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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1Perhaps the condition $n>3$ is necessary? Because for $n=2, 3$ the example $a_i=1\forall i$ contradicts the inequality. – 2012-12-04
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0Yes Sorry I forget to include that $n>3$ – 2012-12-06