Show that $\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$ 
Let $n>2$ an integer and $x_k>0$ with  $x_1\cdot x_2\cdots x_n=1$
Show that $$\sum_{k=1}^{n}\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$$

I tried an induction without succeed, I do not really have idea to approach this inequality.
Any suggestion is appreciated.

it is true that $n\ge 6$ the inequality is always true because 
$ \frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1$ is always true when $x_k \ge 1$, proof:
$\frac{x_k^2}{x_k^2-2x_k\cos(\frac{2\pi}{n})+1}\geq 1 \iff x_k \ge \dfrac{1}{2\cos(\frac{2\pi}{n})} \iff 1 \ge \dfrac{1}{2\cos{(\frac{2\pi}{n})}} \iff \cos{(\frac{2\pi}{n})} \ge \dfrac{1}{2} \iff \dfrac{2\pi}{n} \le \dfrac{\pi}{3} \iff n \ge 6$
so it is remain $n=3,4,5$
$n=3$ is already proved.
$f_n(x)=\frac{x^2}{x^2-2x\cos(\frac{2\pi}{n})+1}$ is mono increasing function when $n=3,4,5$.then the result is easy to induct.
