Where did I go wrong with this inequality involving absolute value function?

Question

Question: Find all $x \in \mathbb R$ such that the inequality $4<|x+2|+ |x-1|<5$ is satisfied.

This is my attempt at solving the problem:

Case (i): If $x+2 \geq 0 $ and $ x-1\geq0$, then

$4<x+2+x-1<5$

$\implies 4<2x+1<5$

$\implies 3<2x<4$

$\implies \dfrac{3}{2}<x<2$

$\implies x\in (\dfrac{3}{2},2)$

Case (ii): If $x+2 < 0 $ and $ x-1<0$, then

$4<-x-2-x+1<5$

$\implies 4<-2x-1<5$

$\implies 5<-2x<6$

$\implies -3<x<\dfrac{-5}{2}$

$\implies x\in (-3,\dfrac{-5}{2})$

Case (iii): If $x+2 \geq 0$ and $x-1<0$, then

$4<x+2-x+1<5$

$\implies 4<3<5$, which is not true.

Case (iv): If $x+2<0$ and $x-1 \geq 0$, then $x<-2$ and $x \geq 1$, which is not possible.

So, from cases (i) and (ii), we have $x \in (-3,\dfrac{-5}{2}) \cup (\dfrac{3}{2},2).$

But $-2.75 \in (-3,\dfrac{-5}{2})$, which does not satisfy the inequality. Where did I go wrong in my calculation?

Also, what do I do if there are more than two absolute value functions? It would be quite tedious to check all cases. Is there an altogether different way to go about such problems involving the absolute value function without considering case-by-case?

Answer 1

It seems to me that your solution is completely correct. Look here for a diagram of the map $x \mapsto |x+2|+|x-1|$. It is clear that the solution set coincides with the set you found.

Answer 2

Your answer is correct. For $x=-2.75$, $|-2.75+2|+|-|-2.75-1|=|-0.75|+|-3.75|=4.5$

So it does satisfy the inequality you came up with. Ironic though, your work was correct but you checking wasn't... funny how that goes.