<s>[INST] Double module in a inequality Can somebody explain to me (or link me a site which does) how to solve this?
$$
    ||x+1| -1| \geq 3
$$
I have no idea how to work out this double absolute value sign.
 [/INST] Here is a solution with squaring technique
$$
\begin{align}
||x+1|-1|\geq 3&\Longleftrightarrow (|x+1|-1)^2\geq 3^2\quad\text{ since both sides of inequality are non negative}\\
&\Longleftrightarrow (|x+1|-1)^2-3^2\geq 0\\
&\Longleftrightarrow ((|x+1|-1-3)((|x+1|-1)+3)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\
&\Longleftrightarrow (|x+1|-4)(|x+1|+2)\geq 0\\
&\Longleftrightarrow |x+1|-4\geq 0\quad\text{ since }\quad\quad|x+1|+2>0\quad\text{for all }x\\
&\Longleftrightarrow |x+1|\geq 4\\
&\Longleftrightarrow |x+1|^2\geq 4^2\quad\text{ since both sides of inequality are non negative}\\
&\Longleftrightarrow (x+1)^2\geq 4^2\quad\text{ since }\quad |a|^2=a^2\\
&\Longleftrightarrow (x+1)^2 - 4^2\geq 0\\
&\Longleftrightarrow ((x+1)-4)((x+1)+4)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\
&\Longleftrightarrow (x-3)(x+5)\geq 0\\
&\Longleftrightarrow x\in(-\infty,-5]\cup[3,+\infty)
\end{align}
$$
</s>