Points in unit square Let $n$ points be given in the unit square. How to prove or disprove: the points can be labeled $x_1,\ldots,x_n$ to satisfy the inequality $$\|x_1-x_2\|^2 +\|x_2-x_3\|^2+\cdots+\|x_n-x_1\| ^2 \le 4,$$ where $\|\cdot\| $ is the Euclidian distance? 

[I've seen this problem before, and I'm guessing in Mathematical Gems by Ross, but I'm not too sure. I'm also lazy with the notation, and am treating them as points, instead of coordinates.]
Lemma: Let $ABC$ be a right triangle with $\angle ABC = 90^\circ $. THere are $n$ points in/on the triangle. Then, it is possible to find a sequence of points such that
$$ |Ax_1^2 | + |x_1 x_2|^2 + \ldots + |x_k B|^2 + |Bx_{k+1}|^2 + \ldots + |x_nC|^2 \leq |AC|^2$$
Proof: We proceed by induction on the number of points.
Let $BD$ be the perpendicular to side $AC$. Apply the induction hypothesis, to get
$$ |Ax_1^2 | + |x_1 x_2|^2 + \ldots + |x_i D|^2 + |Dx_{i+1}|^2 + \ldots + |x_jB|^2 \leq |AB|^2$$
$$ |Bx_{j+1}^2 | + |x_{j+1} x_{j+2}|^2 + \ldots + |x_k B|^2 + |Bx_{k+1}|^2 + \ldots + |x_nC|^2 \leq |BC|^2$$
Now, it remains to show that $|x_i D|^2 + |Dx_{i+1}|^2 \leq |x_ix_{i+1}|^2$, but this is obvious because $\angle x_i D x_{i+1}$ is at most $90^\circ$. Ditto for $|x_k D|^2 + |Dx_{k+1}|^2 \leq |x_k x_{k+1}|^2$.
Add up both inequalities and apply that $|AB|^2 + |BC|^2 = |AC|^2$.
Note: There is a small issue of when $ABD$ or $BCD$ is empty, which prevents us from applying the induction hypothesis. However, what we simply do is to continue cutting up the right triangles till we get a split of the points.
Corollary: Your question
Proof:

 Let the square be $ABCD$. Throw in points $A, B, C, D$ if they are not already in the list. Apply the Lemma to triangle $ABC$ and $BCD$, and we are done. You will need to apply a step of the Lemma's proof to remove $A, B, C, D$ if need be.

Followup: Can you classify all equality cases? Consider how the Lemma is proven.
