If $\sum\limits_{i=1}^{2017}\frac{1}{x_i+2017}=\frac{1}{2017}$ for $x_i>0$, then prove that $\sqrt[2017]{x_1x_2...x_{2017}}\ge2016\cdot2017$

Question

Let $x_1,x_2,\cdots{},x_{2017}$ be positive reals such that $$\dfrac{1}{x_1+2017}+\cdots{}+\dfrac{1}{x_{2017}+2017}=\dfrac{1}{2017}.$$ Prove that: $$\sqrt[2017]{x_1x_2\cdots{}x_{2017}}\ge2016\cdot{}2017.$$

Progress: By the AM-HM inequality, I've managed to show that $$x_1+x_2+\cdots{}+x_{2017}\ge 2017^2(2016).$$ I'm not sure how to proceed further.

Answer 1

By AM-GM $\frac{1}{2017}-\frac{1}{x_i+2017}=\frac{x_i}{2017(x_i+2017)}=\sum\limits_{k\neq i}\frac{1}{x_k+2017}\geq\frac{2016}{\sqrt[2016]{\prod\limits_{k\neq i}(x_k+2017)}}$ and product of these