Stars of Mathematics Competition

We prove that, for all $a_1,a_2,\dots ,a_n>0$, the following inequality holds $${\left(\frac{a_1}{a_2}\right)}^{n-1}+{\left(\frac{a_2}{a_3}\right)}^{n-1}+\cdots +{\left(\frac{a_{n-1}}{a_n}\right)}^{n-1}+{\left(\frac{a_n}{a_1}\right)}^{n-1}\ge \frac{a_1^2+a_2^2+\cdots +a_n^2}{\sqrt[n]{a_1^2{a}_2^2\dots a_n^2}}.$$ This inequality is obtained by adding the inequality below with its analogues obtained by cyclic permutation of the variables: $$\begin{array}{rl}& (n-1)\cdot {\left(\frac{a_1}{a_2}\right)}^{n-1}+(n-2)\cdot {\left(\frac{a_2}{a_3}\right)}^{n-1}+\cdots +{\left(\frac{a_{n-1}}{a_n}\right)}^{n-1}\ge \\ & \ge \frac{n(n-1)}{2}\cdot \left({\left(\frac{a_1}{a_2}\right)}^{n-1}{\left(\frac{a_2}{a_3}\right)}^{n-2}\dots {\left(\frac{a_{n-1}}{a_n}\right)}^{\frac{2}{n}}\right)=\\ & =\frac{n(n-1)}{2}{\left(\frac{a_1^n}{a_1{a}_2{a}_3\dots a_n}\right)}^{\frac{2}{n}}.\end{array}$$ Equality holds if all the numbers are equal to $1$.