, 2017

Solution:
Let $n=2016$. Define a sequence of real numbers $\{p_{k}\}$ by $p_{1}=0$, and for all $k \geq 1$,
$$p_{k+1}=\frac{1}{4\left(1-p_{k}\right)}$$
Note that, for every $i \geq 1$,
$$\left(1-p_{i}\right) \cdot x_{i}^{2}+x_{i} x_{i+1}+p_{i+1} x_{i+1}^{2}=\left(\frac{x_{i}}{2 \sqrt{p_{i+1}}}+\sqrt{p_{i+1}} x_{i+1}\right)^{2} \geq 0$$
Summing from $i=1$ to $n$ gives
$$\sum_{i=1}^{n} x_{i}\left(x_{i}+x_{i+1}\right) \geq-p_{n+1} x_{n+1}^{2}$$
One can show by induction that $p_{k}=\frac{k-1}{2 k}$. Therefore, our answer is $-p_{2017}=-\frac{1008}{2017}$.