jensen's Inequality is flipped. Thus, the inequality sign in <math>(1)</math> is flipped, but as <math>k_2<0</math>, <math>x^\frac{1}{k_2}</math> is a decreasing function, the inequality sign is flipped again after applying <math>x^{\frac{1}{k_2}}</math>, resulting in <math>M(k_1)\ge M(k_2)</math> as desired.
  
−
[[Category:Inequality]]
+
[[Category:Algebra]]
−
[[Category:Theorems]]
+
[[Category:Inequalities]]

Latest revision as of 12:57, 23 August 2024

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 $\sum_{i=1}^n w_i=1$, the power mean with exponent $t$, where $t\in\mathbb{R}$, is defined by \[M(t)= \begin{cases} \prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\ \left(\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise} \end{cases}.\]

($M(0)$ is the weighted geometric mean.)

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 $\lim_{t\rightarrow +\infty} M(t)=\max\{a_i\}$, $\lim_{t\rightarrow -\infty} M(t)=\min\{a_i\}$ and $\lim_{t\rightarrow 0} M(t)= M(0)$.

The Power Mean Inequality follows from 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 are flipped 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 Inequality is flipped. Thus, the inequality sign in $(1)$ is flipped, but as $k_2<0$, $x^\frac{1}{k_2}$ is a decreasing function, the inequality sign is flipped again after applying $x^{\frac{1}{k_2}}$, resulting in $M(k_1)\ge M(k_2)$ as desired.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Power_Mean_Inequality&oldid=225600"