How to prove specific inequality, assuming $\prod\limits_{i=1}^{n}(a_{i}-1)=1$ 
Let $n\ge 2$ be postive integer and $a_{i},(i=1,2,\cdots,n)$ be real numbers such that $a_{i}>1,(i=1,2,\cdots,n),\prod_{i=1}^{n}(a_{i}-1)=1$. Show that
  $$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\sum_{k=1}^{n-1}ka_{i+k-1}}\le\dfrac{1}{n-1}$$
  where $a_{n+m}=a_{m},\forall m\ge 1$

It seem use AM-GM inequality. I have try all methods can't solve this problem. 
$$\sum_{k=1}^{n-1}ka_{i+k-1}=a_{i}+2a_{i+2}+3a_{i+3}+\cdots+(n-1)a_{i+n-1-1}$$
$$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\sum_{k=1}^{n-1}ka_{i+k-1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+2a_{i+2}+3a_{i+3}+\cdots+(n-1)a_{i+n-1-1}}\le\dfrac{1}{n-1}$$
 A: This problem was proposed by Dmitriy Maximov on XXIV Russian Festival of Math Youth. Here is the official solution.
Denote $x_k^{n-1}=a_k-1$. The product of new positive variables $x_i$ equals 1. Consider the denominator of the first fraction:
\begin{align*}
a_1+2a_2+3a_3+ \ldots +(n-1)a_{n-1}&=x_1^{n-1}+2x_2^{n-1}+ \ldots (n-1)x_{n-1}^{n-1}+
1+2+\ldots+(n-1)=
\\
=(n-1)+\bigl(x_{n-1}^{n-1}+(n-2)\bigr)&+\bigl(x_{n-1}^{n-1}+x_{n-2}^{n-1}+(n-3)\bigr)+
\bigl(x_{n-1}^{n-1}+x_{n-2}^{n-1}+x_{n-3}^{n-3}+(n-4)\bigr)+\\ \ldots+
&
\bigl(x_{n-1}^{n-1}+x_{n-2}^{n-1}+\ldots+x_1^{n-1}\bigr).
\end{align*}
Now we estimate all summands (except the first) by AM-GM for
$n-1$ numbers (some of them equal to 1). This makes denominators lesser, thus fractions greater.
The denominator of the first fraction becomes
$$
(n-1)x_{n-1}+(n-1)x_{n-1}x_{n-2}+(n-1)x_{n-1}x_{n-2}x_{n-3}+\ldots+(n-1)x_{n-1}x_{n-2}\ldots
x_{1}+(n-1).
$$
Now it suffices to prove that the cyclic sum of the fractions
$$
\frac{1}{x_{n-1}+x_{n-1}x_{n-2}+\ldots+x_{n-1}x_{n-2}\ldots x_1+1}
$$
does not exceed 1. Actually this sum equals 1. For proving this we multiply the second fraction by $x_{n-1}x_{n-2} \ldots x_1$ (I mean, multiply both the numerator and denominator), the third by $x_{n-2}x_{n-3} \ldots x_1$ and so on (the last fraction by $x_1$). The denominators become equal, and the sum of numerators becomes equal to the denominator.
