Solve the inequality : $ ||x|-1|<|1-x| $ Solve the following inequality:
$$||x|-1|<|1-x|$$
My Attempt:
I tried expanding this inequality by considering $8$ cases, but I am having trouble finding the range of the each of the solutions I got in the $8$ cases.
I am sorry I couldn't explain what I did properly. This is my third question on Math.SE  and am still learning about how to properly frame questions.
 A: If $x>0$ there is no solution since $0$ is not strictly less than $0$.
Suppose $x<0$. You have to solve then:
$|-x-1|=|x+1|<1-x$ since $1-x>0$. Squaring both side yields:
$$(x+1)^2<(1-x)^2.$$
I will let you finish.
A: Well, if $x\geqslant 0$, we have $|x|=x$, so the inequality is $|x-1|<|1-x|$,  but as $|t|=|-t|$ always, there can be no solution here.
Remaining case is $x< 0$, where the inequality is $|1+x|<|1-x|$, which says the distance of $x$ from $-1$ is less than the distance of $x$ from $1$, which is true for all negatives, so all such $x$ is in the solution set. 
A: The best I think to solve such problems is to draw the graphs. I am attaching an image below stating the steps to draw the LHS. Leaving out drawing the RHS as an exercise to you and figuring out the answer.

Hope this helps. 
Cheers! 
A: It can be interesting to notice that , in case x>0 or equal to 0, the two numbers inside the absolute value function are opposites. 

We know that, as a general rule: it is impossible for the absolute value of a number to be less than the absolute value of its opposite ( additive inverse). If A and B are opposites, then , their absolute value is necessarily equal. ( In other words, two opposite numbers are necessarily at the same distance from 0). 
We also know as a general rule that : a-b and b-a are opposites. For example : 9-3 and 3-9. So (x-1) and (1-x) are opposites, in other words (1-x) = - (x-1) 
Suppose x>0 or x=0. In that case |x| = x , and therefore 
| |x|-1 | < | 1- x | 
means 
|x-1| < |1-x| 
which implies
|x-1| < |- (x-1) | 
The case " x>0 or x = 0 "  leads to an impossibility for all values of x. The reason is that it is impossible the absolute value of A and of -A not to be equal.
So only the case in which x<0 has to be treated ( with its subordinate cases). 
A: In this case the best strategy is the graphic one(knowing that absolute value is a reflection of the negative $y$ semiplane on the positive one). Here there is the right side in blue and the leftside in red:

We can easily notice that the equality is true only for $x \geq 0$
:)
