I am trying to minimize $f(x,y)= \exp(-xy)+\alpha \exp\left(-\frac{1}{x}\right)$ with reespect to $x\in[0,\gamma]$ for a given $y\geq 0$. Here $\alpha$ and $\gamma$ are positive constants.
Differnetiating $f(x,y)$ and solving does not provide any closed-form solution. I need to find a simple solution which is in some way related to the optimal value.
So, I tried replacing $e^{-1/x}$ with $\ln(x)$ as they both have similar behavior for certian range of $x$. I scaled $\ln(x)$ in a way that they both have same value at $\gamma$.
\begin{equation} g(x,y)= \exp(-xy)+\alpha e^{-\frac{1}{\gamma}} \ln(x) / \ln(\gamma) \end{equation}
g(x,y) has a closed-form solution in terms of LambertW function. For cerain range of $x$, using plots, I see it to be very close to the optimal value of $f(x.y)$.
Now, I am trying to mathmatically show that they are indeed close.
Is there any good way to show this?
I tried the following
1) Derivative of $f(x.y)$ is close to zero at $x^{*}$ that minimizes $g(x.y)$
2) Show that both $f(x.y)$ an $g(x.y)$ are close for that range of $x^*$