How to prove the following statement using direct proofs. I'm currently in Discrete Mathematics and I'm currently on the chapters about direct proofs. I'm working on some practice problems and got stuck on this currently one. Any help would be much appreciated! Thanks in advance.
Problem:
Proof by using the cases, if $x\in\mathbb R$, then $| x + 3 |  - x > 2$.
I believe that I can separate this problem into three cases: when $x > 0$, $x = 0$, and $x < 0$.  But I'm unsure how to go on from there.
***Had fixed a mistake to the problem, sorry for any inconveniences.
 A: Splitting this into the cases $x>0, x<0$, and $x=0$ is not useful for this problem. The reason you should use cases for this problem is because the absolute value function behaves differently for positive, negative, and zero inputs, so your cases should be the cases of the input to the absolute value function being positive, negative, or zero. In this case, that means you should split on the cases $x+3>0, x+3<0$, and $x+3=0$. In each of those three cases, you can use the definition of absolute value to simplify the left hand side of your inequality.
A: The important thing is the sign of $x+3$, not the sign of $x$. So if $x<-3$ then $|x+3|=-(x+3)$ and $|x+3|-x=-3-2x>-3-2\times(-3)=3$. 
Otherwise (if $x\geq -3$) you get $|x+3|-x=(x+3)-x=3$.
(You don't really have three cases; you have two cases that overlap. This is generally the case with the modulus function: $|z|=z$ for $z\geq 0$ and $|z|=-z$ for $z\leq 0$. Since these two cases both cover $z=0$, you don't need to deal with it separately.)
