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Difference between revisions of "Power Mean Inequality"

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M(t)=
 
M(t)=
 
\begin{cases}
 
\begin{cases}
−
\prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\
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\displaystyle\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}
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\left(\displaystyle\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise}
 
\end{cases}.
 
\end{cases}.
 
</cmath>
 
</cmath>

Revision as of 19:12, 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 \[M(t)= \begin{cases} \displaystyle\prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\ \left(\displaystyle\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise} \end{cases}.\]

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

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