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? 
 A: [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.
A: Edit: I didn't notice the squares above the norms. With those in place, it is true that there is a permutation of vertices such that 
$$\|x_1-x_2\|^2 +\|x_2-x_3\|^2+\cdots+\|x_n-x_1\| ^2 \leq 4. \tag{$\spadesuit$}$$
The approach is to create a problem instance (i.e. a set of points) that maximizes the shortest (in the sense of inequality $(\spadesuit)$) cycle and the key point is the combination of:


*

*Pythagorean theorem: $a^2+b^2=c^2$ in any right-angled triangle, 

*the hypotenuse of any right triangle makes the diameter of circumcircle.


Consider the following pictures.

Any point that lies in the red circle (like $X$) may be skipped thus making the cycle longer (it stays the same if the point lies on the circle, like $X'$). Also, for any point that lies beyond the red line (like $Y$), the intermediate point $B$ maybe skipped for the same reason. In other words, for any problem instance that maximizes the shortest cycle, there may be no points like $X$ or like $Y$ (otherwise the instance could be modified so that shortest cycle could be made longer). This implies that for any instance with shortest cycle of length $L$, there exists an instance of $4$ vertices which shortest cycle is at least of length $L$. This (I will skip cases of $n < 4$ here) implies $(\spadesuit)$.
I hope this helps ;-)
Edit:
Some more explanation. For example the corollary from the second picture is this: let $A$ be a set of points in question and $x_1,x_2,\ldots,x_n$ a sequence that is the shortest cycle (in the sense of inequality $(\spadesuit)$). Then if there is an $x_i$ such that $|\angle x_{i-1}x_ix_{i+1}| > \frac{\pi}{2}$, then for $B = A \setminus \{x_i\}$ the shortest cycle is longer than the shortest cycle in $A$.
A: I've seen this in a blog, together with the idea for the solution. Here it is.
