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 ?

Hint: $x$ cannnot be integer, otherwise the denominator of the RHS would be $0$.
So $x$ is not integer, which gives $\lfloor x \rfloor+\lfloor -x \rfloor=-1$
Also $\lfloor x \rfloor\in\{-1,0,1\}$ since it must be integer and $1-\lfloor x\rfloor^2>0$ which is equivalent to $\lfloor x\rfloor^2<1$.
Then you solve on each interval $-1<x<0$, $0<x<1$, and $1<x<2$
EDIT: For a geometric solution, I would plot $x^2-2x$ and I would observe that the RHS is either $0$ or $-1$.
