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 $r\ne0$ , 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 $$ M_w^r(x_1,x_2,\ldots,x_n)=(w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{1/r}. $$ 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 $$ M_w^r(x_1,x_2,\ldots,x_n)\le M_w^s(x_1,x_2,\ldots,x_n) $$ 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 $1\le i\le n$ ; this implies $y_i^t=x_i^s$ . Consider the function \begin{eqnarray*} f\colon(0,\infty)&\to&(0,\infty)\\ x&\mapsto&x^t. \end{eqnarray*}Since $t>1$ , the second derivative of $f$ satisfies $f''(x)=t(t-1)x^{t-2}>0$ for all $x>0$ , so $f$ is a strictly convex function. Therefore, according to Jensen's inequality, \begin{eqnarray*} (w_1y_1+w_2y_2+\cdots+w_ny_n)^t&=&f(w_1y_1+w_2y_2+\cdots+w_ny_n)\\ &\le&w_1f(y_1)+w_2f(y_2)+\cdots+w_nf(y_n)\\ &=&w_1y_1^t+w_2y_2^t+\cdots+w_ny_n^t, \end{eqnarray*}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 $$ (w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{s/r}\le w_1x_1^s+w_2x_2^s+\cdots+w_nx_n^s $$ 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: $$ (w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{1/r}\le (w_1x_1^s+w_2x_2^s+\cdots+w_nx_n^s)^{1/s}, $$ 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 $M_w^r\le M_w^s$ 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 $M_w^r\le M_w^0$ for all $r<0$ . We conclude that for all real numbers $r$ and $s$ such that $r<s$ , $$ M_w^r(x_1,x_2,\ldots,x_n)\le M_w^s(x_1,x_2,\ldots,x_n). $$