More precise way of solving inequality I need to solve this function:
$$
\lvert x^2-1\rvert\ge 2x-2\\
$$
I solved this equation:  
For $x<0$, the solution is non existing, here I got negative root, when I tried to solve quadratic function and for $x\ge 0$ I got points $x_1=-1$ and $x_2=3$.
My question is:
How do I set the solution of equation. Is there any procedure, with wich I can determine is equation valid for $[-1,3]$ or $[-\infty, -1]\lor [3, +\infty]$.
I know that I can just set the numbers and see the result, but I just want to now is there any other different way to do this.
Thanks.
 A: Before trying an “automated” method, it's better to have a close look at the problem. 
You can first of all observe that if $2x-2<0$, then the inequality obviously holds, because $|x^2-1|\ge0$. So all values $x<1$ will satisfy the inequality.
Now suppose $x\ge 1$. In this case also $x^2\ge1$, so we can write the inequality as
$$
x^2-1\ge 2x-2
$$
which becomes
$$
x^2-2x+1\ge0
$$
or
$$
(x-1)^2\ge0
$$
which is surely true for all $x$, in particular for $x\ge 1$.
Conclusion: the inequality holds for all real $x$.

The "automated" method has been treated in other answers; to summarize it, you can divide the inequality into two and you'll put together the solution sets at the end
$$
\begin{cases}
x^2-1\ge0\\
x^2-1\ge 2x-2
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x^2-1<0\\
1-x^2\ge 2x-2
\end{cases}
$$
Simple algebraic manipulation transforms them into
$$
\begin{cases}
x\le -1 \text{ or }x\ge1\\
(x-1)^2\ge0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
-1<x<1\\
x^2+2x-3\le0
\end{cases}
$$
and, going on,
$$
x\le -1\text{ or }x\ge 1
\qquad\text{or}\qquad
\begin{cases}
-1<x<1\\
-3\le x\le 1
\end{cases}
$$
whence the same conclusion as before. (Note: the brace is another way of saying “and”.)
A: As $\displaystyle |x|=\begin{cases} x &\mbox{if }  x\ge0 \\-x  & \mbox{if } x<0 \end{cases} $
If $x^2-1\ge0\iff x\ge1$ or $x\le-1,$
we get $$x^2-1\ge2x-2\iff x^2-2x+1\ge0\iff (x-1)^2\ge0$$ which is true
If $x^2<1\iff -1<x<1\  \ \  \  (1),$
we get $$-(x^2-1)>2x-2\iff x^2+2x-3<0$$
$$\iff (x+3)(x-1)<0\iff -3<x<1\  \ \ \ (2)$$ 
Now using $(1),(2)-1<x<1$
So, $x$ can assume any real value
A: You have two cases :
Case 1 : When $x^2-1\ge0$, you have $x^2-1\ge 2x-2.$
Case 2 : When $x^2-1\lt0$, you have $-(x^2-1)\ge 2x-2.$
The answer is $$"x^2-1\ge0\ \text{and}\ x^2-1\ge 2x-2"\ \text{or}\ "x^2-1\lt0\ \text{and}\ -(x^2-1)\ge 2x-2."$$
A: Hints:
$$\begin{align*}|x|<1\implies x^2-1<0\implies |x^2-1|=1-x^2\\
|x|\ge 1\implies x^2-1\ge 0\implies |x^2-1|=x^2-1\end{align*}$$
So, for example, for  $\;x=-1\;$ you get
$$|(-1)^2-1|\ge 2(-1)-2\iff2\ge-4\;\color{green}{\checkmark}$$
A: Clearly $x=1$ is a solution.  Dividing by $|x-1|$ yields to
$$|x+1|\leq2\frac{x-1}{|x-1|},$$
which is obviously true both for $x<1$ and $x>1$.
