One, (40 points) Given $\sum_{i=1}^{n} a_{i} \leqslant n, \sum_{j=1}^{n} b_{j} \leqslant n$. Find
$$
\sqrt{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{1}{a_{i}}+\frac{1}{b_{j}}}-\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{a_{i}^{2}+b_{j}^{2}}{a_{i}^{2} b_{j}+a_{i} b_{j}^{2}}}
$$

the minimum value.
$$
\text { I. Let } \frac{1}{a_{i}}=A_{i}, \frac{1}{b_{j}}=B_{j} \text {. }
$$

Then the required expression is
$$
=\sqrt{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{A_{i}+B_{j}}-\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{A_{i}^{2}+B_{j}^{2}}{A_{i}+B_{j}}} \text {. }
$$

Notice that,
$$
\begin{array}{l}
\sqrt{2}\left(A_{i}+B_{j}\right)=\sqrt{2} \sqrt{\left(A_{i}^{2}+B_{j}^{2}\right)+2 A_{i} B_{j}} \\
\geqslant \sqrt{A_{i}^{2}+B_{j}^{2}}+\sqrt{2 A_{i} B_{j}} \\
\Rightarrow \sqrt{2} \sqrt{A_{i}+B_{j}} \geqslant \sqrt{\frac{A_{i}^{2}+B_{j}^{2}}{A_{i}+B_{j}}}+\sqrt{\frac{2 A_{i} B_{j}}{A_{i}+B_{j}}} . \\
\text { Then expression (1) } \geqslant \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{2 A_{i} B_{j}}{A_{i}+B_{j}}} \\
=\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{2}{a_{i}+b_{j}}} \\
=\sqrt{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{1}{a_{i}+b_{j}}} . \\
\text { Also, } \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\frac{1}{a_{i}+b_{j}}} \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}} \\
\geqslant\left(n^{2}\right)^{2}=n^{4},
\end{array}
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
2 n^{4}=2 n^{2} \cdot n^{2} \geqslant \sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i}+b_{j}\right) n^{2} \\
\geqslant\left(\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}}\right)^{2} \\
\Rightarrow \sqrt{2} n^{2} \geqslant \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}} . \\
\text { Therefore, expression (2) } \geqslant \frac{\sqrt{2} n^{4}}{\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}}} \\
\geqslant \frac{n^{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}}}{\sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{a_{i}+b_{j}}}=n^{2} .
\end{array}
$$

Thus, the minimum value of the required expression is \( n^{2} \) when and only when
$$
a_{1}=a_{2}=\cdots=a_{n}=b_{1}=b_{2}=\cdots=b_{n}=1
$$