Prove that since $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$, we have
$$\begin{aligned} & \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n-1}\right) \\ \geqslant & (1+1)(1+2) \cdots[1+(n-1)] \\ = & a_{1} a_{2} \cdots a_{n} . \end{aligned}$$

Thus,
$$\begin{aligned} & \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n} \\ = & \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \\ = & \frac{1}{a_{1}}+\frac{1+a_{1}}{a_{2}}+\frac{1+a_{2}}{a_{3}}+\cdots+\frac{1+a_{n-1}}{a_{n}} \\ \geqslant & n \sqrt[n]{\frac{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n-1}\right)}{a_{1} a_{2} \cdots a_{n}}} \geqslant n \end{aligned}$$

Since $n=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}\right)$, we have
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}} \geqslant \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n}$$