An inequality on sequences with each term dividing sum of two neighbouring terms Positive integers $x_1, x_2, \dots, x_n$ ($n \ge 4$) are arranged in a circle such that each $x_i$ divides the sum of the neighbors; that is  $$\frac{x_{i-1}+x_{i+1}}{x_i} = k_i $$ is an integer for each $i$, where $x_0 = x_n$, $x_{n+1} = x_1$.  Prove that $$ 2n \le k_1 + k_2 + \dots + k_n < 3n. $$
This is a problem from some olympiad but I don't remember. 
What I did was like this :

Let $x_1$ be the smallest. Then obviously, $k_1\ge 2$. I tried to proceed inductively, but then I have no idea. I know it is not something useful, but I lack in any kind of idea. If I try induction, three new terms come replacing three previous terms. Which seems complicated and I can't do anything with it.

Sorry for such a bad try, but I seriously don't have any idea about it. Maybe something ingenious is required here. I hope someone can help. Thanks. 
EDIT : Removed meaningless ramblings of mine. Apologies.

The lower bound only:
Update: I just realized there is an easier argument. The terms in the
sum $\sum_i x_{i+1}/x_i$ are just the reciprocals of the terms in the sum 
 $\sum_i x_{i-1}/x_i$. Therefore, our sum $\sum_i k_i$ can be expressed as the sum of 
$n$ terms of the form $y+{1\over y}.$ Since $y+{1\over y}\geq 2$ for all positive $y$,
we get $\sum_i k_i\geq 2n.$

This is exercise 5.7(a) from The Cauchy-Schwarz Master Class by J. Michael Steele.
Let $y_i$ be your values $x_i$ only put in order so that $y_1\leq y_2\leq \cdots \leq y_n$.
Then the reciprocals are in the opposite order $1/y_n\leq 1/y_{n-1}\leq \cdots \leq 1/y_1$.
The lower bound in the rearrangement inequality says that 
$$n={y_1\over y_1}+{y_2\over y_2}+\cdots+{y_n\over y_n}
\leq {y_1\over y_{\sigma(1)}}+{y_2\over y_{\sigma(2)}}+\cdots +{y_n\over y_{\sigma(n)}},$$
for any permutation $\sigma$. 
For the appropriate permuations $\sigma$ we get $n\leq \sum_i x_{i-1}/x_{i}$ and  $n\leq \sum_i x_{i+1}/x_{i}$ 
where the index runs cyclically: $\dots,n,1,2,\dots,n-1,n,1,\dots$. Adding these together gives
$$2n\leq \sum_i {x_{i-1}+x_{i+1}\over x_i}.$$ 
