$Thesetofrealvaluesofxforwhichtheinequality|x-1|+|x+1|<4alwaysholdsgoodis$

$\begin{array}{l}GivenEquation|x-1|+|x+1|<4\\ case\left(i\right):ifx<-1\\ \Rightarrow -(x-1)-(x+1)<4\Rightarrow -x+1-x-1<4\\ \Rightarrow x>-2\\ \therefore x\in (-2,-1)\\ case\left(ii\right):if-1\le x<1\\ \Rightarrow -(x-1)+x+1<4\\ \Rightarrow -x+1+x+1<4\\ \Rightarrow 2<4\\ \Rightarrow x\in [-1,1)\\ case\left(iii\right):ifx\ge 1\\ \Rightarrow x-1+x+1<4\\ \Rightarrow 2x<4\Rightarrow x<2\\ x\in [1,2)\\ \therefore x\in (-2,2)\end{array}$