Take any two strictly positive vectors $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ and any vector $(\alpha_1,\dots,\alpha_n)$ such that $0<\alpha_i < 1$ for all $i=1,\dots,n$.
Consider:
$$A=\frac{\sum_i x_i \alpha_i y_i}{\sum_i x_i\alpha_i} $$
$$B=\frac{\sum_i x_i y_i}{\sum_i x_i} $$
$$C=\frac{\sum_i x_i (1-\alpha_i) y_i}{\sum_i x_i(1-\alpha_i)} $$
My conjecture is that B is always between A and C. That is, either $A\geq B \geq C$ or $C \geq B \geq A$.
Does anyone know how to prove this or can provide a counterexample?
[First edit: note that if $y_i=y_j$ for all $i,j$ then $A=B=C=y_i$ ; while if $x_i=x_j$ and $\alpha_i=\alpha_j$ for all $i,j$ then $A=B=C=\sum_i y_i /n$.]
You can show that if $A \geq B$ then it has to be that $B \geq C$ (and the other way around), as follows:
Say $A \geq B$, that is
$$\frac{\sum_i x_i \alpha_i y_i}{\sum_i x_i\alpha_i} \geq \frac{\sum_i x_i y_i}{\sum_i x_i} $$
Now, let's show that this implies $B \geq C$. We can rewrite $C$ as:
$$C=\frac{\sum_i x_i (1-\alpha_i) y_i}{\sum_i x_i(1-\alpha_i)} = \frac{\sum_i x_i y_i - \sum_i x_i \alpha_i y_i}{\sum_i x_i - \sum x_i \alpha_i} = \frac{B\sum_i x_i - A\sum_i x_i \alpha_i}{\sum_i x_i - \sum_i x_i \alpha_i}$$
We need to show that $B$ is greather than this expression:
$$ B \geq \frac{B\sum x_i - A\sum_i x_i \alpha_i}{\sum_i x_i - \sum x_i \alpha_i} \implies -B\sum_i x_i \alpha_i \geq -A\sum_i x_i \alpha_i $$
which is only true whenever $A \geq B$, which in turn was our initial guess.
The same procedure works for the case $A \leq B$.