9 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 $M$.
9. Let $A=a_{1} a_{2} \cdots a_{n}$, then $M=\frac{1}{A} \prod_{i=1}^{n}\left(a_{i} b_{i}+1\right)$. From $\left(a_{i} b_{i}+1\right)^{2} \leqslant\left(a_{i}^{2}+1\right)$ $\left(b_{i}^{2}+1\right)$, we know that equality holds $\Leftrightarrow a_{i}=b_{i}$. This leads to $M \leqslant \frac{1}{A} \prod_{i=1}^{n}\left(1+a_{i}^{2}\right)$, and equality holds $\Leftrightarrow$ $a_{i}=b_{i}(i=1,2, \cdots, n)$. Therefore, when $b_{1}=a_{1}, b_{2}=a_{2}, \cdots, b_{n}=a_{n}$, $M$ attains its maximum value.