TITLE: Prove $\sum_{i=1}^{n}\frac{x_{i}^3}{x_{i+1}^2}\geq \sum_{i=1}^{n}\frac{x_{i}^2}{x_{i+1}}$
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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
$$(\sum x_i)(\sum\frac{x_i^2}{x_{i+1}})\geq (\sum x_i)^2$$