9.45 Given $0<a_{1}<a_{2}<\cdots<a_{n}$, for any permutation $b_{1}, b_{2}, \cdots, b_{n}$ of $a_{1}, a_{2}, \cdots, a_{n}$, let
$$
M=\prod_{i=1}^{n}\left(a_{i}+\frac{1}{b_{i}}\right) .
$$

Find the permutation $b_{1}, b_{2}, \cdots, b_{n}$ that maximizes $M$.
[Solution] Let $A=a_{1} a_{2} \cdots a_{n}$, then
$$
M=\frac{1}{A} \prod_{i=1}^{n}\left(a_{i} \cdot b_{i}+1\right) \text {. }
$$

Since $\left(a_{i} \cdot b_{i}+1\right)^{2}=a_{i}^{2} \cdot b_{i}^{2}+2 a_{i} \cdot b_{i}+1 \leqslant a_{i}^{2} b_{i}^{2}+a_{i}^{2}+b_{i}^{2}+1$,
we have
$$
\left(a_{i} b_{i}+1\right) \leqslant \sqrt{\left(a_{i}^{2}+1\right)\left(b_{i}^{2}+1\right)},
$$

and the equality holds if and only if $a_{i}=b_{i}$. Therefore, we get
$$
M \leqslant \frac{1}{A} \prod_{i=1}^{n}\left(1+a_{i}^{2}\right),
$$

and the equality holds $\Leftrightarrow a_{i}=b_{i}, i=1,2, \cdots, n$, that is, $M$ reaches its maximum value if and only if $b_{1}=a_{1}, b_{2}=a_{2}$, $\cdots, b_{n}=a_{n}$.