Example 5 Let $n \in \mathbf{N}_{+}, a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers, and $\sum_{i=1}^{n} a_{i}=1, \sum_{i=1}^{n} b_{i}=1$. Find the minimum value of $\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}$.
(2004, French Team Selection Exam)
Solution: By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}\right) \\
\geqslant\left[\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}+b_{i}}} \sqrt{a_{i}+b_{i}}\right]^{2} \\
=\left(\sum_{i=1}^{n} a_{i}\right)^{2}=1 . \\
\text { Then } \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} .
\end{array}
$$
$$
\text { When } a_{1}=a_{2}=\cdots=a_{n}=b_{1}=\cdots=b_{n}=\frac{1}{n}
$$

the equality holds, at this time, $\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}$ takes the minimum value $\frac{1}{2}$.