Balkan Shortlist, 2012

We have $f(A) = 1/(\sqrt{2} + 1) = \sqrt{2} - 1$. We will prove that this value is the minimum of function $f$, or in other words,
$$PA + PB \geq (\sqrt{2} - 1)(PC + PD).$$
for all $P$.
Applying Ptolemy's inequality for the points $P$, $A$, $B$, $C$, we have $PA + \sqrt{2} PB \geq PC$, that is
$$PA + \sqrt{2} PB \geq PC.$$
Applying Ptolemy's inequality for the points $P$, $A$, $B$, $D$ we have $\sqrt{2} PA + PB \geq PD$, that is
$$\sqrt{2} PA + PB \geq PD.$$
Adding these inequalities we get
$$(\sqrt{2} + 1)(PA + PB) \geq PC + PD,$$
hence the desired inequality.
By Ptolemy's Theorem it follows that the minimum is attained if and only if the point $P$ belongs to the arc $AB$ of the circumcircle of the square.
If $P'$ is the symmetric of $P$ with respect to the center of the square, then $f(P) = 1/f(P')$. It follows that the maximum of the function is $1/(\sqrt{2} - 1) = \sqrt{2} + 1$ and it occurs exactly at the points on the arc $CD$ of the circumcircle of the square.

Place the square in the Cartesian plane with the vertices at $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$. Use the polar coordinates to obtain
$$f(P) = \frac{\sqrt{r^2 + 1 - 2r \cos \theta} + \sqrt{r^2 + 1 - 2r \sin \theta}}{\sqrt{r^2 + 1 + 2r \cos \theta} + \sqrt{r^2 + 1 + 2r \sin \theta}}$$
The change of variables $u = 2r \cos \theta/(r^2 + 1)$, $v = 2r \sin \theta/(r^2 + 1)$ reduces the problem to showing that
$$\sqrt{2} - 1 \leq \frac{\sqrt{1-u} + \sqrt{1-v}}{\sqrt{1+u} + \sqrt{1+v}} \leq \sqrt{2} + 1$$
for $u^2 + v^2 \leq 1$.
Note that this expression is decreasing in $u$ for constant $v$, and decreasing in $v$ for constant $u$. On the unit circle $u^2 + v^2 = 1$, the tangent half-angle substitutions $u = 2t/(1+t^2)$, $v = (1-t^2)/(1+t^2)$ give
$$\frac{\sqrt{1-u} + \sqrt{1-v}}{\sqrt{1+u} + \sqrt{1+v}} = \frac{|1-t| + \sqrt{2}|t|}{|1+t| + \sqrt{2}}$$
which reduces to $\sqrt{2} - 1$ in the first quadrant (that is, for $0 \leq t \leq 1$) and to $\sqrt{2} + 1$ in the third quadrant (that is, for $t \leq -1$) finishing the proof.