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Difference between revisions of "Power Mean Inequality"

(→‎Proof)
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== Inequality ==
 
== Inequality ==
−
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.
+
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>
  
−
Algebraically, <math>k_1\ge k_2</math> implies that
+
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>
−
\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}}
+
\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>  
  
−
which can be written more concisely as
+
The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> together with [[Jensen's Inequality]].
−
<cmath>
 
−
\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}}
 
−
</cmath>  
 
  
−
The Power Mean Inequality follows from the fact that <math>\frac{\partial M(t)}{\partial t}\geq 0</math> (where <math>M(t)</math> is the <math>t</math>th power mean) 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{a_{i}^{k_1}}{n}\right)^{\frac{1}{k_1}}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{1}{k_2}}
+
\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>
−
Raising both sides to the <math>k_1</math>th power, we get
+
By [[Cauchy-Schwarz Inequality]],
 
<cmath>
 
<cmath>
−
\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\ge \left(\sum_{i=1}^n \frac{a_{i}^{k_2}}{n}\right)^{\frac{k_1}{k_2}}
+
\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:
  
−
We can see that the function <math>f(x)=x^{\frac{k_2}{k_1}}</math> is concave for all <math>x > 0</math>, and so we can apply [[Jensen's Inequality]]. Therefore,
+
Note that
 
<cmath>
 
<cmath>
−
\left(\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\right)^{\frac{k_2}{k_1}}= f\left(\sum_{i=1}^n \frac{a_{i}^{k_1}}{n}\right)\geq \sum_{i=1}^n \frac{1}{n}f\left(a_i^{k_1}\right)= \sum_{i=1}^n \frac{a_{i}^{k_2}}{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>
−
{{stub}}
+
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 $n$ positive real numbers $a_i$ and $n$ positive real weights $w_i$ with sum $\sum_{i=1}^n w_i=1$, define the function $M:\mathbb{R}\rightarrow\mathbb{R}$ with \[M(t)= \begin{cases} \prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\ \left(\sum_{i=1}^n w_ia_i^t \right)^{\frac{1}{t}} &\text{otherwise} \end{cases}.\]

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. $w_i=\frac{1}{n}$), if $k_1>k_2$, then \[\left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_1} \right)^{\frac{1}{k_1}} \ge \left( \frac{1}{n} \sum_{i=1}^n a_{i}^{k_2} \right)^{\frac{1}{k_2}}\]

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. $M(t)\ge M(-t)$ for $t>0$

2. $M(t)\ge M(0)\ge M(-t)$ for $t>0$

3. $M(k_1)\ge M(k_2)$ 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 \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*} By Cauchy-Schwarz Inequality, \[\left(\sum_{i=1}^n w_i a_i^{t}\right)\left(\sum_{i=1}^n w_i a_i^{-t}\right) \ge \left( \sum_{i=1}^n\sqrt{w_ia_i^t}\sqrt{w_ia_i^{-t}} \right)^2 = \left( \sum_{i=1}^n w_i \right)^2 = 1\] This concludes case 1.

Case 2:

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 3:

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 $0>k_1\ge k_2$, the inequality sign in $(1)$ is flipped, but $f(x)$ becomes convex as $|k_1|\le |k_2|$, and thus the inequality sign when applying Jensen's Inequality is also flipped.

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