Showing $x_n-x_nx_1+\sum_{k=1}^{n-1} (x_k-x_kx_{k+1})\leq\left\lfloor\frac{n}{2}\right\rfloor$, with $x_i\in[0,1]$ 
Let $x_1, x_2,\ldots, x_n$ be arbitrary numbers from the interval $[0,1]$ with $n>1$.
Show that $$x_n-x_nx_1+\sum_{k=1}^{n-1} (x_k-x_kx_{k+1})\leq\left\lfloor\frac{n}{2}\right\rfloor$$

I tried to factor out the $x_k$ from each term to show that if the coefficient $x_k$ of $x_k(1-x_{k+1})$ is larger than $\dfrac{1}{2}$, then the term $x_{k-1}(1-x_k)$ must be smaller than $\dfrac12$, but I don't know where to go from here or if it is even the right approach.

*This is not an official answer.
Edit
I apologise for my inconsistency in my previous answer, hence its been removed. While, this question already has an accepted answer/correct answers, this method serves to confirm the answers/inequality.

The following are for $n = 2,3,4$ respectively and display the inequality approaching values $[0, $floor$(\frac{n}{2})]$ but never exceed  floor$(\frac{n}{2})$ computationally. Higher values of $n$ were not computed as they pose time restrictions in computation due to probability restrictions.
$n = 2$" />
$n = 3$" />
$n = 4$" />
