proof of general means inequality

Let $w_{1}$ , $w_{2}$ , …, $w_{n}$ be positive real numbers such that $w_{1}+w_{2}+\cdots+w_{n}=1$ . For any real number $r\neq 0$ , the weighted power mean of degree $r$ of $n$ positive real numbers $x_{1}$ , $x_{2}$ , …, $x_{n}$ (with respect to the weights $w_{1}$ , …, $w_{n}$ ) is defined as

$M_{w}^{r}(x_{1},x_{2},\ldots,x_{n})=(w_{1}x_{1}^{r}+w_{2}x_{2}^{r}+\cdots+w_{n% }x_{n}^{r})^{1/r}.$

The definition is extended to the case $r=0$ by taking the limit $r\to 0$ ; this yields the weighted geometric mean

$M_{w}^{0}(x_{1},x_{2},\ldots,x_{n})=x_{1}^{w_{1}}x_{2}^{w_{2}}\ldots x_{n}^{w_% {n}}$

(see derivation of zeroth weighted power mean). We will prove the weighted power means inequality, which states that for any two real numbers $r<s$ , the weighted power means of orders $r$ and $s$ of $n$ positive real numbers $x_{1}$ , $x_{2}$ , …, $x_{n}$ satisfy the inequality

$M_{w}^{r}(x_{1},x_{2},\ldots,x_{n})\leq M_{w}^{s}(x_{1},x_{2},\ldots,x_{n})$

with equality if and only if all the $x_{i}$ are equal.

First, let us suppose that $r$ and $s$ are nonzero. We distinguish three cases for the signs of $r$ and $s$ : $r<s<0$ , $r<0<s$ , and $0<r<s$ . Let us consider the last case, i.e. assume $r$ and $s$ are both positive; the others are similar. We write $t=\frac{s}{r}$ and $y_{i}=x_{i}^{r}$ for $1\leq i\leq n$ ; this implies $y_{i}^{t}=x_{i}^{s}$ . Consider the function

	$\displaystyle f\colon(0,\infty)$	$\displaystyle\to$	$\displaystyle(0,\infty)$
	$\displaystyle x$	$\displaystyle\mapsto$	$\displaystyle x^{t}.$

Since $t>1$ , the second derivative of $f$ satisfies $f^{\prime\prime}(x)=t(t-1)x^{t-2}>0$ for all $x>0$ , so $f$ is a strictly convex function. Therefore, according to Jensen’s inequality,

$\displaystyle(w_{1}y_{1}+w_{2}y_{2}+\cdots+w_{n}y_{n})^{t}$	$\displaystyle=$	$\displaystyle f(w_{1}y_{1}+w_{2}y_{2}+\cdots+w_{n}y_{n})$
	$\displaystyle\leq$	$\displaystyle w_{1}f(y_{1})+w_{2}f(y_{2})+\cdots+w_{n}f(y_{n})$
	$\displaystyle=$	$\displaystyle w_{1}y_{1}^{t}+w_{2}y_{2}^{t}+\cdots+w_{n}y_{n}^{t},$

with equality if and only if $y_{1}=y_{2}=\cdots=y_{n}$ . By substituting $t=\frac{s}{r}$ and $y_{i}=x_{i}^{r}$ back into this inequality, we get

$(w_{1}x_{1}^{r}+w_{2}x_{2}^{r}+\cdots+w_{n}x_{n}^{r})^{s/r}\leq w_{1}x_{1}^{s}% +w_{2}x_{2}^{s}+\cdots+w_{n}x_{n}^{s}$

with equality if and only if $x_{1}=x_{2}=\cdots=x_{n}$ . Since $s$ is positive, the function $x\mapsto x^{1/s}$ is strictly increasing, so raising both sides to the power $1/s$ preserves the inequality:

$(w_{1}x_{1}^{r}+w_{2}x_{2}^{r}+\cdots+w_{n}x_{n}^{r})^{1/r}\leq(w_{1}x_{1}^{s}% +w_{2}x_{2}^{s}+\cdots+w_{n}x_{n}^{s})^{1/s},$

which is the inequality we had to prove. Equality holds if and only if all the $x_{i}$ are equal.

If $r=0$ , the inequality is still correct: $M_{w}^{0}$ is defined as $\lim_{r\to 0}M_{w}^{r}$ , and since $M_{w}^{r}\leq M_{w}^{s}$ for all $r<s$ with $r\neq 0$ , the same holds for the limit $r\to 0$ . The same argument shows that the inequality also holds for $s=0$ , i.e. that $M_{w}^{r}\leq M_{w}^{0}$ for all $r<0$ . We conclude that for all real numbers $r$ and $s$ such that $r<s$ ,

$M_{w}^{r}(x_{1},x_{2},\ldots,x_{n})\leq M_{w}^{s}(x_{1},x_{2},\ldots,x_{n}).$

Title	proof of general means inequality
Canonical name	ProofOfGeneralMeansInequality
Date of creation	2013-03-22 13:10:26
Last modified on	2013-03-22 13:10:26
Owner	pbruin (1001)
Last modified by	pbruin (1001)
Numerical id	5
Author	pbruin (1001)
Entry type	Proof
Classification	msc 26D15
Related topic	ArithmeticMean
Related topic	GeometricMean
Related topic	HarmonicMean
Related topic	RootMeanSquare3
Related topic	PowerMean
Related topic	WeightedPowerMean
Related topic	ArithmeticGeometricMeansInequality
Related topic	JensensInequality