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Prove that $\forall x\in\Bbb R:\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1\iff|r|\le2$.

> Prove that the inequality $\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1$ holds true for all real numbers $x$ if and only if $|r|\le2$.

My (proposed) solution, which took me around 2 hours:
> First, we reframe the inequality to $\lfloor x^2\rfloor-\lfloor rx\rfloor+1\ge0$.Suppose $r=a$ satisfies the inequality for all real numbers $x$. Consequently, it also does for all real numbers $-x$:
$$\lfloor(-x)^2\rfloor-\lfloor a(-x)\rfloor+1\ge0$$
$$\lfloor x^2\rfloor-\lfloor (-a)x\rfloor+1\ge0$$
And we get that $r=-a$ satisfies the condition if and only if $r=a$ does too. Then, we only need to consider cases where $r\ge0$.

> Furthermore, let us consider cases where $r\gt2$. The inequality holds for all $x$, so we test it for $0\lt x=\frac2r\lt1$.
$$\lfloor x^2\rfloor-\lfloor2\rfloor+1\ge0$$
$$0-2+1\ge0$$, contradiction. We are left with $r\in[0,2]$.

> Another observation. If $a\ge0$ and $r=a$ fulfills the condition,
$$\lfloor x^2\rfloor-\lfloor rx\rfloor+1\ge0$$
For $x\ge0$, we find that $r\lt a$ also fulfills the condition. All three terms are positive when $x\lt0$, and therefore for $a\ge0$, all $r\lt a$ fulfill the condition if $r=a$ does. Therefore, all that's left to do is to prove the case for $r=2$, and we are done.

> Notice that $\lfloor a+b\rfloor-1\le\lfloor a\rfloor+\lfloor b\rfloor$, and $(x-1)^2\ge0\implies\lfloor x^2-2x+1\rfloor=\lfloor x^2-2x\rfloor+1\ge0$.
Combining the two,
$$0\le\lfloor x^2-2x\rfloor-1+2\le\lfloor x^2\rfloor+\lfloor-2x\rfloor+2$$.
And as long as $a\notin\Bbb Z$,
$$\lfloor-a\rfloor=-\lfloor a\rfloor-1$$. Applying that, for $2x\notin\Bbb Z$,
$$0\le\lfloor x^2\rfloor+\lfloor-2x\rfloor+2=\lfloor x^2\rfloor-\lfloor2x\rfloor+1\ \blacksquare$$
And if $n=2x\in\Bbb Z$,
$$0\le\left\lfloor\frac14n^2\right\rfloor-n+2$$.
Notice that $n\equiv1\pmod4$ or $n\equiv3\pmod4\iff n^2\equiv1\pmod4$. So,
$$0\le\frac14(n^2-1)-n+2$$
$$0\le\frac14n^2-n+\frac74$$
$$0\le\left(\frac12n-1\right)^2+\frac34\ \blacksquare$$.

Is my proposed solution correct? And better yet, is there a faster (and shorter) way to find this proof?

Thank you in advance, and feel free to correct any errors I made!