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 $1\le c \le ab$.
 A: Let $a=b=2$ and $c=1$.  Then $f(0)=1/4$ but $f(1)=1/3$.
A: Let 
$$c=t+(1-t)ab,\text{  } 0\le t\le 1,\text{   } \tag{A}$$, 
then 
$$f(x,a,b,t)=\frac{t+(1-t)ab+2x}{(a+x)(b+x)}$$
Now consider $g(x)=f(x,a,a,t)$.
$$g'(x)=\frac{dg}{dx}=\frac{2(a + a^2(-1 + t) - t - x)}{(a + x)^3}$$
Thus $g'(x)$ has a zero at $x_0=(a-1)(a(1-t)-t)$.  
(1) If $a>t=1$, then this zero is positive. In this case g(x) increase when $x$ varies from 0 to $x_0>0$ and decreases afterwards.
(2) If $t=0$, then $x_0<0$. So $g(x)$ is a decreasing function.
(3) If $a\gt=\frac{t}{1-t}$ and $0<t<1$, then $x_0>0$. So this case is similar to (1).
We have not considered that the restriction on $t$ because $c$ in (A) is a positive integer.
A: 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:
$$
x = \frac{-2c \pm \sqrt{(4c^2-8c(a+b)+16ab)}}{4}\\
= -\frac12 \left(c \pm \sqrt{(c-2a)(c-2b)} \right)
$$
A: Check $\frac{1}{f}$ ! 
$$\frac{1}{f(x)}=(c+2x)q(x)+r$$ 
$r$ is a constant and does not contribute to the growth of the function.
$(c+2x)$ is increasing and positive. So analyze the linear function $q(x)$.
Some Maxima output:

(%i2) f(x):=(c+2*x)/((x+a)*(x+b));
(%o2) f(x):=(c+2*x)/((x+a)*(x+b))
(%i3) makelist(f(i),i,[1,2,3]),a=10,b=10,c=100,numer;
(%o3) [0.84297520661157,0.72222222222222,0.62721893491124]
(%i4) makelist(f(i),i,[1,2,3]),a=10,b=10,c=1,numer;
(%o4) [0.024793388429752,0.034722222222222,0.041420118343195]

