Inequality $\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$

Question

Show that for all $n\ge 2$ $$\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$$ where $x_i$ are real positive numbers

I was going to use $\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}x_{3}}=\frac{1}{1+\frac{x_{2}x_{3}}{x_{1}^{2}}}\le \frac{x_{1}}{2}\cdot \frac{1}{\sqrt{x_{2}x_{3}}}\le \frac{x_{1}}{4}\left( \frac{1}{x_{2}}+\frac{1}{x_{3}} \right)$

$\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3}x_{4}}+...+\frac{x_{n-2}^{2}}{x_{n-2}^{2}+x_{n-1}x_{n}}+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n}x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1}x_{2}}$

$=\frac{1}{1+\frac{x_{2}x_{3}}{x_{1}^{2}}}+\frac{1}{1+\frac{x_{3}x_{4}}{x_{2}^{2}}}+...+\frac{1}{1+\frac{x_{n-1}x_{n}}{x_{n-2}^{2}}}+\frac{1}{1+\frac{x_{n}x_{1}}{x_{n-1}^{2}}}+\frac{1}{1+\frac{x_{1}x_{2}}{x_{n}^{2}}}$

$\le \frac{x_{1}}{4}\left( \frac{1}{x_{2}}+\frac{1}{x_{3}} \right)+\frac{x_{2}}{4}\left( \frac{1}{x_{3}}+\frac{1}{x_{4}} \right)+...+\frac{x_{n-2}}{4}\left( \frac{1}{x_{n-1}}+\frac{1}{x_{n}} \right)+\frac{x_{n-1}}{4}\left( \frac{1}{x_{n}}+\frac{1}{x_{1}} \right)+\frac{x_{n}}{4}\left( \frac{1}{x_{1}}+\frac{1}{x_{2}} \right)$

$=\frac{1}{4}\left( \left( \frac{x_{1}}{x_{2}}+\frac{x_{1}}{x_{3}} \right)+\left( \frac{x_{2}}{x_{3}}+\frac{x_{2}}{x_{4}} \right)+\left( \frac{x_{3}}{x_{4}}+\frac{x_{3}}{x_{5}} \right)+...+\left( \frac{x_{n-2}}{x_{n-1}}+\frac{x_{n-2}}{x_{n}} \right)+\left( \frac{x_{n-1}}{x_{n}}+\frac{x_{n-1}}{x_{1}} \right)+\left( \frac{x_{n}}{x_{1}}+\frac{x_{n}}{x_{2}} \right) \right)$

$=\frac{1}{4}\left( \left( \frac{x_{1}+x_{2}}{x_{3}} \right)+\left( \frac{x_{2}+x_{3}}{x_{4}} \right)+\left( \frac{x_{3}+x_{4}}{x_{5}} \right)+...+\left( \frac{x_{n-3}+x_{n-2}}{x_{n-1}} \right)+\left( \frac{x_{n-1}+x_{n-2}}{x_{n}} \right)+\left( \frac{x_{1}+x_{n}}{x_{2}} \right)+\left( \frac{x_{n-1}+x_{n}}{x_{1}} \right) \right)$

....I thought about using Cauchy's inequality, but that would only increase the problem

Answer 1

Let $\frac{x_2x_3}{x_1^2}=\frac{a_1}{a_2}$,... and similar, where $a_i>0$ and $a_{n+1}=a_1$.

Thus, we need to prove that: $$\sum_{i=1}^n\frac{1}{1+\frac{a_i}{a_{i+1}}}\leq n-1$$ or $$\sum_{i=1}^n\left(\frac{1}{1+\frac{a_i}{a_{i+1}}}-1\right)\leq-1$$ or $$\sum_{i=1}^n\frac{a_i}{a_i+a_{i+1}}\geq1,$$ which is true because $$\sum_{i=1}^n\frac{a_i}{a_1+a_{i+1}}\geq\sum_{i=1}^n\frac{a_i}{a_1+a_2+...+a_n}=1$$