Balkan 2012 shortlist

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, $$PA+PB\ge (\sqrt{2}-1)(PC+PD).$$ for all $P$. Applying Ptolemy's inequality for the points $P$, $A$, $B$, $C$, we have $PA+\sqrt{2}PB\ge PC$, that is $$PA+\sqrt{2}PB\ge PC.$$ Applying Ptolemy's inequality for the points $P$, $A$, $B$, $D$ we have $\sqrt{2}PA+PB\ge PD$, that is $$\sqrt{2}PA+PB\ge PD.$$ Adding these inequalities we get $$(\sqrt{2}+1)(PA+PB)\ge PC+PD,$$ 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 $P^{\prime }$ is the symmetric of $P$ with respect to the center of the square, then $f(P)=1/f(P^{\prime })$. 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.

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Alternative solution.

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 $$\sqrt{2}-1\le \frac{\sqrt{1-u}+\sqrt{1-v}}{\sqrt{1+u}+\sqrt{1+v}}\le \sqrt{2}+1$$ for $u^2+v^2\le 1$. 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 $0\le t\le 1$) and to $\sqrt{2}+1$ in the third quadrant (that is, for $t\le -1$) finishing the proof.