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!

$\lfloor 2x + 1\rfloor$ takes on two possible values for $-1 < x < 0$: $-1$ and $0$.
It takes value $-1$ if $2x + 1 < 0$, which happens when $x < -\frac12$; in this case, $\lfloor 2x + 1\rfloor$ cannot be larger than $x^2$, since $x^2$ is positive and $-1$ is not.
It takes value $0$ if $0 \leq 2x + 1 < 1$, which happens when $x \geq -\frac12$ and $x < 0$. In that case, $\lfloor 2x + 1\rfloor$ cannot be larger than $x^2$, since $x^2$ is positive and $0$ is not.
In other words, there are no solutions to the inequality in the interval $(0,1)$.
