Let $f$, $g$, and $h$ be continuous real functions with strictly positive values. $f$ is strictly increasing. $M, N>0$ are constants, and $a\in(0,1)$.
Can you prove (or disprove) the following inequality?
$$\frac{M+\int_0^a f(x)g(x) dx}{N+\int_0^a g(x)dx} < \frac{M+\int_a^1 f(x)h(x) dx}{N+\int_a^1 h(x)dx}. $$
Here is my guess: on each side, it is a reweighted average of two components. The first component is $\frac{M}{N}$, which is the same on both sides. For the second component, we know that
$$\frac{\int_0^a f(x)g(x) dx}{\int_0^a g(x)dx}<f(a)<\frac{\int_a^1 f(x)h(x) dx}{\int_a^1 h(x)dx}.$$
So the reweighted average of the left hand side should also be smaller.
Simply choose $f = g = h = 1$, except $f$ which you let grow at an epsilon rate, and consider the case where $a$ is essentially equal to 1. Your inequality then becomes $$ \frac{M + 1}{N + 1} < \frac{M}{N} . $$ This fails for e.g. $M = 1$ and $N = 2$.