Solving an Absolute Value Inequality The Question: 

$|x^2+x-1|-|3x+1|<x^2-2$

I don't really know how to work with absolute values, besides basic ones. 
Here is my attempt at simplifying, can someone tell me where I messed up and give some hints on how to continue?
$|x^2+x-1|-|3x+1|<x^2-2$
$|x^2+x-1|<x^2-2+|3x+1|$
$x^2+x-1<x^2-2+3x+1$ or $x^2+x-1>-x^2+2-3x-1$
$2x>1$ or $x^2+2x-1>0$
Answer:

 $x<-1 $ or $x>-1+\sqrt{2}$

 A: We have four cases which depend on the sign of $(x^2+x-1)$ and $(3x+1)$ (the arguments of the absolute values).

$1)$ If $x\geq -1/3$ then you have to solve
  $$|x^2+x-1|-(3x+1)<x^2-2$$
  or $|x^2+x-1|<x^2+3x-1$.

Two subcases:

$1.1)$ For $-1/3\leq x< (\sqrt{5}-1)/2$, it becomes
  $$-(x^2+x-1)<x^2+3x-1$$
  or $0<2x^2+4x-2$ which holds in this interval for $\sqrt{2}-1<x$.
$1.2)$ For $x\geq (\sqrt{5}-1)/2$, we have
  $$(x^2+x-1)<x^2+3x-1$$
  or $0<2x$, $x>0$ which always satisfied in this interval.



$2)$ If $x< -1/3$ then you have to solve
  $$|x^2+x-1|+(3x+1)<x^2-2$$
  or $|x^2+x-1|<x^2-3x-3$.

Two subcases:

$2.1)$ Now for $-(\sqrt{5}+1)/2<x<-1/3 $, it becomes
  $$-(x^2+x-1)<x^2-3x-3.$$
  or $0<2x^2-2x-4$ which holds in this interval for $x<-1$.
$2.2)$ For $x\leq -(\sqrt{5}+1)/2$, we have
  $$(x^2+x-1)<x^2-3x-3.$$
  or $0<-2x-2$, $x<-1$ which always satisfied in this interval.


Finally, putting all together, we have that the solution is:
$$\mbox{$x>\sqrt{2}-1$ (1.1 and 1.2) or $x<-1$ (2.1 and 2.2).}$$
