# Inequality involving nested absolute values

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: • Following “So then I have” the same inequality is displayed twice. The solution of $2x-1 < 1$ is not $x \ge 1/2$. Sep 21 at 9:03
• What did you enter into Mathematica? The blue line is definitely not the graph of $\vert x - \vert 1-x\vert \vert$. Sep 21 at 9:19
• Wolfram Alpha seems to give a sensible graph Sep 22 at 9:48

$$x< 1$$ 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].
• Thank you, yet Mathematica displays a plot in which $x = 2$ seems legit... Sep 21 at 9:12
"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 $$-1<2x-1<1\\ 0<2x<2\\ 0 Hence the solution is $$x\in (0,1)$$
Your attempt works, it is only on the very last yard, or metre, after the last case distinction, where you fail: The second row $$\,1-2x < 1\,\land\, x <\frac12\,$$ and the first row $$\,2x-1 < 1\,\land\, x\geq\frac12\,$$ yield $$\,0 and $$\,\frac12 \leqslant x < 1$$, respectively, whence any $$\,0 fulfils the inequality.
Your attempt starts by resolving the inner absolute value first. Alternatively peel the onion, i.e., first resolve the exterior one to obtain \begin{align}-1 & \;<\; x-|1-x| \;<\; 1\qquad\big|-x\quad\big|\cdot(-1)\\[2ex] \implies\quad x- 1 & \;<\; |1-x| \;<\; x+1 \end{align} The first "$$<$$" yields $$\,x-1<0\iff x<1$$. By the second "$$<$$" you then get $$\,1-x. And all is resumed by $$\,x\in (0,1)\,$$.