I'm attempting to show that
$$F(x;a,b,c) = \frac{1}{x}\log\left( \frac{c-b}{c-b-xa} \right)$$
is increasing in $x$ for $0<x<1$, $\lbrace a,b,c\rbrace>0$, and $c>b+xa$. I have attempted to show that its derivative is positive, but have failed so far. The presence of log in the derivative makes it hard for me to see how to proceed:
$$ F'_x(x;a,b,c) = \frac{1}{x^2}\left[ \frac{xa}{c-b-xa} -\log\frac{c-b}{c-b-xa} \right] $$
I have tried to find a counter example in Python, but have failed to do so as well.
This is from a problem in economics. I do not know with certainty that the function is increasing in x, yet economic intuition tells me that it should be.
All you have to realize is that $\frac{xa}{c - b - xa} - \log\frac{c - b}{c - b - xa} > 0$. We can rewrite this as: $$\frac{xa}{c - b - xa} - \log\left(1 + \frac{xa}{c - b - xa}\right)$$ Since $a,b,c,x > 0$ and $c > b + ax$, we know that $\frac{xa}{c - b - xa}$ is always greater than zero. So if we can show that $u - \log(1 + u) > 0$ for $u > 0$, we are done! A quick way to see that is to note that when $u = 0$, $u - \log(1+u) = 0$, and that $$\frac{\partial}{\partial u}(u - \log (1+u)) = 1 - \frac{1}{1+u}$$ is positive for $u > 0$. So $u - \log(1+u) > 0$ for $u > 0$!