15 Given a natural number $n \geqslant 2$, find the smallest positive number $\lambda$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, and any $n$ numbers $b_{1}, b_{2}, \cdots, b_{n}$ in $\left[0, \frac{1}{2}\right]$, if
$$\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1,$$

then
$$\prod_{i=1}^{n} a_{i} \leqslant \lambda \sum_{i=1}^{n} a_{i} b_{i}$$
15. By Cauchy's inequality, we have \(1 = \sum_{i=1}^{n} b_{i} \leqslant \left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}} \left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}}\), thus \(\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\). Let \(M = \prod_{i=1}^{n} a_{i}, A_{i} = \frac{M}{a_{i}}, i=1,2, \cdots, n\), then \(\frac{M}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} b_{i} A_{i}\). Assume \(b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n}\), by the rearrangement inequality, we get \(\sum_{i=1}^{n} b_{i} A_{i} \leqslant b_{1} A_{1} + (1 - b_{1}) A_{2}\). Since \(0 \leqslant b_{1} \leqslant \frac{1}{2}, A_{1} \geqslant A_{2}\), we have \(\sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2}(A_{1} + A_{2}) = \frac{1}{2}(a_{1} + a_{2}) a_{3} \cdots a_{n}\). By the AM-GM inequality, we get \(\sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1}\), so \(\lambda \leqslant \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1}\). On the other hand, when \(a_{1} = a_{2} = \frac{1}{2(n-1)}\), \(a_{3} = \cdots = a_{n} = \frac{1}{n-1}\), \(b_{1} = b_{2} = \frac{1}{2}\), \(b_{3} = \cdots = b_{n} = 0\), we have \(\prod_{i=1}^{n} a_{i} = \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1} \sum_{i=1}^{n} a_{i} b_{i}\), so \(\lambda \geqslant \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1}\). Therefore, \(\lambda_{\min} = \frac{1}{2} \left(\frac{1}{n-1}\right)^{n-1}\).