A square of side 1 is given, and 10 points are inside the square.
If we divide the square into 9 smaller squares, and apply Dirichlet principle, we can prove that there are 2 of these 10 points whose distance is at most $\sqrt2/3$.
Can this statement be improved, in the sense that $\sqrt2/3$ can be replaced with lower value?
(This question came to my mind while reading some other similar questions here. I have no idea what would be a good approach.)