Let $A, B, C, D \in \mathbb{R^*_+}$.
Is it possible to solve $$ \max_{ \substack{0 \leq x\leq A \\ 0\leq y\leq B}} \frac{1+x+y}{(1+Cx)(1+Dy)} $$
The KKT conditions give for an extrema $(x^*,y^*)$
If $\frac{1}{C} > (1+y^*)$ then $x^*=A$ otherwise $x^*=0$,
If $\frac{1}{D} > (1+x^*)$ then $y^*=B$ otherwise $y^*=0$.
Thus it would solve the problem (if one could show concavity in $(x,y)$) when $1/C>1+B$ and $1/D>1+A$.
I have no idea for the case $1/C\leq 1+B$ or $1/D\leq 1+A$
The gradient is $0$ iff $[x,y] = [1/D-1, 1/C-1]$, at which point the objective value is $1/(C+D-CD)$. If this is in the feasible region $[0, A] \times [0,B]$, it may be optimal. Compare to the optimal values on the edges $\{0\} \times [0,B]$, $\{A\} \times [0,B]$, $[0,A] \times \{0\}$, $[0,A] \times \{B\}$.
Parametrize the borders and look for the roots of the gradient in the interior of the function domain.
Moreover, analize the limits in the singularities. I did not make the details, but if the root of $(1+Cx)(1+Dy)$ are in $[0,A]\times[0,B]$, it is possible that an infite limit appears.
