Suppose $w_i>0$ are normalized weights, and $\sum_{i}^{n}w_i x_i>\sum_{i}^{n}w_i y_i > \frac{1}{n}\sum_{i}^{n} y_i$. Does this imply the following:
$$\sum_{i=1}^{n}x_i>\sum_{i=1}^{n}y_i$$.
This is my trial. Suppose $w_n = \sup\{w_1,...,w_n\}$. Then,
$$\sum_{i=1}^{n}w_n x_i=w_n\sum_{i=1}^{n} x_i>\sum_{i=1}^{n}w_i x_i>\sum_{i=1}^{n}w_i y_i>\sum_{i=1}^{n} y_i.$$
Because all weights are strictly positive and normalized between 0 and 1, then the following should hold:
$$\sum_{i=1}^{n}x_i>w_n\sum_{i=1}^{n} x_i>\sum_{i=1}^{n}y_i.$$
Is my solution correct?
This is not true.
Take $n=2$, $x_1=1$, $x_2=10$, $y_1=3$, $y_2=9$, $w_1=0.1$ and $w_2=0.9$.
Then $ \displaystyle \sum_{i}^{n}w_i x_i = 9.1$ ; $ \displaystyle \sum_{i}^{n}w_i y_i = 8.4$ ; and $\dfrac{1}{n} \displaystyle \sum_{i}^{n} y_i = 6$, so one has $$ \displaystyle \sum_{i}^{n}w_i x_i > \sum_{i}^{n}w_i y_i > \dfrac{1}{n} \sum_{i}^{n} y_i $$
but $\displaystyle \sum_{i}^{n} x_i =11$ and $\displaystyle \sum_{i}^{n} y_i =12$.