In a $\triangle ABC$, prove that $\;$ $r_1 \cdot r_2 \cdot r_3 = r^3 \cot^2 \left(\dfrac{A}{2}\right) \cot^2 \left(\dfrac{B}{2}\right) \cot^2 \left(\dfrac{C}{2}\right)$ $\;$ where the symbols have their usual meanings.
Escribed radius $\;$ $r_1 = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)$
Escribed radius $\;$ $r_2 = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right)$
Escribed radius $\;$ $r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$
Inradius $\;$ $r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$
where $\;$ $R$ is the circumradius of $\triangle ABC$.
$\begin{aligned}
LHS & = r_1 \cdot r_2 \cdot r_3 \\\\
& = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \\
& \hspace{1cm} \times 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \\
& \hspace{2cm} \times 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \\\\
& = 64 R^3 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \cos^2 \left(\dfrac{A}{2}\right) \cos^2 \left(\dfrac{B}{2}\right) \cos^2 \left(\dfrac{C}{2}\right) \\\\
& = \left[4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right]^3 \times \dfrac{\cos^2 \left(\dfrac{A}{2}\right) \cos^2 \left(\dfrac{B}{2}\right) \cos^2 \left(\dfrac{C}{2}\right)}{\sin^2 \left(\dfrac{A}{2}\right) \sin^2 \left(\dfrac{B}{2}\right) \sin^2 \left(\dfrac{C}{2}\right)} \\\\
& = r^3 \cot^2 \left(\dfrac{A}{2}\right) \cot^2 \left(\dfrac{B}{2}\right) \cot^2 \left(\dfrac{C}{2}\right) \\\\
& = RHS
\end{aligned}$
Hence proved.