Is this function decreased with $x$? Given three positive integers $a,b,c\ge 1$, I am wondering if the following $f(x)$ is decreased with $x$ ?
$$f(x)=\frac{c+2x}{(a+x)(b+x)}, \quad x \in Z^+ \cup \{0\}$$
where $1\le c \le ab$.

since my calculations are error-prone, i will merely show some working which someone might check. if this result is correct then at present i cannot see see how it can be non-trivially related to the given condition $1 \le c \le ab$. might one expect a condition involving $c^2$ rather than $c$? apologies in advance
$$
\frac{df}{dx} = \frac{2(a+x)(b+x) - (c+2x)(a+b+2x)}{(a+x)^2(b+x)^2}
$$
the numerator is
$$
2(ab+(a+b)x+x^2) - c(a+b) - 2cx -2x(a+b)-4x^2 \\
= 2ab - 2x^2 -2cx - c(a+b)\\
= - ( 2x^2+2cx + [c(a+b)-2ab])
$$
the critical values of $x$ are thus the solutions, if any, of:
$$
x = \frac{-2c \pm \sqrt{(4c^2-8c(a+b)+16ab)}}{4}\\
= -\frac12 \left(c \pm \sqrt{(c-2a)(c-2b)} \right)
$$
