# Show that for all n>=2 (x_1^2)/(x_1^2+x_2x_3)+(x_2^2)/(x_2^2+x_3x_4)+...+(x_(n-1}^2)/(x_(n-1)^2+x_nx_1)+(x_n^2)/(x_n^2+x_1x_2)<= n-1 where x_i are real positive numbers

Inequality $\frac{{x}_{1}^{2}}{{x}_{1}^{2}+{x}_{2}{x}_{3}}+\frac{{x}_{2}^{2}}{{x}_{2}^{2}+{x}_{3}{x}_{4}}+\cdots +\frac{{x}_{n-1}^{2}}{{x}_{n-1}^{2}+{x}_{n}{x}_{1}}+\frac{{x}_{n}^{2}}{{x}_{n}^{2}+{x}_{1}{x}_{2}}\le n-1$
Show that for all $n\ge 2$
$\frac{{x}_{1}^{2}}{{x}_{1}^{2}+{x}_{2}{x}_{3}}+\frac{{x}_{2}^{2}}{{x}_{2}^{2}+{x}_{3}{x}_{4}}+\cdots +\frac{{x}_{n-1}^{2}}{{x}_{n-1}^{2}+{x}_{n}{x}_{1}}+\frac{{x}_{n}^{2}}{{x}_{n}^{2}+{x}_{1}{x}_{2}}\le n-1$
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
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Kyle Delacruz
Let $\frac{{x}_{2}{x}_{3}}{{x}_{1}^{2}}=\frac{{a}_{1}}{{a}_{2}}$,... and similar, where ${a}_{i}>0$ and ${a}_{n+1}={a}_{1}$
Thus, we need to prove that:
$\sum _{i=1}^{n}\frac{1}{1+\frac{{a}_{i}}{{a}_{i+1}}}\le n-1$
or
$\sum _{i=1}^{n}\left(\frac{1}{1+\frac{{a}_{i}}{{a}_{i+1}}}-1\right)\le -1$
or
$\sum _{i=1}^{n}\frac{{a}_{i}}{{a}_{i}+{a}_{i+1}}\ge 1,$
which is true because
$\sum _{i=1}^{n}\frac{{a}_{i}}{{a}_{1}+{a}_{i+1}}\ge \sum _{i=1}^{n}\frac{{a}_{i}}{{a}_{1}+{a}_{2}+...+{a}_{n}}=1$