How to solve $|x-1|+|x+1|<2$-- my method does not work for this one I've just successfully solved the following exercise:
Find $x$ such that $|x-1| + |x-2| >1$.
I did it by making four cases depending on whether the two absolute value expressions are positive or negative.
I moved on the next exercise expecting to solve it using the same method but what I got is completely wrong (not even close!?).

Please could someone explain to me why my method does not work here?

Here is the exercise and what I did:
Find $x$ with $|x-1| + |x+1| < 2$.
We have $|x-1| + |x+1| < 2$ if and only if one of the four cases is satisfied:
Case 1: $x - 1 + x + 1 < 2$ if and only if $2x < 2$ if and only if $x < 1$.
Case 2: $x-1 - x - 1 < 2$ if and only if $0 < 4$ which is always true. This case would imply that the inequality is true for all $x \in \mathbb R$.
Case 3: $-x + 1 + x + 1 < 2$ if and only if $2<2$ which is always false.
Case 4: $-x + 1 - x - 1 < 2$ if and only if $x > -1$.

You need to consider 3 regions not 4 cases (considering 2 roots for each absolute value expression) $x<-1$ , $-1\leq x\leq 1$, $x> 1$.
