9.61 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
$$
a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=1 \text {, }
$$

then $\quad a_{1} a_{2} \cdots a_{n} \leqslant \lambda\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right)$.
【Solution】By Cauchy-Schwarz inequality, we have
$$
1=\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} \frac{\sqrt{b_{i}}}{\sqrt{a_{i}}} \sqrt{a_{i} b_{i}} \leqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{\frac{1}{2}}\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{\frac{1}{2}} \text {. }
$$

Thus,
$$
\frac{1}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} \frac{b_{i}}{a_{i}}
$$

Let $M=a_{1} a_{2} \cdots a_{n}, A_{i}=\frac{M}{a_{i}}, i=1,2, \cdots, n$, then
$$
\frac{M}{\sum_{i=1}^{n} a_{i} b_{i}} \leqslant \sum_{i=1}^{n} b_{i} A_{i}
$$

By the rearrangement inequality, without loss of generality, assume $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n}$. Thus,
$$
\sum_{i=1}^{n} b_{i} A_{i} \leqslant b_{1} A_{1}+\left(1-b_{1}\right) A_{2},
$$

Since $0 \leqslant b_{1} \leqslant \frac{1}{2}, A_{1} \geqslant A_{2}$, we have
$$
\sum_{i=1}^{n} b_{i} A_{i} \leqslant-\frac{1}{2}\left(A_{1}+A_{2}\right)=\frac{1}{2}\left(a_{1}+a_{2}\right) a_{3} \cdots a_{n} \text {. }
$$

By the AM-GM inequality, noting that $\sum_{i=1}^{n} a_{i}=1$, we get
$$
\sum_{i=1}^{n} b_{i} A_{i} \leqslant \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} \text {. }
$$

Therefore, the required $\lambda \leqslant \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}$.
On the other hand, when $a_{1}=a_{2}=\frac{1}{2(n-1)}, a_{3}=\cdots=a_{n}=\frac{1}{n-1}$,
$$
\begin{array}{l}
b_{1}=b_{2}=\frac{1}{2}, b_{3}=\cdots=b_{n}=0 \text {, } \\
a_{1} a_{2} \cdots a_{n}=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1} \sum_{i=1}^{n} a_{i} b_{i},
\end{array}
$$

Thus, $\lambda \geqslant \frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}$.
In conclusion, $\lambda=\frac{1}{2}\left(\frac{1}{n-1}\right)^{n-1}$.