How to solve inequalities which have absolute value inside absolute value? I know how to solve if I have something like this: $$|2x + 3| - |1 - x| \geq 3$$
But I don't know how to solve something like this: $$|x + |3x + 9|| \geq 3$$
Anyone can show me how to start solving this type of absolute value in inequalities?
 A: One simple way: split it into cases.
$$|x+|3x+9||\geqslant 3$$
Case 1
If $x+|3x+9| \geqslant 0$ then $x+|3x+9|\geqslant 3$.
$$|3x+9| \geqslant -x \implies x\in(-\infty;-\tfrac92\rangle \cup \langle -\tfrac94;+\infty)=A$$
$$|3x+9|\geqslant 3 - x \implies x\in(-\infty;-6\rangle \cup \langle -\tfrac32;+\infty)=B$$
$$\Downarrow$$
$$x\in A\cap B=(-\infty;-6\rangle \cup \langle -\tfrac32;+\infty) = S_1$$
The set above contains the solutions from this case.
Case 2
If $x+|3x+9| < 0$ then $-x-|3x+9|\geqslant 3$.
$$|3x+9| < -x \implies x\in(-\tfrac92;-\tfrac94)=A$$
$$-x-|3x+9|\geqslant 3 \implies x\in\{-3\}=B$$
$$\Downarrow$$
$$x\in A\cap B=\{-3\}=S_2$$
Summary
$$x\in S_1\cup S_2=(-\infty;-6\rangle \cup \{-3\}\cup \langle -\tfrac32;+\infty)$$
A: It is the same as you do it when you have only one absolute sign. I would do the following;
$x + |3x+9| \geq 3$ or $x + |3x+9| < -3$
$|3x+9| \geq 3-x$ or $|3x+9| < -3-x$
case 1. $|3x+9| \geq 3-x$
case 2. $|3x+9| < -3-x$
Let's consider case 1 first,
case 1.
$|3x+9| \geq 3-x$
$3x+9 \geq 3-x$  or $3x+9 < -(3-x)$
$4x \geq -6$  or $ 2x < -12$
$x \geq -3/2$ or $x < -6$.
Now consider case 2.
Case 2
$|3x+9| < -3-x$
$3x+9 < -3-x$ or $3x-9 \geq 3+x$
$x < -3$ or $x \geq -3$
Now for the final answer,  you have to consider 4 regions of $x$ values;
$x \geq -3/2$ or $x <-6$ or $x <-3$ or $x >= -3$
That is, the answer is :
$(-\infty , -6) \cup [-3, \infty)$
Hope this helps you.
Thanks.
