KKT example nonlinear programming Say you have the function 
$$ f(x,y)=n(q_{u}\ln{x} +q_{m}\ln{y} -\ln{(P_{u}(x-1)+P_{m}(y-1)+1)})$$ where $P_{u}, P_{m}, q_{u}, q_{m} \in (0,1]$. 
I need to solve the following optimization problem:
\begin{equation}
\begin{aligned}
& \underset{x, y}{\text{max}}
& &\mathcal f (x, y) \\
& \text{s.t.} & &  x\leq{y} \\
%& & &  x,y \geq{1} \\
\end{aligned}
\end{equation}

We write the Lagrangian as
$$L(x,y,\lambda)=n(q_{u}\ln{x} +q_{m}\ln{y} -\ln{(P_{u}(x-1)+P_{m}(y-1)+1)}) - \lambda (x-y)$$
We can then write the KKT conditions for the problem as:
\begin{equation}
\frac{\partial f(x,y)}{\partial y}= \frac{-n P_{m}}{(P_{u}(x-1)+P_{m}(y-1)+1)} + \frac{nq_{m}}{y}+\lambda=0
\end{equation}
\begin{equation}
\frac{\partial f(x,y)}{\partial x}= \frac{-n P_{u}}{(P_{u}(x-1)+P_{m}(y-1)+1)} + \frac{nq_{u}}{x}-\lambda=0
\end{equation}
We also have complementary slackness:
\begin{equation}
\lambda(x-y)=0
\end{equation}
and finally $\lambda \geq{0}$.
My question is how to start playing so I can get some candidates. Should I start considering the cases from the complementary condition? Any help is welcome. 
 A: Since $n > 0$, I will ignore it in the objective. Furthermore, $\sum P_i = \sum q_i = 1$ implies that the objective function simplies to $f(x,y) = q_u\ln{x} + q_m\ln{y} - \ln{(P_u x + P_m y)}$. 
Let us consider the following two cases:
Case 1: $q_u \geq P_u$. 
From the concavity of $\ln(\cdot)$ and the fact that $\sum P_i = 1$, we have $$\ln{(P_u x + P_m y)} \geq P_u \ln{x} + P_m \ln{y}.$$ Therefore, $f(x,y) \leq (q_u - P_u) \ln{x} + (q_m - P_m) \ln{y} = (q_u - P_u) (\ln{x} - \ln{y}) \leq 0$ (where we have used $\sum P_i = \sum q_i = 1$ and the constraint $y \geq x$). Since $\bar{x} > 0$ gives us $f(\bar{x},\bar{x}) = 0$, we have that the maximum objective value is zero at any $(\bar{x},\bar{x}) > 0$.
Case 2: $q_u < P_u \implies q_m > P_m$. 
We have that setting $y$ an arbitrarily small value greater than any $x > 0$ increases the objective value in this case since $\nabla_y f(\bar{x},\bar{x}) = \frac{q_m}{\bar{x}} - \frac{P_m}{P_u \bar{x} + P_m \bar{x}} = \frac{q_m - P_m}{\bar{x}} > 0$. Therefore, the inequality constraint will not be active at the solution and we can simplify the KKT equation corresponding to $y$ to be $\frac{q_m}{y^*} - \frac{P_m}{P_u x^* + P_m y^*} = 0$ at the KKT point $(x^*,y^*)$, which yields $y^* = \frac{q_m P_u}{P_m q_u}x^*$. You can plug this solution in the objective to verify that the optimal objective value amounts to $$q_m\ln\left(\frac{q_m P_u}{P_m q_u}\right) + \ln\left( \frac{q_u}{P_u} \right).$$
