If $\prod x_k\neq0$ and $\sum\frac{x_k}{x_{k+1}}=0$, then $|\sum x_kx_{k+1}|\le (\max|x_k|- \min|x_k|)\sum x_k$ Question:

Suppose that $x_{1},x_{2},\cdots,x_{n}$ are real numbers, such that
  $$x_{1}x_{2}\cdots x_{n}\neq 0$$
  and
  $$\dfrac{x_{1}}{x_{2}}+\dfrac{x_{2}}{x_{3}}+\cdots+\dfrac{x_{n}}{x_{1}}=0$$
show that
  $$|x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{n}x_{1}|\le \Bigg(\max_{1\le k\le n}|x_{k}|-
\min_{1\le k\le n}|x_{k}|\Bigg)(|x_{1}|+|x_{2}|+\cdots+|x_{n}|)$$

This problem is from  a math contest.  I have already solved the following similar problem :(can see Mitrinovic D.S Analytic inequalitys Page 346 )

let $a_{i},x_{i}\in R$, such that $$\sum_{i=1}^{n}|x_{i}|=1,\sum_{i=1}^{n}x_{i}=0$$
  show that :$$\Bigg|\sum_{i=1}^{n}a_{i}x_{i}\Bigg|\le\dfrac{1}{2}\Bigg(\max_{1\le i\le n}a_{i}-\min_{1\le i\le n}a_{i}\Bigg)$$
  proof : Put $S_{k}=x_{1}+x_{2}+\cdots+x_{k}$. We assume wlog that 
  $$a_{1}\ge a_{2}\ge\cdots\ge a_{n}$$
  using Abel parts, we obtain
  $$I=\Bigg|\sum_{i=1}^{n}a_{i}x_{k}\Bigg|=\Bigg|S_{n}a_{n}+\sum_{k=1}^{n-1}S_{k}(a_{k}-a_{k+1})\Bigg|=\Bigg|\sum_{k=1}^{n-1}S_{k}(a_{k}-a_{k+1})\Bigg|$$
  Now $|S_k|$ is smaller than $A=\sum_{i=1}^k |x_i|$, and smaller than $B=\sum_{i=k+1}^n |x_i|$. Since $A+B=1$, one of $A$ or $B$ is smaller than
  $\frac{1}{2}$. It is clear then that
  $$|S_{k}|\le \dfrac{1}{2}$$
  so
  $$I\le \dfrac{1}{2}\sum_{k=1}^{n-1}(a_{k}-a_{k+1})=\dfrac{1}{2}(a_{1}-a_{n})$$

Perhaps a similar method applies here ?
My try: let
$$\dfrac{x_{i}}{|x_{1}|+|x_{2}|+|x_{3}|+\cdots+|x_{n}|}=x'_{i}$$
so
$$\sum_{i=1}^{n}|x'_{i}|=1$$
and
$$\Longleftrightarrow 
\sum_{i=1}^{n}|x'_{i}x_{i}|\le (\max_{1\le k\le n}|x_{k}|-
\min_{1\le k\le n}|x_{k}|)$$
But then I'm stuck. Thank you for any help
 A: Let $Q$ be the product of $\max_{1\le k\le n}|x_{k}|$ and $\min_{1\le k\le n}|x_{k}|$, that is:
$$Q := \max_{1\le k\le n}|x_{k}|\cdot\min_{1\le k\le n}|x_{k}|\,.$$
We can how represent each product of two neighboring (cyclically) numbers like
$$ x_1 x_2 = Q\dfrac{x_1}{x_2} + x_1(x_2 - \dfrac{Q}{x_2})\,.$$
Then we have
$$ x_1 x_2 + x_2 x_3 + \cdots + x_n x_1 = 
x_1(x_2 - \dfrac{Q}{x_2}) + x_2(x_3 - \dfrac{Q}{x_3}) + \cdots + x_n(x_1 - \dfrac{Q}{x_1})
$$
since $n$ terms of the type $Q\dfrac{x_{k-1}}{x_k}$ sum to zero due to the equality condition on our numbers. On the other hand, for each $l = 1\ldots n$ we have:
$$ | x_l - \dfrac{Q}{x_l} | \le \max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\,,$$
that implies:
$$ | x_{l-1}(x_l - \dfrac{Q}{x_l}) | \le |x_{l-1}|\Bigg(\max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\Bigg)\,.$$
When we sum all $n$ such inequalities, we’ll virtually obtain the inequality that has to be demonstrated (up to applying the triangle inequality for the absolute value function at the left-hand side).
