Thank you for reporting, we will resolve it shortly
Q.
If $R \ge r > 0$ and $d > 0,$ then $0<\frac{d^{2}+R^{2}-r^{2}}{2dR} \le1$
Linear Inequalities
Solution:
Given $R \ge r > 0$ and $d > 0$
$\Rightarrow 0<\frac{d^{2}+R^{2}-r^{2}}{2dR} \le1$
$\Rightarrow 0 <\left(d+R-r\right)\left(d+R+r\right) \le2\,dR;$ which is true iff
$\left(d^{2}+R^{2}-r^{2}\right)\le2\,dR,$ which is true iff $d^{2}+R^{2}-2dR \le r^{2}$
$\Rightarrow \left(d-R\right)^{2} \le r^{2}$
$\left|d-R\right|\le r,$ which is also $- r \le\left(d-R\right)\le r$
$\left|d-R\right|\le,$ which is also $-r \le\left(d-R\right)\le r$