A Very Hard Inequality Find the smallest constant $c$ such that for any positive integers $a_1,a_2,\ldots,a_n$ for $n \geq 3$, the following inequality holds: \begin{align} \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\cdots+\frac{a_n}{a_1+a_2}\geq cn. \end{align}
 A: Here is a solution for the highest $c$ that satisfies the inequality.
In general, setting all $a_i$ equal generates $LHS = n/2$. So $c=1/2$ may be a good conjecture which holds in many cases, albeit not in all. More information can be found when noticing that  the inequality you propose here is Shapiro's inequality, see 
https://en.wikipedia.org/wiki/Shapiro_inequality
This is indeed a very hard one.
As the source above states:
$c = 1/2$ if
    n is even and less than or equal to 12, or if
    n is odd and less than or equal to 23.
So for those $n$, this disproves your conjecture $c = \sqrt 2 - 1$.
As the source above further states:
For greater values of $n$, a strict (highest) lower bound is $c= \frac{\gamma}{2}$ with $\gamma \approx 0.9891\dots$. So again your $c = \sqrt 2 - 1$ is too low.
Here are some results for two particular $n$ which can be proved more easily:
For $n=3$, the inequality becomes
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$ which is Nesbitt's inequality, this is known to have a tight lower bound at $3/2$, so $c = 1/2$.
For $n=6$, this has been treated in 
Prove of Nesbitt's inequality in 6 variables
and a tight lower bound was established at $3$, so again $c=1/2$.
However (see the comment in there) you will normally not find this as a "generalization of Nesbitt's inequality" since this terminology is used for a different inequality.
A: I found this problem from "Problems from the Book" by Titu Andreescu and Gabriel Dospinescu and the authors did not attempt to show the solution because of its extreme difficulty. This was solved by Vladimir Drinfeld, a Ukranian Field's medalist as the authors said.
I tried to solve it for $c=\sqrt{2}-1$ using AM-GM. but dont know how for its optimum value.
\begin{align*} \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\cdots+ \frac{a_n}{a_1+a_2} \geq (\sqrt{2}-1)n \end{align*} Using Am-GM, the inequality becomes \begin{align} \frac{a_1+a_2+a_3}{a_2+a_3} \cdot \frac{a_2+a_3+a_4}{a_3+a_4}\cdots \frac{a_n+a_1+a_2}{a_1+a_2}\geq (\sqrt{2})^n \end{align}
