Problem 1322

Proof: Without loss of generality, assume $a_{1} a_{2} \cdots a_{n}=1$ (normalization). Let $a_{1}=\frac{x_{1}}{x_{2}}, a_{2}=\frac{x_{2}}{x_{3}}, \cdots, a_{n-1}=\frac{x_{n-1}}{x_{n}}$, where $x_{1}, x_{2}, \cdots, x_{n}>0$, then $a_{n}=\frac{x_{n}}{x_{1}}$. The inequality becomes
$$\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\cdots+\frac{x_{n-1}}{x_{n}}+\frac{x_{n}}{x_{1}} \geq n$$

Notice that if the sequence $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is increasing, then the sequence $\left(\frac{1}{x_{1}}, \frac{1}{x_{2}}, \cdots, \frac{1}{x_{n}}\right)$ is decreasing. By the rearrangement inequality, we get
$$\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}}=\sum_{i=1}^{n} x_{i} \cdot \frac{1}{x_{i+1}} \geq \sum_{i=1}^{n} x_{i} \cdot \frac{1}{x_{i}}=n$$