Prove that if $x_{1}x_{2}...x_{n}=1$ then $\frac{1}{1+x_{1}+x_{1}x_{2}}+...+\frac{1}{1+x_{n-1}+x_{n-1}x_{n}}+\frac{1}{1 +x_{n}+x_{n}x_{1}}\ge 1$.
$x_{1},x_{2},...,x_{n}$ are positive real numbers, and $n\ge4$
I tried to rewrite this in more sensible notation - perhaps: $\sum_{k=2}^{n}(\frac{1}{1+x_{k-1}+x_{k-1}x_{k}}) + \frac{1}{1+x_{n}+x_{n}x_{1}}$ and $\prod_{k=1}^{n}x_{k}=1$.
Then, we could write the summation with a common denominator like so: $\frac{\frac{n\prod_{k=2}^{n}(1+x_{k-1}+x_{k-1}x_{k})}{\prod_{k=2}^{n}(1+x_{k-1}+x_{k-1}x_{k})}}{\prod_{k=2}^{n}(1+x_{k-1}+x_{k-1}x_{k})}$ = $\frac{n}{\prod_{k=2}^{n}(1+x_{k-1}+x_{k-1}x_{k})}$
I am unsure how to incorporate the last term of the summation into the expression. If I somehow could then given $n\ge4$, $\frac{n}{p}\ge \frac{4}{p}$ and since $\frac{n}{p}$ is given to be $>1$, p has to be strictly smaller than $4$. Then we would have an inequality involving just one product of the original expression. Neither am I sure in what manner would the product condition be relevant.
Would something akin to this be the right approach? Thank you for your help with this question!