Example 3 $n$ is a positive integer, $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$ are $2 n$ positive real numbers, satisfying $a_{1}+a_{2}+\cdots+a_{n}=1, b_{1}+b_{2}+\cdots+b_{n}=1$, find the minimum value of $\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}$.
By Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}\right)\left(\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}\right) \\
\geqslant & \left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=1
\end{aligned}$$

and given that $\quad a_{1}+a_{2}+\cdots+a_{n}+b_{1}+b_{2}+\cdots+b_{n}=2$,
thus $\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}} \geqslant \frac{1}{2}$,
and the equality holds when $a_{1}=a_{2}=\cdots=a_{n}=b_{1}=b_{2}=\cdots=b_{n}=\frac{1}{n}$.
Therefore, the minimum value of $\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}$ is $\frac{1}{2}$.