Thank, arqady !! I hope that I don't make any mistake this time~ First, let $ y_i=nx_i,i=1,2,...,n $ then the inequality becomes $ \sum_{i=1}^n\frac{y_i}{n^2y_{i+1}-y^3_{i+1}} \ge \frac{n}{n^2-1} $ Now using AM-GM inequality we get $ \frac{\sum_{i=1}^n{\frac{y_i}{n^2y_{i+1}-y^3_{i+1}}}}{n} $ $ \ge \sqrt[n]{\frac{1}{\Pi(n^2-{y_i}^2)}} $ Thus, we suffice to show that $ \sqrt[n]{\frac{1}{\Pi(n^2-{y_i}^2)}} \ge \frac{1}{n^2-1} $--- $ (*) $ Notice $ (*) \Leftrightarrow (n^2-1)^n \ge \Pi(n^2-{y_i}^2) \Leftrightarrow nln(n^2-1) \ge \sum_{i=1}^n{ln(n^2-{y_i}^2)} $, which is true! ((Since we have $ f(x)=\frac{1}{n^2-x^2} $ then $ f''(x)=\frac{-2(n^2+x^2)}{(n^2-x^2)^2} \le 0 $

BSJL wrote: That's because $ f(x)=x-x^3 $ is a increasing function on the interval $ [0,1] $ No.

$\sum_{i=1}^n x^2_i\ge \frac{1}{n}\Rightarrow $ $\sum_{i=1}^n\frac{x_i}{x_{i+1}-x^3_{i+1}}\ge n\sqrt[n]{\prod_{i=1}^n\frac{x_i}{x_{i+1}-x^3_{i+1}}}=\frac{n}{\sqrt[n]{\prod_{i=1}^n (1-x^2_i)}}$ $\ge \frac{n^2}{n-\sum_{i=1}^n x^2_i}\ge \frac{n^2}{n-\frac{1}{n}}=\frac{n^3}{n^2-1}.$