In a $\triangle ABC$, prove that $\;$ $\dfrac{r_1}{bc} + \dfrac{r_2}{ca} + \dfrac{r_3}{ab} = \dfrac{1}{r} - \dfrac{1}{2 R}$
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 $= r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
where $\;$ $R$ $\;$ is the circumradius of $\triangle ABC$
Now,
$\begin{aligned}
\dfrac{r_1}{bc} & = \dfrac{4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{2 R \sin B \times 2 R \sin C} \\\\
& \left[\text{Note: } \sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \right] \\\\
& = \dfrac{\sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{R \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right)} \\\\
& = \dfrac{\sin \left(\dfrac{A}{2}\right)}{4 R \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)} \\\\
& = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)} \\\\
& = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{r} \;\;\; \cdots \; (1a)
\end{aligned}$
Similarly,
$\dfrac{r_2}{ca} = \dfrac{\sin^2 \left(\dfrac{B}{2}\right)}{r}$ $\;\;\; \cdots \; (1b)$ $\;$ and $\;$ $\dfrac{r_3}{ab} = \dfrac{\sin^2 \left(\dfrac{C}{2}\right)}{r}$ $\;\;\; \cdots \; (1c)$
$\therefore \;$ In view of equations $(1a)$, $(1b)$ and $(1c)$, we have
$\begin{aligned}
LHS & = \dfrac{r_1}{bc} + \dfrac{r_2}{ca} + \dfrac{r_3}{ab} \\\\
& = \dfrac{1}{r} \left[\sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)\right] \;\;\; \cdots \; (2)
\end{aligned}$
$\left[\text{Note: } \sin^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}\right]$
Now,
$\sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$= \dfrac{1 - \cos A}{2} + \dfrac{1 - \cos B}{2} + \sin^2 \left(\dfrac{C}{2}\right)$
$= 1 - \dfrac{1}{2} \left[\cos A + \cos B\right] + \sin^2 \left(\dfrac{C}{2}\right)$
$\left[\text{Note: }\cos \alpha + \cos \beta = 2 \cos \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right)\right]$
$= 1 - \cos \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$\left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \implies \dfrac{A + B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2} \right.$
$\left. \therefore \; \cos \left(\dfrac{A + B}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \sin \left(\dfrac{C}{2}\right) \right]$
$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$= 1 - \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A - B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$= 1 - \sin \left(\dfrac{C}{2}\right) \left[\cos \left(\dfrac{A - B}{2}\right) - \sin \left(\dfrac{C}{2}\right)\right]$
$\left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \implies \dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right) \right.$
$\left. \therefore \; \sin \left(\dfrac{C}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)\right] = \cos \left(\dfrac{A + B}{2}\right) \right]$
$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$= 1 - \sin \left(\dfrac{C}{2}\right) \left[\cos \left(\dfrac{A - B}{2}\right) - \cos \left(\dfrac{A + B}{2}\right)\right]$
$\left[\text{Note: }\cos \alpha - \cos \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)\right]$
$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$
$= 1 - \sin \left(\dfrac{C}{2}\right) \times 2 \sin \left(\dfrac{A - B + A + B}{4}\right) \sin \left(\dfrac{A - B - A + B}{4}\right)$
$= 1 - 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (3)$
$\therefore \;$ We have from equations $(2)$ and $(3)$,
$\begin{aligned}
LHS & = \dfrac{1}{r} \left[1 - 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \\\\
& \left[\text{Note: In-radius } = r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \right. \\
& \left. \implies \dfrac{r}{2 R} = 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \right] \\\\
& = \dfrac{1}{r} \left[1 - \dfrac{r}{2 R}\right] \\\\
& = \dfrac{1}{r} - \dfrac{1}{2 R} = RHS
\end{aligned}$
Hence proved.
