Stars of Mathematics Competition

We prove that, for all $a_1, a_2, \dots, a_n > 0$, the following inequality holds
$$\left(\frac{a_1}{a_2}\right)^{n-1} + \left(\frac{a_2}{a_3}\right)^{n-1} + \dots + \left(\frac{a_{n-1}}{a_n}\right)^{n-1} + \left(\frac{a_n}{a_1}\right)^{n-1} \geq \frac{a_1^2 + a_2^2 + \dots + a_n^2}{\sqrt[n]{a_1^2 a_2^2 \dots a_n^2}}.$$
This inequality is obtained by adding the inequality below with its analogues obtained by cyclic permutation of the variables:
$$\begin{aligned} & (n-1) \cdot \left(\frac{a_1}{a_2}\right)^{n-1} + (n-2) \cdot \left(\frac{a_2}{a_3}\right)^{n-1} + \dots + \left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq \\ & \geq \frac{n(n-1)}{2} \cdot \left( \left(\frac{a_1}{a_2}\right)^{n-1} \left(\frac{a_2}{a_3}\right)^{n-2} \dots \left(\frac{a_{n-1}}{a_n}\right)^{\frac{2}{n}} \right) = \\ & = \frac{n(n-1)}{2} \left( \frac{a_1^n}{a_1 a_2 a_3 \dots a_n} \right)^{\frac{2}{n}}. \end{aligned}$$
Equality holds if all the numbers are equal to $1$.