I do not know how to prove this inequality: Suppose that $x_i>0$ and $x_1\cdot ...\cdot x_n=1$, show that $\frac{1}{1+x_1+x_1x_2}+...+\frac{1}{1+x_n+x_nx_1}>1$ The hint is to use quotient substituion, I was thinking about substituting in $a_i=\frac{1}{x_i}$, that will transform the original inequality into $\frac{a_1a_2}{1+a_1+a_1a_2}+...+\frac{a_na_1}{1+a_n+a_na_1}>1$ And now I am stuck, I tried the AM-GM on the fraction but it does not work well.
Inequality, with quotient substitution
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real-analysis
inequality
functional-inequalities
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0Yes Sorry I forget to include that n>3 – 2012-12-06
1 Answers
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Let $x_i=\frac{a_{i+1}}{a_i}$, where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, $\sum_{i=1}^n\frac{1}{1+x_i+x_ix_{i+1}}=\sum_{i=1}^n\frac{1}{1+\frac{a_{i+1}}{a_i}+\frac{a_{i+2}}{a_i}}=$ $=\sum_{i=1}^n\frac{a_i}{a_i+a_{i+1}+a_{i+2}}>\sum_{i=1}^n\frac{a_i}{a_1+a_2+...+a_n}=1.$ Done!