Inequalities with weights

Many of the inequalities we have looked at so far have versions in which the terms in a mean can be weighted unequally.


Weighted AM-GM inequality:
If $x_1,\dots,x_n>0$ and $w_1,\dots,w_n \ge 0$and $w_1+\dots+w_n=1$, then

\begin{displaymath}w_1 x_1 + w_2 x_2+\dots+ w_n x_n \ge x_1^{w_1} x_2^{w_2} \dots x_n^{w_n},\end{displaymath}

with equality if and only if all the xi with $w_i \not=0$ are equal.


One recovers the usual AM-GM inequality by taking equal weights $w_1=w_2=\dots=w_n=1/n$.


Weighted power mean inequality:
Fix $x_1,\dots,x_n>0$ and weights $w_1,\dots,w_n \ge 0$with $w_1+\dots+w_n=1$. For any nonzero real number r, define the r-th (weighted) power mean by the formula

\begin{displaymath}P_r := \left( \frac{w_1 x_1^r + \dots + w_n x_n^r}{n} \right)^{1/r}.\end{displaymath}

Also let P0 be the weighted geometric mean (using the same weights):

\begin{displaymath}P_0 := x_1^{w_1} \dots x_n^{w_n}.\end{displaymath}

Then Pr is an increasing function of $r \in {\mathbb R}$. Moreover, if the xi with $w_i \not=0$ are not all equal, then Pr is a strictly increasing function of r.


Weighted Jensen's Inequality:
Let f be a convex function on an interval I. If $x_1,\dots,x_n \in I$, $w_1,\dots,w_n \ge 0$and $w_1+\dots+w_n=1$, then

\begin{displaymath}w_1 f(x_1)+\dots+ w_n f(x_n) \ge f\left(w_1 x_1 + \dots + w_n x_n\right).\end{displaymath}

If moreover f is strictly convex, then equality holds if and only if all the xi with $w_i \not=0$ are equal.