How would you solve an inequality in the form: $|f(x)| < g(x)$?

Question

An inequality such as $|x + 1| < x + 3$ was given to be solved. I attempted to used the theorem:

$|x + c| < \delta \implies c-\delta < x < c+\delta$

But $x \in \Bbb{R}$ so $x$ could be less than $-3$. EDIT: And it is required for $\delta > 0$

What is the correct way to approach this question?

Answer 1

$$x+3 \ge 0$$

$$x \ge -3 ~~~~(\text I)$$

since ($x+3 \ge 0$):

$$-(x+3)<x+1<x+3 ~~~~(\text{II})$$

and continue as usual

Answer 2

If $x \geq -1$, then $|x+1| = x+1 < x+3$; if $x < -1$, then $|x+1| = -x-1$, which is $< x+3$ iff $2x > -4$, namely iff $x > -2$. Do you then know how to conclude?