The sequence ${a_1, a_2, ..., a_{2019}}$ satisfies the following condition. $a_1=1, a_{n+1}=2019a_{n}+1$ Now let $x_1, x_2, ..., x_{2019}$ real numbers such that $x_1=a_{2019}, x_{2019}=a_1$ (The others are arbitary.) Prove that $\sum_{k=1}^{2018} (x_{k+1}-2019x_k-1)^2 \ge \sum_{k=1}^{2018} (a_{2019-k}-2019a_{2020-k}-1)^2$