The weighted mean results in smaller value than the simple mean. 
If weights attached to larger items are smaller and those attached to smaller items are larger, then the weighted mean results in smaller value than the simple mean.

My try: Let the weights be $w_1<w_2<\dots<w_n $ and let the items be $x_1>x_2>\dots>x_n$,
Weighted mean: $$\frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}$$ and mean :$$\frac{\sum_{i=1}^nx_i}{n}.$$ How to show that $$\frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}<\frac{\sum_{i=1}^nx_i}{n}?$$
 A: 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 
$$ \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} < \frac{\sum_{i=1}^nx_i}{n}.$$
A: I'm posting two answers because one (this one) is very general, and the other only works nicely when the weights are positive.
Rewrite the desired inequality through cross multiplication as
$$n \sum_{i=1}^nw_ix_i < \sum_{i,\ j=1}^nw_ix_j.$$
On the right, every $w_i$ is paired with each $x_j$ exactly once. If we allow the indices to cycle for convenience, so that $x_{n+k} = x_k$, then we can rewrite this as
$$\sum_{j = 1}^n \left(\sum_{i=1}^nw_ix_i\right) < \sum_{j=1}^n\left(\sum_{i = 1}^nw_ix_{i+j}\right).$$
This slight reordering of the terms makes it clear that we can use the rearrangement inequality $n$ times to compare the $n$ parentheses-enclosed summands on the left with the $n$ on the right. The left is the minimal ordering, according to the Rearrangement inequality, so each of the $n$ copies on the left are strictly less than all but one of the summands on the right (and equal to one). Thus we get the desired equality.

Note that we implicitly require that $\sum_i w_i > 0$, since otherwise cross-multiplying as we did above flips the inequality and the statement breaks. This is because replacing $w_i$ by $-w_i$ does not change the sum on the left, as
$$ \frac{\sum_{i=1}^n(-1)w_ix_i}{\sum_{i=1}^n(-1)w_i} = \frac{-\sum_{i=1}^nw_ix_i}{-\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i},$$
but it does flip the ordering of the $w_i$ so that now we get the maximal weighting according to the rearrangement inequality. In other words, very small changes show that if $\sum_i w_i < 0$, then we have 
$$\sum_{j = 1}^n \left(\sum_{i=1}^nw_ix_i\right) > \sum_{j=1}^n\left(\sum_{i = 1}^nw_ix_{i+j}\right).$$
A quick example to confirm this suspicion is as follows: Take $\{x_i\} = 5,3$ and $\{w_i\} = -5, -3$, so that $x_1 > x_2$ and $w_1 < w_2$. But then the weighted mean is $\frac{34}{8} > 4$ and the arithmetic mean is $4$.
