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.

There are 3 cases: 
a) $x\leq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |\leq x+1\implies -x+ 2-x\leq x+1$$
$$-3x\leq -1$$
$$3x\geq1$$
$$x\geq\frac{1}{3}$$
b) $0<x\leq 2$:
$$|x| +|2-x |\leq x+1\implies x+2-x\leq x+1$$
$$x\geq 1$$
c) $x > 2$:
$$|x| +|2-x |\leq x+1\implies x+x-2\leq x+1$$
$$2x\leq x+3\implies x\leq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1\leq x\leq 3\implies x\in [1;3]$
