Answer
$\left( -\infty, -2 \right] \cup \left[ 0,\infty \right)$
Work Step by Step
For any $a\gt0$, $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$. (The symbol $\gt$ may be replaced with $\ge$.)
Using the concept above, the solutions to the given inequality, $
|x+1|\ge1
,$ is
\begin{array}{l}\require{cancel}
x+1\ge1
\\\\
x\ge1-1
\\\\
x\ge0
,\\\\\text{ OR }\\\\
x+1\le-1
\\\\
x\le-1-1
\\\\
x\le-2
.\end{array}
In interval notation, the solution set is $
\left( -\infty, -2 \right] \cup \left[ 0,\infty \right)
.$
