<s>[INST] Solve: $|-(x + 1)^2+1|\geq 1$ I tried to figure out this question. Can any one help? I know the answer but I keep getting the wrong one. Thanks in advance.
Find all values of x that satisfy $|-(x + 1)^2+1|\geq 1$
 [/INST] $$|-(x + 1)^2+1|\geq 1 >0$$
Since both sides of the inequlity are positive squaring the inequality we get,
$$((x + 1)^2-1)^2 \geq 1 $$
$$\Leftrightarrow ((x + 1)^2-1)^2-1^2\geq 0 \tag {difference of squares} $$ 
$$\Leftrightarrow ((x + 1)^2-1-1)((x + 1)^2-1+1)\geq 0  $$
$$\Leftrightarrow ((x + 1)^2-2)(x+1)^2 \geq 0 \tag {with $(x+1)^2>0$ iff $x \neq -1$} $$ 
Hence,  
Case 1 $x = -1$ 
Then, $((x + 1)^2-2)(x+1)^2 =0 \geq 0$ as required
Case 2 $x \neq -1$ 
Then $(x + 1)^2-2\geq 0$ (dividing the inequality by $(x + 1)^2>0$)
$$\Leftrightarrow(x + 1- \sqrt{2})(x + 1+ \sqrt{2})\geq 0$$
$$ \Leftrightarrow x \leq -1-\sqrt{2}  \ or  \ x \geq -1+\sqrt{2}$$ 
Therefore the final solution is $$x \in (- \infty,-1-\sqrt{2} ] \cup \{-1 \} \cup [-1+\sqrt{2}, +\infty)$$
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