Question 11 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2},$ $\cdots, b_{n}$ satisfying the conditions $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+$ $\cdots+b_{n}=1$. Find
$$
\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}
$$

the minimum value.
(2003, Belarus Mathematical Olympiad)
Solution: By Cauchy-Schwarz inequality, we have
$$
\sum_{t=1}^{n} \frac{a_{t}^{2}}{a_{t}+b_{t}} \geqslant \frac{\left(\sum_{t=1}^{n} a_{t}\right)^{2}}{\sum_{t=1}^{n}\left(a_{t}+b_{t}\right)}=\frac{1}{2},
$$

with equality if and only if $a_{1}=a_{2}=\cdots=a_{n}=b_{1}=b_{2}=\cdots=$ $b_{n}$, in which case the minimum value $\frac{1}{2}$ is achieved.