Let $P(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots +a_1x+a_0,$ we have \begin{align*} P\left(\frac{x_1}{x_2} \right)^2+P\left(\frac{x_2}{x_3} \right)^2+ ... +P\left(\frac{x_n}{x_1} \right)^2 &\ge \frac{\left(P\left(\frac{x_1}{x_2} \right)+P\left(\frac{x_2}{x_3} \right)+ ... +P\left(\frac{x_n}{x_1} \right)\right)^2}{n} \\ &=\frac{\left(\displaystyle\sum_{k=0}^m a_k\left(\left(\frac{x_1}{x_2}\right)^k+\left(\frac{x_2}{x_3}\right)^k+\cdots +\left(\frac{x_n}{x_1}\right)^k\right)\right)^2}{n} \\ &\ge \frac{\left(n\cdot \displaystyle\sum_{k=0}^m a_k\sqrt[n]{\left(\frac{x_1}{x_2}\right)^k\cdot \left(\frac{x_2}{x_3}\right)^k\cdot \ldots \cdot \left(\frac{x_n}{x_1}\right)^k}\right)^2}{n} \\ &=\frac{n^2\cdot P(1)^2}{n}\\ &= n\cdot P(1)^2. \ \blacksquare \end{align*}Equality holds when $x_1=x_2=\ldots =x_n.$

Keith50 wrote: Equality holds when $x_1=x_2=\ldots =x_n.$ Or when $P$ is constant...

Keith50 wrote: Let $P(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots +a_1x+a_0,$ we have \begin{align*} P\left(\frac{x_1}{x_2} \right)^2+P\left(\frac{x_2}{x_3} \right)^2+ ... +P\left(\frac{x_n}{x_1} \right)^2 &\ge \frac{\left(P\left(\frac{x_1}{x_2} \right)+P\left(\frac{x_2}{x_3} \right)+ ... +P\left(\frac{x_n}{x_1} \right)\right)^2}{n} \\ &=\frac{\left(\displaystyle\sum_{k=0}^m a_k\left(\left(\frac{x_1}{x_2}\right)^k+\left(\frac{x_2}{x_3}\right)^k+\cdots +\left(\frac{x_n}{x_1}\right)^k\right)\right)^2}{n} \\ &\ge \frac{\left(n\cdot \displaystyle\sum_{k=0}^m a_k\sqrt[n]{\left(\frac{x_1}{x_2}\right)^k\cdot \left(\frac{x_2}{x_3}\right)^k\cdot \ldots \cdot \left(\frac{x_n}{x_1}\right)^k}\right)^2}{n} \\ &=\frac{n^2\cdot P(1)^2}{n}\\ &= n\cdot P(1)^2. \ \blacksquare \end{align*}Equality holds when $x_1=x_2=\ldots =x_n.$ Bravo

Clearly $P^2$ is convex and increasing on $\mathbb{R}^+$ (since it is a sum of convex and increasing functions). Therefore by Jensen and AM-GM:$$\sum P^2\left(\frac{x_i}{x_{i+1}}\right) \geqslant n P^2 \left(\frac{1}{n} \sum \frac{x_i}{x_{i+1}} \right) \geqslant n P^2 (1).$$Equality takes place if i) $P$ is constant or ii) $x_1=\dots=x_n$.