Inequality involving nested absolute values

Question

I don't understand where I'm wrong in my reasoning about this inequality: $$\big\vert x - \vert 1-x\vert\big\vert < 1$$ Attempts $$\vert 1-x\vert = \begin{cases} 1-x & \text{if}\quad x\leq 1 \\ x-1 & \text{if}\quad x > 1 \end{cases}$$ So then I have $$\begin{cases} \vert x - (1-x)\vert < 1 & \text{if}\quad x\leq 1 \\ \vert x - (x-1)\vert < 1& \text{if}\quad x>1 \end{cases}$$ But the second one is false, since it reads $\vert 1 \vert < 1$, and the first one reduces to $$\vert 2x - 1 \vert < 1$$ Here again I have two cases $$\begin{cases} 2x-1 < 1 & \text{if}\quad x \geq 1/2 \\ 1-2x < 1 & \text{if}\quad x < 1/2 \end{cases}$$ with the condition above of $x \leq 1$. The solution of this is $x \geq 1/2$, hence the final solution reads $x \in\left[\frac12, 1\right]$.

Yet the solution should be $-1 < x < 1$ or $1 < x < 3$.

I don't get it.

Here is what Mathematica says:

Answer 1

$x< 1$ and $|2x-1|<1$. You cannot drop the absolute value sign here. Now consider the cases $2x-1>0$ and $2x-1 \le 0$.

[$-1 < x < 1$ or $1 < x < 3$ is also not the correct answer. For example, $x=2$ is surely not a solution].

Answer 2

"and the first one reduces to" I think you miss that the inequality you're considering is $$|2x-1|<1$$ Here you can go solving with $$-1<2x-1<1\\ 0<2x<2\\ 0<x<1$$ Hence the solution is $x\in (0,1)$

Answer 3

Your attempt works, it is only on the very last yard, or metre, after the last case distinction, where you fail: The second row $\,1-2x < 1\,\land\, x <\frac12\,$ and the first row $\,2x-1 < 1\,\land\, x\geq\frac12\,$ yield $\,0<x<\frac12\,$ and $\,\frac12 \leqslant x < 1$, respectively, whence any $\,0<x<1\,$ fulfils the inequality.

Your attempt starts by resolving the inner absolute value first. Alternatively peel the onion, i.e., first resolve the exterior one to obtain $$\begin{align}-1 & \;<\; x-|1-x| \;<\; 1\qquad\big|-x\quad\big|\cdot(-1)\\[2ex] \implies\quad x- 1 & \;<\; |1-x| \;<\; x+1 \end{align}$$ The first "$<$" yields $\,x-1<0\iff x<1$. By the second "$<$" you then get $\,1-x<x+1\iff 0<x$. And all is resumed by $\,x\in (0,1)\,$.