# If larger values have larger weights, then (weighted AM)>(simple AM)

If larger values have larger weights then show that for $n$ observations $x_1,x_2,...,x_n$ with respective weights $w_1,w_2,...,w_n$,
$\overline{x_w}>\overline{x}$, where $\overline{x_w}=\frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}$ and $\overline{x}=\frac{\sum_{i=1}^nx_i}{n}$

The result is somewhat clear intuitively but I am not being able to show it analytically. I was thinking of using Chebyshev's sum inequality to solve this.

Let $x_1<x_2<...<x_n$, then by the problem it follows that $w_1<w_2<...<w_n$.

Then does it follow from Chebyshev's inequality that $\frac{1}{n}\sum_{i=1}^nw_ix_i>(\frac{1}{n}\sum_{i=1}^nx_i)(\frac{1}{n}\sum_{i=1}^nw_i)$

$\Rightarrow \frac{1}{n}\sum_{i=1}^nw_ix_i>\frac{1}{n}\sum_{i=1}^nx_i$ ?

You are almost there; the weighted mean is $$\frac{\sum w_ix_i}{\sum w_i}$$. Just divide both sides by $\sum w_i$ and you have the result you want.