Solve the equation within 'floor function' I added my solution, but I'm not sure I've got it right.
I'd like to know what you think.
The question:

Solve the equation: $$\lfloor |x+1|-|x| \rfloor \ge x^2.$$


the left and right symbols aren't square brackets, they are 'floors'
more info:
https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
My Solution:

Thanks.
 A: At first notice that for  $x=-1$, we have $\lfloor |-1+1|-|-1| \rfloor = -1$ however, $x^2=1$, so the inequality does not hold.
Thus a better approach is to consider different cases to treat the absolute value. 


*

*If  $x< - 1 \Rightarrow  x+1 <0 \text{ and } x<0  \Rightarrow  \lfloor |x+1|-|x| \rfloor =\lfloor -x-1+x \rfloor = -1 $ and as  $x ^2 \geq 0$, so in this case no $x$ satisfies the inequality.

*If $-1 \leq x <0  \Rightarrow x+1 \geq 0 \text{ and } x<0  \Rightarrow  \lfloor |x+1|-|x| \rfloor =\lfloor x+1 + x \rfloor = \lfloor 2x+1 \rfloor $
. Let $ n =\lfloor 2x+1 \rfloor$ but $-1 \leq x <0  \Rightarrow -1 \leq 2x +1<1   $ so  $n=-1 \text{ or } 0$. If  $n=-1<0$ the inequqlity  clearly does not hold, and if  $n=0$ then $0 \geq  x^2 \Rightarrow x=0 $ , but $x <0$.So in this case also we have no solution.

*If $x  \geq 0 \Rightarrow x+1 \geq 0 \text{ and } x<0  $, then  $\lfloor |x+1|-|x| \rfloor= \lfloor x+1-x \rfloor = \lfloor 1 \rfloor =1$, so  $1 \geq x^2  \Rightarrow  -1 \leq x \leq 1 $ but  in this case  $x \geq  0 $ so the set of solution is $ x \in [ 0,1]$.  


Thus the set of solution is $ x \in [ 0,1]$.
