Difference between revisions of "Power Mean Inequality"

(Proof)
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We prove by cases:
 
We prove by cases:
  
1. <math>M(t)\ge M(-t)</math> for <math>t>0</math>
+
1. <math>M(t)\ge M(0)\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>
+
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>
 
 
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:
 
Case 1:
 
Note that
 
<cmath>
 
\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>
 
By [[Cauchy-Schwarz Inequality]],
 
<cmath>
 
\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>
 
This concludes case 1.
 
 
Case 2:
 
  
 
Note that
 
Note that
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Case 3:
+
Case 2:
  
 
For <math>k_1\ge k_2>0</math>,
 
For <math>k_1\ge k_2>0</math>,

Revision as of 11:16, 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. $M(t)\ge M(0)\ge M(-t)$ 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 $0>k_1\ge k_2$, 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.