#### Answer

$$|x+2|>1$$

#### Work Step by Step

The set of real numbers cover the regions: $(-\infty, -3) ∪ (-1, +\infty)$.
From the graph we can see that it is an open interval; non-strict inequality.
To find the actual inequality, involving absolute value, we can do the steps of solving inequality involving absolute value using reverse steps.
So we have two inequality:
$x-1$
Let's try to give them the form, from where we split normal absolute value inequality.
In general, when solving $|x+a|1$
$-(x+2)>1$
$x+2>1$
Which is given by the inequality: $|x+2|>1$