This problem has really been bothering me and I have no idea whether the statement is true or not. So any help is appreciated.
Suppose $f(x)/g(x)$ satisfies monotone likelihood ratio property and have CDF $F$ and $G$, respectively. Assume $f$ and $g$ have support in $[0,1]$. Let $C \leq 1$ be a constant, and some $1 \geq x_1 \geq x_2\geq 0 $, such that the following condition holds: $$ C\cdot \frac{f(x_1)}{g(x_1)} = \frac{f(x_2)}{g(x_2)}.$$ Prove or give a counterexample $$ C\cdot \frac{1-F(x_1)}{1-G(x_1)} \leq \frac{1-F(x_2)}{1-G(x_2)}$$ If false, is there a simple condition that could be added to $f$ and $g$ such that the statement is true?
Attempt to showing proof: A natural place to start was using the inequality $$\frac{f}{g}(x) \leq \frac{1-F(x)}{1-G(x)}$$ but that ended up going nowhere.
Attempt to construct counterexample If there is a counterexample, it could be that $1-F(x_2)$ is extremely close to $0$. But that means $1-F(x_1)$ is also close to $0$ since $x_1 > x_2$. It's also not immediately clear that the construction is valid. I also attempted a simple case of $F = x^2$ and $G = x$ but it was not a counterexample.
The answer should be false. A counter example can be constructed attacking $G(x)$ instead. Suppose $G$ rises extremely fast initially with a no change in slope until $G(x)$ is close to $1$, while $F$ is relatively flat initially with no change in slope until $G(x)$ changes slope as well.
Then $x_1$ can be picked just after the change in slope such that $f(x_1)/g(x_1)$ is very close to $f(x_2)/g(x_2)$. Pick $x_2$ near $0$, so that $(1-F(x_2))/(1-G(x_2))$ is close to $1$, but by the construction of $F$ and $G$, $(1-F(x_1))/(1-G(x_1)) >> 1$. This then implies the desired inequality is false.