Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}<a_{i}$. Then

$$b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}<\frac{2 a_{i}}{a_{i}}=2.$$

But since $a_{i}$ is maximal, $b_{i}$ is also maximal, so we must have $b_{j}<2$ for all $j \in\{1,2, \ldots, n\}$. However, consider the product $b_{1} b_{2} \ldots b_{n}$. We have

$$\begin{aligned} b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\ & \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\ & =2^{n} \end{aligned}$$

where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.

Since the product of all $b_{i}$ is at least $2^{n}$, at least one of them must be greater than 2, which is a contradiction with the previous conclusion.

Thus, all $a_{i}$ must be equal.