Give a square of side $1$. Show that for each finite set of points of the sides of the square you can find a vertex of the square with the following property: the arithmetic mean of the squares of the distances from this vertex to the points of the set is greater than or equal to $3/4$.
The first thing to show is that for any point on the sides of the square, the sum of the squares of the distances from all the vertices of the squares is always greater than or equal to $3$, which should be easy to do.
This implies that the sum of the distances from each of the $n$ points to each vertex must always be greater than or equal to $3n$.
Now, back to the problem. Suppose there exist $n$ points on the sides of the squares such that for any vertex of the square, the arithmetic mean of the squares of the distances from that vertex is always less than $\tfrac34$.
Then, this means that the sum of the squares of the distances from each of the $n$ points to each vertex is less than $3n$, which is a contradiction, because we already showed that the sum of the distances from each of the $n$ points to each vertex must always be greater than or equal to $3n$.