15 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, find
$$\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}$$

the minimum value.
15. Let $b_{i}=2-a_{i} \geqslant 0(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n} b_{i}=2 n-1$. By the inequality of arithmetic and geometric means, we get $\sum_{i=1}^{n} \frac{a_{i}}{b_{i}}=\sum_{i=1}^{n}\left(\frac{a_{i}}{b_{i}}-1\right)-n=2 \sum_{i=1}^{n} \frac{1}{b_{i}}-n \geqslant \frac{2 n}{\sqrt[n]{b_{1} \cdots b_{n}}}-n \geqslant$ $\frac{2 n^{2}}{b_{1}+\cdots+b_{n}}-n=\frac{2 n^{2}}{2 n-1}-n=\frac{n}{2 n-1}$. When $a_{1}=\cdots=a_{n}=\frac{1}{n}$, $\sum_{i=1}^{n} \frac{a_{i}}{2-a_{i}}=$ $\frac{n}{2 n-1}$, hence the minimum value is $\frac{n}{2 n-1}$.