Ok. I'm posting two answers because one is very general, and the other (this one) only works nicely when the weights are positive - but I wrote this one first because I thought of it first.
Assume the weights are non-negative.
Notice that if I multiply all the weights by a fixed positive number $k$, then
$$ \frac{\sum_{i=1}^nkw_ix_i}{\sum_{i=1}^nkw_i} = \frac{k\sum_{i=1}^nw_ix_i}{k\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i},$$
which is to say that it doesn't change anything. It also doesn't affect the order. So let us assume that we have multiplied by the correct $k$ so that $\sum_i w_i = n$.
Notice also that if I add some constant $M$ to each $x_i$, then we have both
$$\frac{\sum_{i=1}^nw_i(x_i+M)}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} + \frac{\sum_{i=1}^nw_iM}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} + M,$$
and
$$\frac{\sum_{i=1}^n(x_i+M)}{n} = \frac{\sum_{i=1}^nx_i}{n} + M,$$
so that the difference between the weighted mean and the arithmetic mean is the same. So we can assume that all the $x_i$ are positive.
Then
$$ \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{n} < \frac{\sum_{i=1}^nw_1x_i}{\sum_{i=1}^nw_i} = \frac{w_1\sum_{i=1}^nx_i}{n},$$
and since the weights are positive, $w_1 < 1$. (If $w_1 = 1$ and the other weights are all $0$, then $x_1 + 0 + 0 + \ldots < x_1 + x_2 + \ldots$, so only the strictly positive case is interesting).
So we conclude that
$$ \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} < \frac{\sum_{i=1}^nx_i}{n}.$$
Best Answer
$$\sum_{i=1}^{n}w_ix_i-s=\sum_{i=1}^{n}w_ix_i-s\sum_{i=1}^{n}w_i=\sum_{i=1}^{n}w_i(x_i-s)\ge 0.$$