Prove that $\frac{1}{1+a_1+a_1a_2}+\frac{1}{1+a_2+a_2a_3}+\cdots+\frac{1}{1+a_{n-1}+a_{n-1}a_n}+\frac{1}{1+a_n+a_na_1}>1.$

Question

If $n > 3$ and $a_1,a_2,\ldots,a_n$ are positive real numbers with $a_1a_2\cdots a_n = 1$, prove that $$\dfrac{1}{1+a_1+a_1a_2}+\dfrac{1}{1+a_2+a_2a_3}+\cdots+\dfrac{1}{1+a_{n-1}+a_{n-1}a_n}+\dfrac{1}{1+a_n+a_na_1}>1.$$

I find it hard to use any inequalities here since we have to prove $>1$ and most inequalities such as AM-GM and Cauchy-Schwarz use $\geq 1$. On the other hand it seems that if I can prove that each fraction is $>1$ that might help, but I am unsure.

Answer 1

Let $a_i=\frac{x_{i+1}}{x_i}$, where $x_{n+1}=x_1$, $x_{n+2}=x_2$ and all $x_i>0$. Hence, $\sum\limits_{i=1}^{n}\frac{1}{1+a_i+a_ia_{i+1}}=\sum\limits_{i=1}^{n}\frac{x_i}{x_i+x_{i+1}+x_{i+2}}>\sum\limits_{i=1}^{n}\frac{x_i}{x_1+x_2+...+x_n}=1$