Example 1 Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+$ $a_{2}+\cdots+a_{n}=1$, find
$$
\begin{array}{l} 
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots \\
+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}} \text { 's minimum value. }
\end{array}
$$
Solution: Assign a constant 1 to each fraction in the required expression, then find a common denominator. After that, replace $a_{1}+a_{2}+\cdots+a_{n}$ with the constant 1, we get $\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+1=$
$$
\frac{1+\left(a_{1}+a_{2}+\cdots+a_{n}\right)}{2-a_{1}}=\frac{2}{2-a_{1}},
$$

Similarly, $\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+1=\frac{2}{2-a_{2}}, \cdots$, $\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}+1=\frac{2}{2-a_{n}}$,

Let $y=\sum_{i=1}^{n} \frac{a_{i}}{1-a_{i}+\sum_{j=1}^{n} a_{j}}$, then $y+n=$ $\frac{2}{2-a_{1}}+\frac{2}{2-a_{2}}+\cdots+\frac{2}{2-a_{n}}$.

To use the Cauchy-Schwarz inequality, note that $\sum_{i=1}^{n}\left(2-a_{i}\right)$ $=2 n-\sum_{i=1}^{n} a_{i}=2 n-1$,

Then $(2 n-1) \sum_{i=1}^{n} \frac{1}{2-a_{i}}=\sum_{i=1}^{n}\left(2-a_{i}\right) \cdot \sum_{i=1}^{n}$ $\frac{1}{2-a_{i}} \geqslant\left(\sum_{i=1}^{n} \sqrt{2-a_{i}} \cdot \frac{1}{\sqrt{2-a_{i}}}\right)^{2}=n^{2}$.

Therefore, $y+n \geqslant \frac{2 n^{2}}{2 n-1}$, which means $y \geqslant \frac{2 n^{2}}{2 n-1}-n=$ $\frac{n}{2 n-1}$.

The equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$. Thus, the minimum value of $y$ is $\frac{n}{2 n-1}$.