Three, (50 points) For a set $Y$ consisting of a finite number of real numbers, define $\sigma(Y)$ as the sum of all elements in $Y$, $\sigma(Y)=\sum_{y \in Y} y$. Given positive integers $m, n$ and positive real numbers $x_{1}<x_{2}<\cdots<x_{m}$, let $A_{1}, A_{2}, \cdots, A_{n}$ be non-empty subsets of the set $\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}$. Find the minimum value of the expression
$$
\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\sigma\left(A_{i} \cap A_{j}\right)}{\sigma\left(A_{i}\right) \sigma\left(A_{j}\right)}
$$
$$
\begin{array}{l}
=\sum_{k=1}^{m}\left(x_{k}\left(\sum_{\substack{1 \leq i \leq 1 \\ x \leq 1}} \frac{1}{\sigma\left(A_{i}\right)}\right)^{2}\right) . \\
\text { By Cauchy-Schwarz inequality, } \\
\left(\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\sigma\left(A_{i} \cap A_{j}\right)}{\sigma\left(A_{i}\right) \sigma\left(A_{j}\right)}\right)\left(x_{1}+x_{2}+\cdots+x_{m}\right) \\
=\sum_{k=1}^{m}\left(x_{k}\left(\sum_{\substack{1 \leq i \leq n \\ x_{k} \in A_{i}}} \frac{1}{\sigma\left(A_{i}\right)}\right)^{2}\right) \sum_{i=1}^{m} x_{i} \\
\geqslant\left(\sum_{k=1}^{m}\left(x_{k} \sum_{\substack{1 \leq i \leq n \\ x_{k} \in A_{i}}} \frac{1}{\sigma\left(A_{i}\right)}\right)\right)^{2} \\
=\left(\sum_{i=1}^{n} \frac{\sigma\left(A_{i}\right)}{\sigma\left(A_{i}\right)}\right)^{2}=n^{2} \text {. } \\
\text { The last two steps are calculated using Table } 2 \text { . } \\
\text { Table } 2 \\
\text { Then }\left(\sum_{k=1}^{m}\left(x_{k} \sum_{\substack{1 \in i \leqslant i \leqslant \\ x_{k} \in A_{i}}} \frac{1}{\sigma\left(A_{i}\right)}\right)\right)^{2} \\
=\left(\sum_{i=1}^{n} \frac{\sigma\left(A_{i}\right)}{\sigma\left(A_{i}\right)}\right)^{2}=n^{2} \text {, } \\
\left(\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\sigma\left(A_{i} \cap A_{j}\right)}{\sigma\left(A_{i}\right) \sigma\left(A_{j}\right)}\right)\left(x_{1}+x_{2}+\cdots+x_{m}\right) \geqslant n^{2} \\
\Leftrightarrow\left(\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\sigma\left(A_{i} \cap A_{j}\right)}{\sigma\left(A_{i}\right) \sigma\left(A_{j}\right)}\right) \geqslant \frac{n^{2}}{x_{1}+x_{2}+\cdots+x_{m}} \text {. } \\
\end{array}
$$
\begin{tabular}{|c|c|c|c|c|c|}
\hline & $x_{1}$ & $x_{2}$ & $\cdots$ & $x_{m-1}$ & $x_{m}$ \\
\hline$A_{1} \cap A_{1}$ & & & & & \\
\hline$\vdots$ & & & & & \\
\hline$A_{1} \cap A_{n}$ & & & & & \\
\hline$\vdots$ & & & & & \\
\hline$A_{i} \cap A_{j}$ & & & & & \\
\hline$\vdots$ & & & & & \\
\hline$A_{n} \cap A_{n}$ & & & & & \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|c|c|c|}
\hline & $x_{1}$ & $x_{2}$ & $\cdots$ & $x_{m-1}$ & $x_{m}$ \\
\hline$A_{1}$ & & & & & \\
\hline$A_{2}$ & & & & & \\
\hline$\vdots$ & & & & & \\
\hline$A_{n}$ & & & & & \\
\hline
\end{tabular}

Filling principle: If $x_{k} \in A_{i}$, then fill $\frac{x_{k}}{\sigma\left(A_{i}\right)}$ in the corresponding position; if $x_{k} \notin A_{i}$, then fill 0 in the corresponding position. Therefore, the sum of each column is $\sum_{k=1}^{m}\left(x_{k} \sum_{\substack{1 \leq i \leq n \\ x_{k} \in A_{i}}} \frac{1}{\sigma\left(A_{i}\right)}\right)$, and the sum of each row is $\sum_{i=1}^{n} \frac{\sigma\left(A_{i}\right)}{\sigma\left(A_{i}\right)}$.

That is $\square$
$$