Minimum of the value $\sum_{1\le k,i,j\le n}\frac{\sigma{(A_{k}\bigcap A_{i}\bigcap A_{j})}}{\sigma{(A_{k})}\cdot\sigma{(A_{i})}\cdot\sigma{(A_{j})}}$ 
Let $m,n$ are given positive integers, and  positive real numbers $x_{1}<x_{2}<\cdots<x_{m}$ are given.
  Define $A=\{x_{1},x_{2},\cdots,x_{m}\}$.
  Find the following minimum of the value
  $$\sum_{k=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{\sigma{(A_{k}\bigcap A_{i}\bigcap A_{j})}}{\sigma{(A_{k})}\cdot\sigma{(A_{i})}\cdot\sigma{(A_{j})}}$$
where $A_{1},A_{2},\cdots,A_{n}\subset A$, and $\sigma{(A)}$ denote the sum of the elements of the set $A$.

I conjecture this answer is $$\color{red}{\dfrac{n^3}{(x_{1}+x_{2}+x_{3}+\cdots+x_{m})^2}}$$
because when $A_{1}=A_{2}=\cdots=A_{n}=A$,we have
\begin{align*}\sum_{k=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{\sigma{(A_{k}\bigcap A_{i}\bigcap A_{j})}}{\sigma{(A_{k})}\cdot\sigma{(A_{i})}\cdot\sigma{(A_{j})}}&=\sum_{k=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{(x_{1}+x_{2}+\cdots+x_{m})}{(x_{1}+x_{2}+\cdots+x_{m})^3}\\
&=\color{red}{\dfrac{n^3}{(x_{1}+x_{2}+x_{3}+\cdots+x_{m})^2}}.\end{align*}
But I can't prove it.
I conjecture： The following for $p\geq 2$,
$$\min{\left(\sum_{x_{1}=1}^{n}\sum_{x_{2}=1}^{n}\cdots\sum_{x_{p}=1}^{n}
\dfrac{\sigma{(A_{x_{1}}\bigcap A_{x_{2}}\bigcap A_{x_{3}}}\cdots A_{x_{p}})}{\sigma{(A_{x_{1}})}\cdot \sigma{(A_{x_{2}})}\cdots \sigma{(A_{x_{p}})}}\right)}=\dfrac{n^p}{(x_{1}+x_{2}+\cdots+x_{m})^{p-1}}.$$
 A: The sum can be written in terms of indicator functions as
$$
\begin{align}
\sum_{i,j,k=1}^n  \sum_{s\in A} \frac{s \mathbf{1}_{A_i}(s)\mathbf{1}_{A_j}(s)\mathbf{1}_{A_k}(s)}{\sigma(A_i)\sigma(A_j)\sigma(A_k)}&=\sum_{s\in A}  \sum_{i,j,k=1}^n \frac{s \mathbf{1}_{A_i}(s)\mathbf{1}_{A_j}(s)\mathbf{1}_{A_k}(s)}{\sigma(A_i)\sigma(A_j)\sigma(A_k)}\\
&=\sum_{s\in A} s\left( \sum_{i=1}^n \frac{\mathbf{1}_{A_i}(s)}{\sigma(A_i)}\right)^3=\sum_{s\in A} s v_s ^3,
\end{align}
$$
where
$$
v_s = \sum_{i=1}^n \frac{\mathbf{1}_{A_i}(s)}{\sigma(A_i)}, 
$$
On the other hand, we have
$$
\begin{align}
\sum_{s\in A} s v_s &=  \sum_{s\in A} s\sum_{i=1}^n \frac{\mathbf{1}_{A_i}(s)}{\sigma(A_i)}
=\sum_{i=1}^n \frac{\sum_{s\in A} s \mathbf{1}_{A_i}(s)}{\sigma(A_i)}=\sum_{i=1}^n \frac{\sigma(A_i)}{\sigma(A_i)} = n.
\end{align}
$$
Then we have to
$$
\textrm{Minimize } \ \sum_{s\in A} s v_s^3, \ \ \textrm{with a constraint} \ \ \sum_{s\in A} s v_s = n.
$$
Applying Lagrange multipliers on the variables $v_s>0$, $s\in A$, we obtain that
$$
s = \lambda \cdot 3  s v_s^2, \ \  \textrm{for each} \ s\in A.
$$
This shows that the mimimizing $v_s$ occurs when all $v_s$ are identical, say $v_s=v$ for all $s\in A$. 
Now, plugging these in the constraint, we have
$$
\sum_{s\in A} s v = n, \ \ \textrm{equivalently} \  \ v= \frac{n}{\sum\limits_{s\in A} s}. 
$$
Then we have
$$
\sum_{s\in A} s v_s^3 = \sum_{s\in A}s \left( \frac{n}{\sum\limits_{s\in A} s}\right)^3 = \frac{n^3}{\left(\sum\limits_{s\in A} s\right)^2}.
$$
This is what has been conjectured. The same method can be applied to the general conjecture for $p\geq 2$. 
