4. Solve the inequality $\frac{|x+3|+|1-x|}{x+2016}<1$.

Answer. $x \in(-\infty ;-2016) \cup(-1009 ; 1007)$.
Solution If $x+20160$, then we multiply both sides by $x+2016$ and get $|x+3|+|x-1|-1009$. Considering the constraint, the interval that fits is $x \in(-1009 ;-3]$.
3) $-3<x<1$. Then $-x-3+x-1<2016 \Leftrightarrow x \in \mathbf{R}$. Considering the constraint, the interval that fits is $x \in(-3 ; 1)$.

Combining all the results, we get $x \in(-\infty ;-2016) \cup(-1009 ; 1007)$.