I am trying to prove this claim but it seems the math somehow does not work out...
Let $f_1,f_2,g_1$ and $g_2$ be real-valued, strictly positive, continuously differentiable and strictly decreasing functions on $\mathbb{R}_{++}$. If $\frac{f_1(x)}{g_1(x)}$ and $\frac{f_2(x)}{g_2(x)}$ are strictly increasing (resp. decreasing) in $x$, then $$\frac{f_1(x)+f_2(x)}{g_1(x)+g_2(x)}$$ $\text{ is strictly increasing (resp. decreasing) in $x$.}$
The intuition behind this claim is convincing: if $f_1$ and $f_2$ are decreasing faster than $g_1$ and $g_2$ respectively, since the functions are positive, then it should be that$f_1+f_2$ are decreasing faster than $g_1+g_2$.
I wonder if anyone has a clue how to prove it. Or, a counter-example would be greatly appreciated.
Thank you very much!
Here's a thought for the counter-example. By taking the derivatives of $\frac{f_1+f_2}{g_1+g_2}$, $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$, we essentially need to prove that $$(g_1+g_2)(f_1'+f_2')-(f_1+f_2)(g_1'+g_2')>0$$ given that $g_1f_1'-g_1'f_1>0$ and $g_2f_2'-g_2'f_2>0$ where $g_1,g_2,f_1,f_2$ are positive and $g_1',g_2',f_1',f_2'$ are negative.
However, it essentially boils down to whether it must be true by imposing these conditions on real numbers. Here is an example of these numbers where all the premises are fulfilled but the proposed inequality does not hold.
https://desmos.com/calculator/hc7okrx6gn
Therefore, the claim is incorrect.