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Difference between revisions of "Power Mean Inequality"

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== Inequality ==
 
== Inequality ==
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For [[real number]]s <math>k_1,k_2</math> and [[positive]] real numbers <math>a_1, a_2, \ldots, a_n</math>, <math>k_1\ge k_2</math> implies the <math>k_1</math>th [[power mean]] is greater than or equal to the <math>k_2</math>th.
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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 with exponent <math>t</math>, where <math>t\in\mathbb{R}</math>, is defined by
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<cmath>
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M(t)=
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\begin{cases}
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\prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\
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\left(\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise}
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\end{cases}.
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</cmath>
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(<math>M(0)</math> is the [[AM-GM_Inequality#Weighted_AM-GM_Inequality|weighted geometric mean]].)
  
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Algebraically, <math>k_1\ge k_2</math> implies that
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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>
−
\sqrt[k_1]{\frac{a_{1}^{k_1}+a_{2}^{k_1}+\cdots +a_{n}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{a_{1}^{k_2}+a_{2}^{k_2}+\cdots +a_{n}^{k_2}}{n}}
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\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>  
  
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which can be written more concisely as
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Considering the limiting behavior, we also have <math>\lim_{t\rightarrow +\infty} M(t)=\max\{a_i\}</math>, <math>\lim_{t\rightarrow -\infty} M(t)=\min\{a_i\}</math> and <math>\lim_{t\rightarrow 0} M(t)= M(0)</math>.
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The Power Mean Inequality follows from [[Jensen's Inequality]].
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== Proof ==
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We prove by cases:
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1. <math>M(t)\ge M(0)\ge M(-t)</math> for <math>t>0</math>
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2. <math>M(k_1)\ge M(k_2)</math> for <math>k_1 \ge k_2</math> with <math>k_1k_2>0</math>
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Case 1:
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Note that
 
<cmath>
 
<cmath>
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\sqrt[k_1]{\frac{\sum\limits_{i=1}^n a_{i}^{k_1}}{n}}\ge \sqrt[k_2]{\frac{\sum\limits_{i=1}^n a_{i}^{k_2}}{n}}
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\begin{align*}
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</cmath>  
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&& \left(\sum_{i=1}^n w_ia_i^{t} \right)^{\frac{1}{t}} &\ge \prod_{i=1}^n a_i^{w_i} \\
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\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} \\
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\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
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\end{align*}
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</cmath>
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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 are flipped after multiplication by <math>-\frac{1}{t}</math>.
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 +
 
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Case 2:
  
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The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> (where <math>M(x)</math> is the <math>t</math>th power mean) together with [[Jensen's Inequality]].
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For <math>k_1\ge k_2>0</math>,
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<cmath>
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\begin{align}
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&& \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 \\
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\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}
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\end{align}
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</cmath>
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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]],
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<cmath>
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\left(\sum_{i=1}^n w_i a_i^{k_1} \right)^{\frac{k_2}{k_1}}
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= f\left(\sum_{i=1}^n w_i a_i^{k_1} \right)
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\geq \sum_{i=1}^n w_i f\left(a_i^{k_1}\right)
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=\sum_{i=1}^n w_i a_{i}^{k_2}
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</cmath>
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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.
  
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{{stub}}
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[[Category:Algebra]]
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[[Category:Inequality]]
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[[Category:Inequalities]]
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[[Category:Theorems]]
 

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.

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