Syler wrote: Real numbers $ x_{1},x_{2},..,x_{n}$ are positive. Prove the inequality: $ \frac {x_{1}}{x_{2} + x_{3}} + \frac {x_{2}}{x_{3} + x_{4}} + ... + \frac {x_{n - 1}}{x_{n} + x_{1}} + \frac {x_{n}}{x_{1} + x_{2}} > (\sqrt {2} - 1)n$ Q.E.D $ \Leftrightarrow \sum \frac{a_1+a_2+a_3}{a_2+a_3}\geq \sqrt{2}n$ we need to prove that:$ \prod \frac{a_1+a_2+a_3}{a_2+a_3}\geq {\left( \sqrt{2} \right)}^{n}$(by AM-GM) we have:$ \prod {\left( {a_1+a_2+a_3}\right)}^{2}\geq \prod {\left( a_1+\frac{a_2}{2}+\frac{a_2}{2}+a_3\right)}^{2}$ $ \geq 4\prod \left( a_1+\frac{a_2}{2}\right)\left( \frac{a_2}{2}+a_3\right)=\prod \left( \left( 2a_1+a_2\right)\left( 2a_3+a_2\right) \right)$ $ \geq \prod \left(2{\left( a_1+a_2\right)}^{2} \right)={2}^{n}{\left( \prod\left( a_1+a_2\right) \right)}^{2}$ So we have Q.E.D

trungk42sp wrote: Syler wrote: Real numbers $ x_{1},x_{2},..,x_{n}$ are positive. Prove the inequality: $ \frac {x_{1}}{x_{2} + x_{3}} + \frac {x_{2}}{x_{3} + x_{4}} + ... + \frac {x_{n - 1}}{x_{n} + x_{1}} + \frac {x_{n}}{x_{1} + x_{2}} > (\sqrt {2} - 1)n$ Q.E.D $ \Leftrightarrow \sum \frac {a_1 + a_2 + a_3}{a_2 + a_3}\geq \sqrt {2}n$ we need to prove that:$ \prod \frac {a_1 + a_2 + a_3}{a_2 + a_3}\geq {\left( \sqrt {2} \right)}^{n}$(by AM-GM) we have:$ \prod {\left( {a_1 + a_2 + a_3}\right)}^{2}\geq \prod {\left( a_1 + \frac {a_2}{2} + \frac {a_2}{2} + a_3\right)}^{2}$ $ \geq 4\prod \left( a_1 + \frac {a_2}{2}\right)\left( \frac {a_2}{2} + a_3\right) = \prod \left( \left( 2a_1 + a_2\right)\left( 2a_3 + a_2\right) \right)$ $ \geq \prod \left(2{\left( a_1 + a_2\right)}^{2} \right) = {2}^{n}{\left( \prod\left( a_1 + a_2\right) \right)}^{2}$ So we have Q.E.D Thank you. This is my solution: Let the LHS be S. $ S\geq\ (\sqrt{2}-1)n| *(\sqrt {2}+1)$ $ S(\sqrt{2}+1)\geq\ n$ $ \frac{S}{n}\geq\ \frac{1}{\sqrt{2}+1}$ It enough to prove that: $ \prod\frac{x_{1}+x_{2}+x_{3}}{x_{2}+x_{3}}\geq\ \frac{1}{(\sqrt{2}+1)^n} (AM-GM)$ It is obvious that $ \prod(x_{1}+x_{2}+x_{3})\geq\ \prod(x_{2}+x_{3})$ and $ (\sqrt{2}+1)^n\geq\ 1$ then our inequality is true.