Existence of a global minimum for a specific potential Consider $n$ points in the plane, and a potential energy given by
$$V(x_1, \ldots, x_n)=\sum_{1\leq i<j \leq n} f( \| x_i - x_j\|)$$
where $x_i \in \mathbb R^2$ are the positions of the particles, and 
$f(r)=\frac{1}{r^2} - \frac{1}{r}$. Note that $f$ has a unique global minimum on $\mathbb R_+^*$ and tends to $+\infty$ as $r \to 0^+$, meaning it corresponds to weak long range attraction and strong short range repulsion.
Question: how to prove that there is (at least one) global minimum on $U:=(\mathbb R^2)^n \backslash \bigcup_{i \neq j} \{x_i=x_j \}$ for $V$? If possible, I would like a geometric/combinatorial solution. 
It is easy to find explicit such minima for $n$ up to $3$, because then you can just arrange the particles to be each at the optimal distance of each other, so that each term in the sum is minimized. Things get more complicated for $n \geq 4$.
Clearly $V$ is continuous, and clearly $V \to +\infty$ as $x=(x_1, \ldots, x_n)$ approaches the boundary of $U$. So the only problem is what happens at infinity, namely: is it possible to prove that there is $R=R_n>0$ such that   $\inf_U V= \inf_{\mathbb{B}(0,R) \cap U} V$?
At first I thought there would be some easy combinatorial way to prove so: for instance, if $n=4$ and you have a configuration of say 3 particles in a ball of radius 5 and the fourth one is at distance at least 20 from the others, you can just bring back the one far away to say distance 10 and decrease the potential energy of each interaction term (while staying far enough away to avoid the short range repulsion effect). 
However, it is possible to come up with configurations where you have two pairs of particles, each pair far from the other pair, and where you cannot decrease $V$ by moving just one particle. That shows that some sort of naive approach by induction is not so easy after all, I think.
Some context: I encountered this question a few years ago as a grad student, and the professor upon realizing it was not as easy as he thought, just moved on without giving a solution. I thought about it for a while and planned to eventually get around to work out a solution myself, but now I just want to know! I just hope I did not miss anything obvious...
 A: The function $f(r)$ has the unique global minimum on $\mathbb R_+^*$ at a point $r_0=2$ and $f$ increases at $[r_0,\infty)$. For each $i$ let $x_i=(x_{i,1}, x_{i,2})$.  Let $x=(x_1,\dots, x_n)$ be any instance of distinct particles. Rearrange particles achieving $x_{1,1}\le x_{2,1}\le\dots  \le x_{n,1}$. If there exists an index $l$ such that $\Delta=x_{l+1,1}-x_{l,1}-r_0>0$ then consider an instance $x’=(x’_1,\dots, x’_n)$ where $x’_{i,2}= x_{i,2}$ for each $i$, $x’_{i,1}= x_{i,1}$ when $i\le l$, and $x’_{i,1}= x_{l,1}-\Delta$ when $i>l$. Let $i<j$ be arbitrary indices. Then $\|x’_i-x’_j\|=\|x_i-x_j\|$ if  $i,j\le l$ or  $i,j\ge l+1$, and $r_0\le\|x’_i-x’_j\|\le \|x_i-x_j\|$, otherwise. Anyway, $f(\|x’_i-x’_j\|)\le f(\|x_i-x_j\|)$, yielding $V(x’)\le V(x)$. Similarly dealing with the other gaps of the first coordinates and with the gaps of the second coordinates, we can show that for each instance $x$ of distinct particles there exists an instance $x^*$ which is contained in a square with a side $(n-1)r_0$ such that $V(x^*)\le V(x)$.
A: Let $u$ be a unit-vector  and consider the projections in the $u$ direction:
$ a_i = u\cdot x_i, i=1,...,n $.
Say that there is a 'gap' of size $\delta$ in the $u$-direction if you may partition the index set $\{1,...,n\}=I \sqcup J$ into disjoint, non-empty subsets such that $a_j - a_i \geq \delta$ for all $i\in I,j\in J$.
If there is a gap of size larger than 2, the minimum radius of your potential,  then you may decrease the energy strictly by displacing one (or both) of the sets of particles in the $u$-direction, e.g. 
$ \;\;r_i' = r_i + t u , i\in I$ and $r_j'=r_j-tu, j\in J$. The energy decrease is strict as long as $0<t<(\delta-2)/2$.
So there can be no gaps greater than 2 in any direction if one is to have a minimum. The diameter of a minimal configuration is consequently  not greater than $2(n-1)$.
