In a $\triangle ABC$, prove that $\;$ $\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} - \dfrac{1}{r} = 0$
Escribed radius $\;$ $r_1 = \dfrac{\Delta}{s - a}$
Escribed radius $\;$ $r_2 = \dfrac{\Delta}{s - b}$
Escribed radius $\;$ $r_3 = \dfrac{\Delta}{s - c}$
In-radius $\;$ $r = \dfrac{\Delta}{s}$
where $\;$ $\Delta$ $\;$ is the area of $\triangle ABC$
and $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$
$\implies$ $2 s = a + b + c$
$\begin{aligned}
LHS & = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} - \dfrac{1}{r} \\\\
& = \dfrac{s - a}{\Delta} + \dfrac{s - b}{\Delta} + \dfrac{s - c}{\Delta} - \dfrac{s}{\Delta} \\\\
& = \dfrac{1}{\Delta} \left[s - a + s - b + s - c - s\right] \\\\
& = \dfrac{1}{\Delta} \left[2 s - \left(a + b + c\right)\right] \\\\
& = \dfrac{1}{\Delta} \left[\left(a + b + c\right) - \left(a + b + c\right)\right] \\\\
& = 0 = RHS
\end{aligned}$
Hence proved.