If $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$. 
If $a, b, c, d, e, f, g, h$ are positive numbers satisfying $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$ and $b+f>d+h$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.

I thought it is easy to prove. But I could not. How to prove this? Thank you.
The question is a part of a bigger proof I am working on. There are two strictly concave, positive valued, strictly increasing  functions $f_1$ and $f_2$ (See Figure 1).  Given 4 points $x_1$, $x_2$, $x_3$ and $x_4$ such that  $x_1< x_i$, $i=2, 3,4$ and  $x_4> x_i$, $i=1, 2, 3$, let $d=x_2-x_1$,  $b=x_4-x_3$ $c=f_1(x_2)-f_1(x_1)$,  $a=f_1(x_4)-f_1(x_3)$. And  given 4 points $y_1$, $y_2$, $y_3$ and $y_4$ such that  $y_1< y_i$, $i=2, 3,4$ and  $y_4> y_i$, $i=1, 2, 3$, let $h=y_2-y_1$,  $f=y_4-y_3$ $g=f_2(y_2)-f_2(y_1)$,  $e=f_2(y_4)-f_2(y_3)$.
Since the functions are concave, we have $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$.  And I thought in this setting, it is true that $\frac{a+e}{b+f} < \frac{c+g}{d+h}$ even without the restriction $b+f>d+h$.


The updated question (with the additional constraint $b+f>d+h$) is also false. For example,
$\frac{1}{1}<\frac{3}{2}$ and $\frac{9}{4}<\frac{5}{2}$, but $\frac{1+9}{1+4} = \frac{10}{5} = \frac{8}{4} = \frac{3+5}{2+2}$.
