Example 5 Let non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$. Find
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+ \\
\cdots+\frac{a_{n}}{1+a_{1}+\cdots+a_{n-1}}
\end{array}
$$

the minimum value.
Explanation: Assign a constant 1 to each fraction in the expression, then find a common denominator. After that, replace $a_{1}+a_{2}+\cdots+a_{n}$ with the constant 1, and we get
$$
\begin{array}{l}
\frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+1 \\
=\frac{1+\left(a_{1}+a_{2}+\cdots+a_{n}\right)}{2-a_{1}}=\frac{2}{2-a_{1}} .
\end{array}
$$

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

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
$$
\sum_{i=1}^{n}\left(2-a_{i}\right)=2 n-\sum_{i=1}^{n} a_{i}=2 n-1 \text {, }
$$

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

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

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