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 ?

RHS is defined only for $-2< x<2$ and $x\ne0$.. 
If $-1<x<0$ or $0<x<1$, then RHS is $\frac{1}{-1}=-1$. Solve $x^2-2x+1\le0$ and only keep solutions in the initial interval.
If $1\le |x|<2$, RHS is $0$, solve $x^2-2x\le 0$ and only keep solutions in intervals $\left]-2,-1\right]$ and $\left[1,2\right[$.
Remember, when you encounter something like $\lfloor x\rfloor$ or $\left|x\right|$, try first to give it a value by making hypothesis on the positions of $x$.
