Example 6 Given a natural number $n \geqslant 2$, find the smallest positive number $\lambda$, such that for any $a_{i} \geqslant 0,0 \leqslant b_{i} \leqslant \frac{1}{2}(i=1,2, \cdots$, $n)$, and $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, we have $a_{1} a_{2} \cdots a_{n} \leqslant \lambda \sum_{i=1}^{n} a_{i} b_{i}$.
Solution: If there exists some $j(j=1,2, \cdots, n)$ such that $a_{j}=0$, then the original inequality holds for any $\lambda \geqslant 0$. Therefore, we can assume that for all $i$, $a_{i}>0$.
$\because f(x)=\frac{1}{x}$ is a convex function on $(0,+\infty)$ and $\sum_{i=1}^{n} b_{i}=1$, hence by Jensen's inequality we have
$$
\sum_{i=1}^{n} b_{i} f\left(a_{i}\right) \geqslant f\left(\sum_{i=1}^{n} a_{i} b_{i}\right),
$$

which means $\square$
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{b_{i}}{a_{t}} \geqslant \frac{1}{\sum_{i=1}^{n} a_{t} b_{i}}, \\
\prod_{i=1}^{n} a_{i} b_{i} \geqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{-1} .
\end{array}
$$

By the rearrangement inequality and the form of the last inequality, we can assume without loss of generality that $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}$ and $a_{1} \leqslant$ $a_{2} \leqslant \cdots \leqslant a_{n}$, and let $M=a_{1} a_{2} \cdots a_{n}, A_{i}=\frac{M}{a_{i}}(i=$ $1,2, \cdots, n)$. Then
$$
\begin{array}{l}
A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n} \geqslant 0 . \\
\therefore \sum_{i=1}^{n} b_{1} A_{i} \leqslant b_{1} A_{1}+b_{2} A_{2}+\cdots+b_{n} A_{2} \\
=b_{1} A_{1}+\left(1-b_{1}\right) A_{2} \\
=\frac{1}{2} A_{1}+\frac{1}{2} A_{2}-\left(\frac{1}{2}-b_{1}\right)\left(A_{1}-A_{2}\right) \\
\leqslant \frac{1}{2}\left(A_{1}+A_{2}\right) \\
=\frac{1}{2}\left(a_{1}+a_{2}\right) a_{3} a_{4} \cdots a_{n} \\
\leqslant \frac{1}{2}\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n-1}\right)^{n-1} \\
=\frac{1}{2(n-1)^{n-1}},
\end{array}
$$

which means $M \leqslant \frac{1}{2(n-1)^{n-1}} \cdot\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{-1}$. 
Thus, from $\sum_{i=1}^{n} a_{i} b_{i} \geqslant\left(\sum_{i=1}^{n} \frac{b_{i}}{a_{i}}\right)^{-1}$ we get
$$
a_{1} a_{2} \cdots a_{n} \leqslant \frac{1}{2(n-1)^{n-1}} \sum_{i=1}^{n} a_{i} b_{i} \text {. }
$$

Furthermore, when $b_{1}=b_{2}=\frac{1}{2}, b_{3}=b_{4}=\cdots=b_{n}=0$;
$$
a_{1}=a_{2}=\frac{1}{2(n-1)}, a_{3}=a_{4}=\cdots=a_{n}=\frac{1}{n-1}
$$

we have
$$
\sum_{i=1}^{n} a_{i}=\searrow_{i=1}^{n} b_{i}=1 .
$$
G. $a_{1} a_{2} \cdots a_{n}=\frac{1}{4(n-1)^{n}}$

and $\frac{1}{2(n-1)^{n-1}} \sum_{i=1}^{n} a_{i} b_{i}$
$$
\begin{array}{l}
=\frac{1}{2(n-1)^{n-1}} \cdot\left[\frac{1}{4(n-1)}+\frac{1}{4(n-1)}\right] \\
=\frac{1}{4(n-1)^{n}} .
\end{array}
$$

In summary, the smallest positive number is
$$
\lambda=\frac{1}{2(n-1)^{n-1}} .
$$