Inequality for Distribution of points in space Consider the space $\mathbb{R}^2.$ We are given $n$ mutually different points $x_1,..,x_n \in \mathbb R^2.$

We can then introduce expressions

$$f_i(x_1,...,x_n) =  n^2 \vert x_i \vert^2 + \sum_{j \neq i } \frac{1}{\vert x_j-x_i \vert^2 }.$$

I would like to know if there exists an explicit constant $c&gt;0$, independent of the points, (as large as possible) such that 
$$ n^2+ \sum_{i=1}^n f_i(x_1,...,x_n) \ge c n^3.$$

I will do two cases by hand to give you a feeling:

$n=1$: This case is clear, as

$$1+ f_1(x_1) \ge 1 =c1^3$$ with $c=1.$

$n=2$: This case is already more tricky, however 

$$4 + f_1(x_1,x_2) + f_2(x_1,x_2) =4+ 4(\vert x_1 \vert^2+\vert x_2 \vert^2) + \frac{2}{\vert x_1-x_2 \vert^2}$$
and thus using the parallelogram identity we find 
$$ 4 + f_1(x_1,x_2) + f_2(x_1,x_2)=4+ 2 (\vert x_1-x_2 \vert^2 + 1/\vert x_1-x_2 \vert^2) + 2 \vert x_1+x_2 \vert^2.$$ 

Now, we may use that $t+1/t  \ge 2$ for $t &gt; 0$ to infer that

$$ 4 + f_1(x_1,x_2) + f_2(x_1,x_2)\ge 4+ 4 = 2^3.$$  

So somehow one could conjecture that $c=1$ is possible, but I don't know whether this is true in general.
No, this is not true, here is the sketch of a sequence of distributions where your sum asymptotically grows slower than $n^3$. In the following I am using the symbol $\asymp$ for expressions whose ratio is bounded above and below by some positive constants independent of $n$, and similarly $\lesssim$ for inequalities which hold up to a positive constant.

Start with the following construction: Let $R&gt;1$ be an integer, and let $x_{k,l} = (k,l)$ be the points with integer coordinates $k,l \in \{-R,\ldots,R\}$. These are $n=(2R+1)^2 \asymp R^2$ points with 
$$
\sum_{k,l=-R}^R \|x_{k,l}\|^2 = \sum_{k,l=-R}^R (k^2+l^2) \asymp R^4 \asymp n^2.
$$
Furthermore, for any fixed $k_0,l_0 \in \{-R,\ldots,R\}$ we have
$$
\sum_{{k,l=-R}\atop{(k,l) \ne (k_0,l_0)}}^R \frac{1}{\|x_{k,l} - x_{k_0,l_0}\|^2} \le
\sum_{{k,l=-2R}\atop{(k,l) \ne (0,0)}}^{2R} \frac{1}{\|x_{k,l}\|^2} \le
\sum_{m=1}^{2R} \,\, \sum_{\max(|k|,|l|)=m} \frac{1}{\|x_{k,l}\|^2},
$$
which is rearranging the sum into sums over boundaries of concentric squares of radius $m$. The number of integer points on such a boundary (i.e., the number of terms of the inner sum, given the radius $m$) is $\asymp m$, and each term in that sum is $\frac{1}{\| x_{k,l} \|^2} \asymp \frac{1}{m^2}$, so we get that the total sum above is
$$ 
\asymp \sum_{m=1}^{2R} \frac{1}{m} \asymp \ln R \asymp \ln n.
$$
Summing this up over all lattice points $(k_0,l_0)$, we get that
$$
\sum_{{k,l,k_0,l_0=-R}\atop{(k,l) \ne (k_0,l_0)}}^R \frac{1}{\|x_{k,l} - x_{k_0,l_0}\|^2} \lesssim n \ln n.
$$
For your functions $f_i$, where $i$ now runs through all pairs $(k,l)$ as above, we then get
$$
\sum_{k,l=-R}^R f_{k,l}(x_{-R,-R},\ldots,x_{R,R}) \lesssim n^4 + n \ln n,
$$
which obviously is not quite a counterexample yet. However, from the homogeneity of the terms it is immediate that scaling by a factor $\lambda &gt; 0$ leads to
$$
\sum_{k,l=-R}^R f_{k,l}(\lambda x_{-R,-R},\ldots,\lambda x_{R,R}) \lesssim \lambda^2 n^4 + \lambda^{-2} n \ln n.
$$
A choice of $\lambda = n^{-3/4} (\ln n)^{1/4}$ (so that both terms have the same asymptotic behavior) leads to
$$
\sum_{k,l=-R}^R f_{k,l}(\lambda x_{-R,-R},\ldots,\lambda x_{R,R}) \lesssim n^{2.5} \sqrt{\ln n}.
$$
Somehow I suspect that this is actually the lower bound for the asymptotic behavior of your sum for general $(x_i)$, but I don't have a proof of that.
It is true and follows from expanding the following expression: 
Consider a vector $x=(x_1,...,x_N)$, and define vectors $X,Y$ as 

$X_i:=\frac{2}{N} x_i$ and $Y_i:=-\frac{2}{N^2} \sum_{j \neq i } \frac{x_i-x_j}{\vert x_i- x_j \vert^2}.$

It follows then that 

$$ \vert X+Y \vert^2= \frac{4}{N^2} \vert x \vert^2 + \frac{4}{N^4} \sum_{i=1}^N \vert \sum_{j \neq i } \frac{x_i-x_j }{\vert x_i-x_j \vert^2} \vert^2+ -\frac{4}{N^3} \sum_{i \neq j } \left( x_i-x_j \right) \frac{\left(x_i-x_j\right)}{\vert x_i-x_j \vert^2}$$

Thus, $$ \vert X+Y \vert^2= \frac{4}{N^2} \vert x \vert^2 + \frac{4}{N^4}  \sum_{j \neq i } \frac{1}{\vert x_i-x_j \vert^2}  -\frac{4(N-1)}{N^2}.$$