In a $\triangle ABC$, prove that $\;$ $r_1 + r_2 + r_3 - r = 4R$ $\;$ 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 $\triangle ABC$.
$\begin{aligned}
LHS & = r_1 + r_2 + r_3 - r \\\\
& = \left(r_1 - r\right) + \left(r_2 + r_3\right) \\\\
& = 4 R \sin \left(\dfrac{A}{2}\right) \left[\cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) - \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \\
& \hspace{1cm} + 4 R \cos \left(\dfrac{A}{2}\right) \left[\sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) + \cos \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \;\;\; \cdots \; (1)
\end{aligned}$
Now, $\;$ $\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ $\;\;\; \cdots \; (2a)$
$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,
$LHS = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B + C}{2}\right) + 4 R \cos \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B + C}{2}\right)$ $\;\;\; \cdots \; (3)$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
$\implies$ $B + C = \pi - A$
i.e. $\;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$
$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (4a)$
Similarly, $\;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (4b)$
In view of equations $(4a)$ and $(4b)$ equation $(3)$ becomes,
$\begin{aligned}
LHS & = 4 R \sin \left(\dfrac{A}{2}\right) \times \sin \left(\dfrac{A}{2}\right) + 4 R \cos \left(\dfrac{A}{2}\right) \times \cos \left(\dfrac{A}{2}\right) \\\\
& = 4 R \left[\sin^2 \left(\dfrac{A}{2}\right) + \cos^2 \left(\dfrac{A}{2}\right)\right] \\\\
& = 4 R = RHS
\end{aligned}$
Hence proved.