In a $\triangle ABC$, prove that $\;$ $\Delta = 4 \; R \; r \; \cos \left(\dfrac{A}{2}\right) \; \cos \left(\dfrac{B}{2}\right) \; \cos \left(\dfrac{C}{2}\right)$
In-radius $\;$ $r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$
Circumradius $\;$ $R = \dfrac{a \; b \; c}{4 \Delta}$
$\sin A = \dfrac{2 \Delta}{b c}$, $\;$ $\sin B = \dfrac{2 \Delta}{c a}$, $\;$ $\sin C = \dfrac{2 \Delta}{a b}$
where $\;$ $\Delta$ $\;$ is the area of $\triangle ABC$.
$\begin{aligned}
RHS & = 4 \; R \; r \; \cos \left(\dfrac{A}{2}\right) \; \cos \left(\dfrac{B}{2}\right) \; \cos \left(\dfrac{C}{2}\right) \\\\
& = 4 R \times 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \times \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\
& \left[\text{Note: } \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) = \dfrac{\sin \theta}{2}\right] \\\\
& = 16 R^2 \times \dfrac{\sin A}{2} \times \dfrac{\sin B}{2} \times \dfrac{\sin C}{2} \\\\
& = 2 R^2 \sin A \sin B \sin C \\\\
& = 2 \times \left(\dfrac{abc}{4 \Delta}\right)^2 \times \left(\dfrac{2 \Delta}{bc}\right) \times \left(\dfrac{2 \Delta}{ca}\right) \times \left(\dfrac{2 \Delta}{ab}\right) \\\\
& = \dfrac{16 \; \Delta^3 \; a^2 \; b^2 \; c^2}{16 \; \Delta^2 \; a^2 \; b^2 \; c^2} \\\\
& = \Delta = LHS
\end{aligned}$
Hence proved.