Solving $\lfloor|x+1|-|x|\rfloor\geq x^2$ I need to solve this inequality:$$\lfloor|x+1|-|x|\rfloor\geq x^2.$$
I checked the domains:


*

*$x\geq0$ resulted in $[0,1]$.

*$x\leq-1$ resulted in a contradiction.


But when trying to solve the inequality for $-1<x<0$ I get stuck:
$$
\lfloor|x + 1| - |x|\rfloor = \lfloor x + 1 - (-x)\rfloor = \lfloor2x + 1\rfloor \geq x^2,
$$
and I don't how to proceed from here…
Your help is appreciated, thanks in advance!

Depending on whether $x<-1/2$ or $-1/2\le x$, when $x\in(0,1)$, $\lfloor 2x+1\rfloor$ is $-1$ or $0$ respectively. But over the same interval, $x^2>0$. So there are no solutions if $x\in(0,1)$.
