Let $a_j$ be non-negative real numbers, try to find the optimal constant $c(n)$, such that $\sum_{j=1}^{n} \frac{a_{j}\left(s_{n}-x_{j}\right)}{x_{j}} \geqslant c(n)\left(\prod_{j=1}^{n} a_{j}\right)^{\frac{1}{n}}$.
Parse left $=\sum_{1 \leqslant i<j \leqslant n}\left(\frac{a_{j} x_{j}}{x_{j}}+\frac{a_{i} x_{i}}{x_{i}}\right) \geqslant 2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{a_{i} a_{j}}$,
$$
\geqslant 2 \cdot \frac{n(n-1)}{2} \cdot \sqrt[\frac{m(n-1)}{2}]{\prod_{i=1}^{n} a_{i}^{\frac{n-1}{2}}}=n(n-1)\left(\prod_{j=1}^{n} a_{j}\right)^{\frac{1}{n}},
$$

Taking $x_{1}=x_{2}=\cdots=x_{n}=1, a_{1}=a_{2}=\cdots=a_{n}=1$, we get $c(n) \leqslant n(n-1)$, hence $c(n)=n(n-1)$ is the optimal constant.