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

Question

I need to solve this and explain every step: $$\left\lfloor\;|x+1|-|x|\;\right\rfloor \geq x^2$$

My try:

$$\lfloor 1\rfloor =\left\lfloor\;|x+1-x|\right\rfloor \geq \left\lfloor\;|x+1|-|x|\;\right\rfloor\geq x^2$$ $$ \lfloor 1\rfloor \geq x^2 $$ I understand it has something to do with $\lfloor\cdot\rfloor$, but I can't find out.

Answer 1

Let's spend some time understanding the function on the left.

By triangle inequality:

$$|x+1| \le |x| + 1$$

$$|x+1|-|x| \le 1$$

The equality holds if and only if $x \ge 0$.

Hence for $x < 0$, $\lfloor |x+1|-|x|\rfloor \le 0$, however, when $x<0$, $x^2 >0$.

For $x \ge 0$, $\lfloor |x+1|-|x| \rfloor =1$, $x^2 \le 1$ when $0 \le x \le 1$.

Hence the solution is $0 \le x \le 1$.

Answer 2

$\lfloor \, |x+1|-|x| \, \rfloor \geq x^2$ which means the RHS is $\geq 0$, so LHS has to be positive as well. ($x \in \mathbb{R}$)

For $x \geq 0$, it reduces to $\lfloor \, 1 \, \rfloor \geq x^2$ which gives us a solution $0 \leq x \leq 1$.

For $-1 \leq x \lt 0$, it reduces to
$\lfloor \, x + 1 - (-x) \, \rfloor \geq x^2 \implies \lfloor \, 2x + 1 \, \rfloor \geq x^2$ which does not give us any solution either.

For $x \lt -1$, it reduces to
$\lfloor \, - x - 1 - (-x) \, \rfloor \geq x^2 \implies \lfloor \, - 1 \, \rfloor \geq x^2$ and hence no solutions.