Prove that because $\sum_{i=1}^{n} x_{i}=1$,
so, by the AM-GM inequality, we have
$$\begin{array}{l} \sqrt{\left(1+x_{0}+x_{1}+\cdots+x_{i-1}\right)\left(x_{i}+x_{i+1}+\cdots+x_{n}\right)} \\ \leqslant \frac{1+x_{0}+x_{1}+\cdots+x_{n}}{2}=1 . \end{array}$$

Thus, $\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+x_{i+1}+\cdots+x_{n}}} \geqslant \sum_{i=1}^{n} x_{i}=1$.
That is, the left inequality we need to prove holds.
Since $0 \leqslant x_{0}+x_{1}+\cdots+x_{i} \leqslant 1, i=1,2, \cdots, n$,
we can let $\theta_{i}=\arcsin \left(x_{0}+x_{1}+\cdots+x_{i}\right), i=1,2, \cdots, n$.
Then, $\theta_{i} \in\left[0, \frac{\pi}{2}\right]$, and we have $0=\theta_{0}<\theta_{1}<\cdots<\theta_{n}=\frac{\pi}{2}$,
thus, we have $x_{i}=\sin \theta_{i}-\sin \theta_{i-1}=2 \cos \frac{\theta_{i}+\theta_{i-1}}{2} \cdot \sin \frac{\theta_{i}-\theta_{i-1}}{2}<2 \cos \theta_{i-1} \cdot \sin \frac{\theta_{i}-\theta_{i-1}}{2}$.
Since, for $x \in\left[0, \frac{\pi}{2}\right]$, we have $\sin x<x$, so we have
$$x_{i}<2 \cos \theta_{i-1} \cdot \frac{\theta_{i}-\theta_{i-1}}{2}=\left(\theta_{i}-\theta_{i-1}\right) \cdot \cos \theta_{i-1} \text {, }$$

Therefore, $\frac{x_{i}}{\cos \theta_{i-1}}<\theta_{i}-\theta_{i-1}, i=1,2, \cdots, n$.
Summing both sides of the above inequality from $i=1$ to $n$, we get
$$\sum_{i=1}^{n} \frac{x_{i}}{\cos \theta_{i-1}}<\theta_{n}-\theta_{0}=\frac{\pi}{2} \text {. }$$

By the definition of $\theta_{i}$, we know $\sin \theta_{i}=x_{1}+x_{2}+\cdots+x_{i}$,
so $\cos \theta_{i-1}=\sqrt{1-\sin ^{2} \theta_{i-1}}=\sqrt{1-\left(x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}\right)^{2}}$
$$=\sqrt{\left(1+x_{0}+x_{1}+\cdots+x_{i-1}\right)\left(x_{i}+x_{i+1}+\cdots+x_{n}\right)} \text {. }$$

Substituting (2) into (1), we obtain the right inequality we need to prove.