Proving $\left|\frac{r-x}{1-xr}\right| \leq r$ for $0\leq x\leq r$.
Solution 1:
Note that for the given conditions $f(x)\ge 0$ so we can get rid of the absolute value:
- for $x\in[0,r]$ you have $r-x\ge 0$
- also $1-xr\ge 1-r^2>0$ for $r\in(0,1)$
We can then study the sign of $\ r-f(x)=r-\dfrac{r-x}{1-xr}=\dfrac{(1-r^2)x}{1-xr}\ge 0$ for the same reasons as above.
Solution 2:
$$ f(x)=\frac{r-x}{1-xr} = \frac 1r + \frac{r^2-1}{r(1-xr)} $$ is decreasing on $[0, r]$, so that $$ r = f(0) \ge f(x) \ge f(r) = 0 $$ on that interval.