Absolute inequality question. I'm stuck with this kind of absolute inequality:
$$|x+1|>x+2.  \tag1$$
Firstly, when I solve this one:
$$|x+1|=x+2, \tag2$$
I make sure the right side of the equation is greater than zero; 
Condition:
$$x+2\ge0;\quad x\ge-2.$$
When I solve this by separating it in two cases and finding a solution, I get, in this example: $x=-1.5$, which belongs to the interval of $[-2,+\infty)$, so it's the right solution.
However, when I solve the first inequality $(1)$, I get the solution:
$$x<-1.5$$.
$-1.5$ is in the interval of the condition for the equation $(1)$ (which is the same for the second one -- the inequality): $[-2,+\infty )$, therefore, the WHOLE interval of $(-∞, -1.5)$ is the solution. Is this the right way of solving this kind of inequality? 
Thanks in advance.
 A: You have some good ideas.
You don't have a problem with the expression on the right-hand side. You need to know though whether $$|x+1|=x+1 \text{ or }-x-1$$ and this depends on whether $x+1\ge 0$ or $x\ge-1$. 
If this is true then the equation becomes $$x+1\gt x+2$$ which is always false.
If $x\lt -1$ the equation becomes $$-x-1\gt x+2 \text{ or, as you computed }x\lt -\frac 32$$
Here there are two conditions, as you observed, and only the values of $x$ which satisfy both are admissible as solutions, and this is true  when $x\lt -\frac 32$
It doesn't actually matter whether the right-hand side is positive - you aren't multiplying or dividing anything by it, or, for example squaring both sides of the equation (when you do have to be careful, so your instinct is a good one). 
A: Yes, you arrived at the correct solution.
One way to approach this is to note that given $|x + 1| > x + 2$, there are two cases to consider: $$x + 1 > x+2 \tag 1$$ $$-(x + 1) > x + 2\tag 2$$
Case $(1)$ never holds for any $x$, since $$x + 1 > x + 2 \iff 1 \gt 2$$ which clearly is false. 
Case $(2)$ gives us $$-(x + 1) > x + 2 \iff -x - 1 > x + 2 \iff 0 > 2x + 3 \iff x \lt -1.5$$
Hence, indeed, we have $$x\in (-\infty, -1.5)$$
A: Well, I think the best method you can used to be clear is separate cases:
$|x+1|=x+1 $ if $x\geq -1\quad  (1)$
$|x+1|=-x-1 $ if $x< -1\quad  (2)$
So, in the first case you have $x>x+1$ thats it's always false.
For the rest of the solution you we look at the second case: $-x-1<x+1 \Rightarrow x<-1.5$ so the another part is $(-\infty,-1.5)$
The complete solution is $$(-\infty,-1.5)$$
