Prove that by Cauchy's inequality,
$$\begin{array}{l} \sum_{i=1}^{n} \sqrt{x_{i}} \sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}}=\sum_{i=1}^{n} \sqrt{x_{i}}\left(\sum_{i=1}^{n} \sqrt{1+x_{i}}-\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{i}}}\right) \leqslant \\ \sum_{i=1}^{n} \sqrt{x_{i}}\left(\sum_{i=1}^{n} \sqrt{1+x_{i}}-\frac{\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}}{\sum_{i=1}^{n} \sqrt{1+x_{i}}}\right) \leqslant \\ \sum_{i=1}^{n} \sqrt{x_{i}}\left(\sqrt{n(n+1)}-\frac{\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}}{\sqrt{n(n+1)}}\right) \end{array}$$

Let $\sum_{i=1}^{n} \sqrt{x_{i}}=y$, then $0<y \leqslant \sqrt{n}$, it suffices to prove
$$\begin{array}{l} y\left(\sqrt{n(n+1)}-\frac{y^{2}}{\sqrt{n(n+1)}}\right) \leqslant \frac{n^{2}}{\sqrt{n+1}} \Leftrightarrow \\ y^{3}-(n+1) n y+n^{2} \sqrt{n} \geqslant 0 \\ (y-\sqrt{n})\left(y^{2}+\sqrt{n} y-n^{2}\right) \geqslant 0 \end{array}$$

The above inequality is obvious, and the proposition is proved! Equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$.