11. Find the largest real number $c$ such that for any 2017 real numbers $x_{1}, x_{2}, \ldots, x_{2017}$, the inequality $\sum_{i=1}^{2016} x_{i} \cdot\left(x_{i}+x_{i+1}\right) \geq c \cdot x_{2017}^{2}$ always holds.
11. Analysis: $\sum_{i=1}^{2016} x_{i} \cdot\left(x_{i}+x_{i+1}\right) \geq c \cdot x_{2017}^{2} \Leftrightarrow x_{1}^{2}+\sum_{i=1}^{2016}\left(x_{i}+x_{i+1}\right)^{2} \geq(2 c+1) x_{2017}^{2}$

By the Cauchy-Schwarz inequality, we have:
$$
(1+1+\mathrm{L}+1)\left(x_{1}^{2}+\sum_{i=1}^{2016}\left(x_{i}+x_{i+1}\right)^{2}\right) \geq\left(x_{1}+\sum_{i=1}^{2016}(-1)^{i}\left(x_{i}+x_{i+1}\right)\right)^{2}=x_{2017}^{2} ;
$$

Thus, $x_{1}^{2}+\sum_{i=1}^{2016}\left(x_{i}+x_{i+1}\right)^{2} \geq \frac{x_{2017}^{2}}{2017}$; which means we only need $\frac{x_{2017}^{2}}{2017} \geq(2 c+1) x_{2017}^{2}$.
Solving this, we get: $c \leq-\frac{1008}{2017}$, with equality holding if and only if $2 x_{1}=-x_{2}, x_{i}+2 x_{i+1}=-x_{i+2}, i=1,2, \mathrm{~L}, 2015$, i.e., taking $x_{n}=(-1)^{n-1} n$ makes the equality hold.