Given parameters $k_1, k_2 > 0$, consider the following optimization problem
$$\begin{array}{ll} \underset{x, y}{\text{minimize}} & k_{1} y + \dfrac{x}{1-x} \left( k_2 x^y - k_{1} + k_{1} x^y \right)\\ \text{subject to} & 0 \leq x < 1, \quad y \geq 1\end{array}$$
All numerical checks show that the optimal solution is attained at the boundary at $y^{*}=1$. I have tried proving this and failed. Intuitively, it seems clear and simple - unless you can come up with a counterexample - I am looking for proof. Can anyone please help?
The objective is convex in $y$ in the feasible region since its second derivative is
$$\frac{(k_1+k_2) \log^2(x) x ^{y+1}}{1-x}>0$$
The first derivative with respect to $y$ is $$k_{1}+\frac{(k_{1}+k_{2}) \log (x)x^{y+1}}{1-x}.$$
Note, that as $x\rightarrow 1$ then the first derivative value is $-k_{2}<0$.
The first derivative with respect to $x$ is $\frac{(k_1+k_2) ((1-x) y+1) x^y-k_1}{(x-1)^2}$.
Anyone? :)