For $k\in N, n-2\ge k\ge2$, and $n \in N, n\ge4$
minimize the function $f(n,k)=(n-1)-\sqrt{(n-1)^2-4(k-1)(n-k-1)}$ over n,k
EDIT: Attempt to solve
First I differentiated it partially w.r.t $k$ and equated that to zero. I got $n=2k$, next I put $n=2k$ in $f$ and observed it is an increasing function of $k$. So k$=2$ and $n=4$ should be point of minimum. Am I correct?
I do not see why you use $n=2k$, but your basic approach is correct.
Fix $n$. Then you want to choose $k$ to maximise the square root and hence to minimise $(k-1)(n-k-1)$. That is a quadratic, so the minimum comes at either $k=2$ (the smallest allowed value) or $k=n-2$ (the largest allowed value). Checking, you find that $k=2$ gives the minimum.
So now you want to choose $n$ to minimise $n-1-\sqrt{n^2-6n+13}$. You find $n=4$ gives approx 0.76. But $n\ge 5$ gives $\sqrt{n^2-6n+13}<n-2$ and hence $n-1-\sqrt{n^2-6n+13}>1$. So the minimum is at $n=4,k=2$.