In a $\triangle ABC$, prove that $\;$ $\dfrac{1}{r^2} + \dfrac{1}{r_1^2} + \dfrac{1}{r_2^2} + \dfrac{1}{r_3^2} = \dfrac{a^2 + b^2 + c^2}{\Delta^2}$
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,
$\begin{aligned}
LHS & = \dfrac{1}{r^2} + \dfrac{1}{r_1^2} + \dfrac{1}{r_2^2} + \dfrac{1}{r_3^2} \\\\
& = \dfrac{s^2}{\Delta^2} + \dfrac{\left(s - a\right)^2}{\Delta^2} + \dfrac{\left(s - b\right)^2}{\Delta^2} + \dfrac{\left(s - c\right)^2}{\Delta^2} \\\\
& = \dfrac{s^2 + s^2 - 2 a s + a^2 + s^2 - 2 b s + b^2 + s^2 - 2 c s + c^2}{\Delta^2} \\\\
& = \dfrac{4 s^2 - 2 s \left(a + b + c\right) + a^2 + b^2 + c^2}{\Delta^2} \\\\
& = \dfrac{4 s^2 - 4 s^2 + a^2 + b^2 + c^2}{\Delta^2} \\\\
& = \dfrac{a^2 + b^2 + c^2}{\Delta^2} = RHS
\end{aligned}$
Hence proved.