Solving inequality involving floor function We have this inequality (over real numbers) :
$$x^2-2x\le \frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x \rfloor + \lfloor -x \rfloor}$$
How we can solve it using both of algebraic and geometric methods ?

$$\begin{cases}x= \lfloor x\rfloor+\{x\}\\-x= \lfloor -x\rfloor+1-\{x\}\end{cases}\Rightarrow\lfloor x\rfloor+\lfloor -x\rfloor=-1$$
$$\sqrt{1-\lfloor x\rfloor^2}=\begin{cases}0 \text{ if } -1\le x\lt0\\1\text{ if }  0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$ Hence
$$\frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x\rfloor+\lfloor -x\rfloor}=\begin{cases}0 \text{ if } -1\le x\lt0\\-1\text{ if }  0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$
It follows that if $f(x)=x^2-2x$ then since $$f((-1,0))=(0,3)\\f([0,1))=[-1,0]\\f([1,2))=[-1,0)$$ the solution is $$1\le x\lt 2$$
