Method to solve $|x| + |2-x| \leq x+ 1$ Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |\leq x+1.$$
The book says that $ x \in [1,3] $.
I first rewrote as $x+(2-x)\leq x+1$
with $x\geq 0$ and $-x-(2-x)\leq x+1$ with $x<0$. Then I solved.
For the first, I got $1\leq x$ and for the 2nd, $-3\leq x$ 
$$x\in ]-3;0[\cup[1;+\infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong? 
Thanks for your help.

The mistake you have made is that,you have taken 
$|2-x| = 2-x \forall x\ge 0$, which is not true. 
For $x \gt 2$, $|2-x|$ is $ = x-2$. 
Hence, you must take intervals as three cases, as George Law points out, 
$ (i)x\le 0$,
$(ii) 0\lt x\le2$ 
$(iii) x>2$
