In a $\triangle ABC$, prove that $\;$ $a \left(r r_1 + r_2 r_3\right) = b \left(r r_2 + r_3 r_1\right) = c \left(r r_3 + r_1 r_2\right)$
Inradius $= r = \dfrac{\Delta}{s}$
Escribed radius $= r_1 = \dfrac{\Delta}{s - a}$
Escribed radius $= r_2 = \dfrac{\Delta}{s - b}$
Escribed radius $= r_3 = \dfrac{\Delta}{s - c}$
where $\;$ $\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ is the area of $\triangle ABC$
i.e. $\;$ $\Delta^2 = s \left(s - a\right) \left(s - b\right) \left(s - c\right)$
and $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$
i.e. $\;$ $2 s = a + b + c$
Now,
$\begin{aligned}
a \left(r r_1 + r_2 r_3\right) & = a \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - a\right)} + \dfrac{\Delta}{\left(s - b\right)} \times \dfrac{\Delta}{\left(s - c\right)}\right] \\\\
& = \dfrac{a \Delta^2 \left[s^2 - sc - s b + b c + s^2 - s a\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\
& = a \left[2 s^2 - s \left(a + b + c\right) + bc\right] \\\\
& = a \left[2 s^2 - 2 s^2 + bc\right] \\\\
& = a b c \;\;\; \cdots \; (1a)
\end{aligned}$
$\begin{aligned}
b \left(r r_2 + r_3 r_1\right) & = b \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - b\right)} + \dfrac{\Delta}{\left(s - c\right)} \times \dfrac{\Delta}{\left(s - a\right)}\right] \\\\
& = \dfrac{b \Delta^2 \left[s^2 - s a - s c + a c + s^2 - s b\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\
& = b \left[2 s^2 - s \left(a + b + c\right) + ac\right] \\\\
& = b \left[2 s^2 - 2 s^2 + ac\right] \\\\
& = a b c \;\;\; \cdots \; (1b)
\end{aligned}$
$\begin{aligned}
c \left(r r_3 + r_1 r_2\right) & = c \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - c\right)} + \dfrac{\Delta}{\left(s - a\right)} \times \dfrac{\Delta}{\left(s - b\right)}\right] \\\\
& = \dfrac{c \Delta^2 \left[s^2 - s b - s a + a b + s^2 - s c\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\
& = c \left[2 s^2 - s \left(a + b + c\right) + ab\right] \\\\
& = c \left[2 s^2 - 2 s^2 + ab\right] \\\\
& = a b c \;\;\; \cdots \; (1c)
\end{aligned}$
$\therefore \;$ We have from equations $(1a)$, $(1b)$ and $(1c)$,
$a \left(r r_1 + r_2 r_3\right) = b \left(r r_2 + r_3 r_1\right) = c \left(r r_3 + r_1 r_2\right)$
Hence proved.