<s>[INST] Logical suite and inequalities This is a related problem (see here)
We have the following inequalities :

For $n=3$ with $a,b,c$ real numbers the following inequality holds.
  $$\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(c-b)^2+(c-a)^2+(a-b)^2\geq \sqrt{54}$$

For $n=4$ with $a,b,c,d$ real numbers
$$\frac{1}{(a-d)^2}+\frac{1}{(d-b)^2}+\frac{1}{(d-c)^2}+\frac{1}{(a-b)^2}+\frac{1}{(c-b)^2}+\frac{1}{(a-c)^2}+(d-b)^2+(c-d)^2+(a-d)^2+(a-b)^2+(b-c)^2+(a-c)^2\geq \sqrt{288}$$
If we continue with $n=5,6,7\cdots$ there is a logical suite but I don't know how to prove this and what is the following numbers .
Thanks.
 [/INST] To complete the answer of achille hui I will show that the lower bound
$$ \frac{n(n-1)}{2}\sqrt{2n}$$
is actually a minimum.
Let $z_j$ be the $n$ zeros of the Hermite polynomial of order n, then we have the so-called  Stieltjes sums (see http://epubs.siam.org/doi/pdf/10.1137/0514028) :
$$ \sum_{j=1}^n \frac{1}{(z_k-z_j)^2} =  \frac{1}{3}(2n-(z_k)^2-2)$$
(everytime the sums have to be understood without the singular terms $1/0$).
Now define
$$ x_j := \alpha\, z_j$$
for some $\alpha >0$. The above formula gives
$$ \sum_{j=1}^n \frac{1}{(x_k-x_j)^2} =  \frac{1}{3\alpha^2}(2n-\alpha^{-2}(x_k)^2-2).$$
Notice that the $x_j$ are symetrics with respect to $x = 0$ (if $z$ is a zero of the Hermite polynomial of order $n$, then $-z$ too), we have
$$\sum_{ 1 \leq i < j \leq n} \frac{1}{(x_i-x_j)^2} = \frac{1}{2} \sum_{k=1}^n \sum_{j=1}^n \frac{1}{(x_k-x_j)^2}
 = \frac{n(n-1)}{3\alpha^2} - \frac{1}{6\alpha^4} \sum_{k=1}^n (x_k)^2.$$
Now still using the symmetry of the $x_j$, it is not hard to show that
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 = n \sum_{k=1}^n (x_k)^2.$$
So one has
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 +  \frac{1}{(x_i-x_j)^2} =  \frac{n(n-1)}{3\alpha^2} + \left(n-\frac{1 }{6\alpha^4}\right) \sum_{k=1}^n (x_k)^2.$$
Now we use another "well-known"
 formula (see Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$?)
$$\sum_{k=1}^n (z_k)^2 = \frac{n(n-1)}2 $$
so one has
$$ \sum_{k=1}^n (x_k)^2 = \alpha^2\frac{n(n-1)}2 $$
and
$$ \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 +  \frac{1}{(x_i-x_j)^2} = \frac{n(n-1)}{4} \frac{2 n \alpha^4 +1}{\alpha^2}.$$
The above quantity is minimal for
$$ \alpha = (2n)^{-1/4}$$
and in this case we have
$$   \sum_{ 1 \leq i < j \leq n} (x_i-x_j)^2 +  \frac{1}{(x_i-x_j)^2} = \frac{n(n-1)}{2}\sqrt{2n} \quad \text{for $x_k = (2n)^{-1/4}  z_k$}.$$
</s>