In $\triangle A B C, A+B+C=\pi$ ...(i)
When $A, B, C$ are the angles of $\triangle A B C$.
Let $R$ be the circumradius
and $s$ be semiperimeter and $\triangle$ be the area of $\triangle A B C$
$\frac{16 R s \Delta \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}}{s-c}$
$=\frac{(4 R \Delta)(4 s)}{s-c} \cdot \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}} \sqrt{\frac{s(s-c)}{a b}}$
$\because \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}, \sin \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{a c}}$
$\cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{a b}}$
$r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$
$\begin{aligned} & \text { and } \Delta=\frac{a b c}{4 R} \\ & \text { or } 4 R \Delta=a b c\end{aligned}$
$\begin{aligned} & =\frac{(a b c)(4 s)}{s-c} \frac{(s-c) \sqrt{s(s-a)(s-b)(s-c)}}{a b c} \\ & =4 \Delta s\end{aligned}$
$\begin{aligned} & =\frac{4 s \cdot \Delta^3}{\Delta^2}=\frac{4 s \Delta^3}{s(s-a)(s-b)(s-c)} \\ & =4 \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}\end{aligned}$
$=4 r_1 r_2 r_3$