# 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, 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, $r<0, and $0. 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 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,

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