0<\lfloor x\rfloor(1-\{x\})+1+1-\{x\}^2=(1-\{x\})\underbrace{(\lfloor x\rfloor +1+\{x\})}_{<0}+1\leq\left(1-\frac{-\lfloor x\rfloor-3}{-\lfloor x\rfloor-1}\right)\left(\lfloor x\rfloor +1+\frac{-\lfloor x\rfloor-3}{-\lfloor x\rfloor-1}\right)+1=\frac{-(\lfloor x\rfloor+2)^2-3}{(\lfloor x\rfloor +1)^2}<0,