What is the trick used in skip lists to minimize $k + \frac{n}{k}$?
The method you mention is based on the following theorem:


  If $f \geq 0$ is non-decreasing, $g \geq 0$ is non-increasing, and $f(t) = g(t) = y$, then
  $$ y \leq \min_x [f(x) + g(x)] \leq 2y. $$


Note that there is (at most) a unique point $t$ such that $f(t) = g(t)$.

In your case, $f(k) = k$ is increasing and $g(k) = n/k$ is decreasing, and $f(\sqrt{n}) = g(\sqrt{n}) = \sqrt{n}$, so $\sqrt{n} \leq \min_k [k + n/k] \leq 2\sqrt{n}$. In fact, calculus shows that the minimum is indeed $2\sqrt{n}$.

The proof of the theorem is very simple: if $x \geq t$ then $f(x) + g(x) \geq f(t) = y$, whereas if $x \leq t$ then $f(x) + g(x) \geq g(t) = y$.