Solving $\lfloor|x+1|-|x|\rfloor\geq x^2$

Question

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!

Answer 1

Hint: Since $-1<x<0$, then $$-1<2x+1<1$$ which implies $$\lfloor 2x+1\rfloor =-1\quad\text{or}\quad \lfloor 2x+1\rfloor =0$$

Answer 2

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 $x\in(0,1)$.

Answer 3

$\lfloor 2x + 1\rfloor$ 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 $(0,1)$.