What am I doing wrong in this inequality? Q. The solution set of the inequality
$$||x|-1|<|1-x|, x \in R$$
My Solution:
Case1: If $x≥0$
$||x|-1|<|1-x|$
=> $|x-1|<|1-x|$
Hence, S1=>$ x \in ∅ $
Case2: If $x<0$
$||x|-1|<|1-x|$
=> $|-x-1|<|1-x|$
=> $|x+1|>|x-1|$
=> $|x+1|-|x-1|>0$
Critical Points are -1 and 1
Subcase1: If $X<-1$
$|x+1|-|x-1|>0$
=>$-(x+1)-[-(x-1)]>0$
=>$-2>0$ 
Hence SubSolution1=> $x \in ∅$
Subcase2: If $-1<x<1$<br
$|x+1|-|x-1|>0$
=>$x+1-[-(x+1)]>0$
=> $x>0$
Hence SubSolution2=> $x \in (0,1)$
Subcase3: If $x>1$
$|x+1|-|x-1|>0$
=>$x+1-(x-1)>0$
=>$2>0$
Hence SubSolution3=> $x \in (1, \infty)$
Hence Solution2=> $Subsolution1 \cup SubSolution2 \cup Subsolution3 $
=> $∅ \cup (0,1) \cup (1, \infty)$
=>$(0, \infty)$
Final solution=> $ S1 \cup S2$
=> $∅ \cup (0, \infty)$
=> $(0,\infty)$
But the actual answer is $(-\infty, 0)$
 A: In your case 2 your implication $\vert -x -1 \vert   \lt \vert 1-x\vert $ implies $\vert x +1 \vert   \gt \vert 1-x\vert $ is wrong as
$$\vert -x -1 \vert  =\vert x+1 \vert \lt \vert 1-x\vert $$
A: Here it is another way to approach it for the sake of curiosity.
Since both sides are nonnegative, we can square them so that we obtain an equivalent inequation:
\begin{align*}
||x| - 1| < |1 - x| & \Longleftrightarrow (|x| - 1)^{2} < (1 - x)^{2}\\\\
& \Longleftrightarrow x^{2} - 2|x| + 1 < 1 - 2x + x^{2}\\\\
& \Longleftrightarrow |x| > x
\end{align*}
which is true iff $x\in(-\infty,0)$, and we are done.
Hopefully this helps!
A: I'm not sure if this is in the spirit of the question but we can also solve the inequality by plotting or just visualising the graph of $|1-x|=|x-1|$ and $||x|-1|$.
First you can plot $|x-1|$ which is $|x|$ curve shifted one unit right. Now, in the other curve $x$ is replaced by $|x|$ in $|x-1|$ which makes its graph different for only negative $x$ where the curve $||x|-1|$ is reflection of $|x-1|$ about $y-axis$.
Now it can easily noticed that the inequality is satisfied for all $x<0$.
