Prove $\sum_{i=1}^{n}\frac{x_{i}^3}{x_{i+1}^2}\geq \sum_{i=1}^{n}\frac{x_{i}^2}{x_{i+1}}$ As in the topic, my question is to prove $$\sum_{i=1}^{n}\frac{x_{i}^3}{x_{i+1}^2}\geq \sum_{i=1}^{n}\frac{x_{i}^2}{x_{i+1}}$$We know that $x_{n+1}=x_{1}$ and $x_{1},x_{2}, x_{3},..., x_{n}\in \mathbb{R}_{+}$.
As I suspect that it would be to cool if all $x_{i}$ were equal, and question is marked as tough one, after a few hours my brain stopped without producing anything reasonable so I ask you for hints how to move it. Everything will be appreciated. Thanks in advance.
 A: Hint:We can derive your inequality by utilizing the Cauchy-Schwarz inequality twice,i.e.,
$$(\sum x_i)(\sum\frac{x_i^3}{x_{i+1}^2})\geq (\sum\frac{x_i^2}{x_{i+1}})^2$$
and
$$(\sum x_i)(\sum\frac{x_i^2}{x_{i+1}})\geq (\sum x_i)^2$$
A: $$3^n\frac{x_1^3}{x_2^2} + 2\cdot 3^{n-1}\frac{x_2^3}{x_3^2} +4\cdot 3^{n-2}\frac{x_3^3}{x_4^2}+\cdots+2^{n-1}\cdot3\frac{x_{n-1}^3}{x_n^2}+2^n\frac{x_n^3}{x_1^2}\ge (3^n-2^n)x_1$$
By Weighted AM-GM. (Note that the sum of the coefficients is the expression for $\frac{3^n-2^n}{3-2} = 3^n-2^n$)
Adding cyclicly and dividing by $3^n-2^n$, we find
$$\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} \ge \sum_{i=1}^n x_i$$
Now note that another application of AM-GM gives $\frac{x_i^3}{x_{i-1}^2}+x_i \ge 2\frac{x_i^2}{x_{i-1}}$, so summing over all $i$,
$$\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} + \sum_{i=1}^n x_i \ge \sum_{i=1}^n\frac{x_i^2}{x_{i-1}}$$
So then, $$2\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} \ge \sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} + \sum_{i=1}^n x_i \ge \sum_{i=1}^n\frac{x_i^2}{x_{i-1}}$$
A: Multiply both sides by 
$$\prod_{i=1}^{n} x_i^2$$
then just use Muirhead inequality.
