I have a function $f(x)$. With simulations, I found that $f(x)$ is increasing w.r.t. $x$. Therefore I computed the partial derivative of the function $f(x)$ w.r.t. $x$, which is $f_x=\frac{1+\sqrt{x+1}}{(x+1)^2}+\frac{A-B+1}{(A-x+1)^2}-\frac{1}{(A-x+1)^{3/2}}$. The following conditions hold among the variables A, B, and x: $0 \leq x \leq B, B \leq A$. How can I show that $f_x$ is positive? I simplified the expression. However, I cannot seem to find a way to show that it's positive.
$f_x=\frac{1+\sqrt{x+1}}{(x+1)^2}+\frac{A-B+1}{(A-x+1)^2}-\frac{1}{(A-x+1)^{3/2}}\\ =\frac{(1+\sqrt{x+1})(A-x+1)^2+(A-B+1)(x+1)^2-(x+1)^2 \sqrt{A-x+1}}{(x+1)^2(A-x+1)^2}\\ =\frac{(A-x+1)^2+(A-x+1)^2\sqrt{x+1}+(A-B+1)(x+1)^2-(x+1)^2 \sqrt{A-x+1}}{(x+1)^2(A-x+1)^2}$
Well this is leading me nowhere.
With the change in the second term from your original posting, it is now no longer true that $f_x$ is always positive. I personally like color plots for the kind of numerical simulation that would indicate that. Alternatively, minimization methods work okay since the structure of $f$ isn't too complex.
As a specific example, take $x=15$, $A=B=18$. I tried to pick nice values to work with, and at those inputs we have $f_x=-\frac{11}{256}$. In general, when $A=B$ and $x$ is large enough, there is a band of $A$ values somewhere between $x$ and $2x$ producing negative values of $f_x$. I haven't looked at it further for $B\neq A$, but continuity gives us that there are instances of $B\neq A$ with negative $f_x$.