Relating fraction to convex combinations of numerators and denominators of two other fractions Consider the following functions
$$h(\alpha,x,x',y,y') = \frac{\alpha x + (1-\alpha)x'}{\alpha(x+y) + (1-\alpha)(x'+y')}$$
$$g(x,y) = \frac{x}{x+y}$$
where all values $x,x',y,y',\alpha$ lie in $[0,1]$.
Notice that $h(0,x,x',y,y') = g(x',y')$ and $h(1,x,x',y,y') = g(x,y)$.
Question: Is it true that $h(\alpha,x,x',y,y')$ always lies between $g(x,y)$ and $g(x',y')$ for all $x,x',y,y',\alpha$ in $[0,1]$?
 A: Suppose
\begin{eqnarray*}
\frac{x}{x+y} < \frac{x'}{x'+y'}
\end{eqnarray*}
then $xy'<x'y$.
Now if 
\begin{eqnarray*}
\frac{x}{x+y} < \frac{ \alpha x+(1-\alpha )x'}{\alpha (x+y)+(1-\alpha)(x'+ y')} \\
\end{eqnarray*}
then
\begin{eqnarray*}
x\alpha (x+y)+x(1-\alpha)(x'+ y')  < (x+y) \alpha x+(1-\alpha )x'(x+y)\\
\end{eqnarray*}
and this is the same as $xy'<x'y$. The upper bound can be shown similarly, so the answer is yes.
A: Assume we want to have $h(\alpha,x,x',y,y')>g(x,y)$
$$\frac{\alpha x + (1-\alpha)x'}{\alpha(x+y) + (1-\alpha)(x'+y')}>\frac{x}{x+y}\\ \Rightarrow (x+y)(\alpha x + (1-\alpha)x')>x(\alpha(x+y) + (1-\alpha)(x'+y'))\\ \Rightarrow  \alpha x^2+\alpha xy+(1-\alpha)xx'+(1-\alpha)x'y>\alpha x^2+\alpha xy +(1-\alpha)xx' +(1-\alpha)xy' \\ \Rightarrow x'y>xy'$$
or $\frac{x'}{y'}>\frac{x}{y}$. Similarly you can show that under the condition $\frac{x'}{y'}>\frac{x}{y}$ we have $h(\alpha,x,x',y,y')<g(x',y')$.
On the other hand if $\frac{x'}{y'}<\frac{x}{y}$ then $g(x,y)<h(\alpha,x,x',y,y')<g(x,y)$.
Only special cases $y=0$ or $y'=0$ or $y=y'=0$ is left which are obvious.
A: The desired result follows from
\begin{align}
&\big[h(\alpha, x, x', y, y') - g(x,y)\big]\big[g(x',y') - h(\alpha, x, x', y, y')\big]\\
=\ & \frac{(xy' - x'y)^2\alpha(1-\alpha)}{(\alpha (x+y) + (1-\alpha)(x'+y'))^2(x+y)(x'+y')}\\
\ge \ & 0.
\end{align}
