# Global minimum for $\frac{2(q - 1)(q^k + 1)}{q^{k+1} + q - 1}$, if $q \geq 5$ and $k \geq 1$

Let $$q$$ be a prime number, and let $$k$$ be an integer.

THE PROBLEM

Does the function $$f(q,k) = \frac{2(q - 1)(q^k + 1)}{q^{k+1} + q - 1}$$ have a global minimum, if $$q \geq 5$$ and $$k \geq 1$$?

MY ATTEMPT

I tried asking WolframAlpha, it was unable to find a global minimum for $$f(q,k)$$ in the domain $$q \geq 5$$ and $$k \geq 1$$.

I then computed the partial derivatives (still using WolframAlpha):

Partial derivative with respect to $$q$$ $$\frac{\partial}{\partial q} f(q,k) = \frac{2q^{k-1}\bigg(q^{k+1} - k(q - 1) + q\bigg)}{\bigg(q^{k+1} + q - 1\bigg)^2} > 0$$

Partial derivative with respect to $$k$$ $$\frac{\partial}{\partial k} f(q,k) = -\frac{2(q-1){q^k}\log(q)}{\bigg(q^{k+1} + q - 1\bigg)^2} < 0.$$

Does this mean that we can have (say) $$f(q,k) \geq f(5,1) = \frac{48}{29} \approx 1.65517?$$

$$\frac{2}{f(q,k)}=\frac{q^{k+1}+q-1}{(q-1)(q^k+1)}=1+\frac{1}{q-1}\left(1-\frac{1}{q^k+1}\right)$$ is strictly increasing in $$k$$, so it can't have a global maximum, hence $$f(q,k)$$ can't have a global minimum.