In $\triangle ABC$, prove that $a b c \left(\cot A + \cot B + \cot C\right) = R \left(a^2 + b^2 + c^2\right)$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
LHS & = a b c \left(\cot A + \cot B + \cot C\right) \\\\
& = a b c \left(\dfrac{\cos A}{\sin A} + \dfrac{\cos B}{\sin B} + \dfrac{\cos C}{\sin C}\right) \\\\
& = a b c \left(\dfrac{\dfrac{b^2 + c^2 - a^2}{2 b c}}{\dfrac{a}{2 R}} + \dfrac{\dfrac{c^2 + a^2 - b^2}{2 c a}}{\dfrac{b}{2 R}} + \dfrac{\dfrac{a^2 + b^2 - c^2}{2 a b}}{\dfrac{c}{2 R}}\right) \\\\
& = R \left(b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2\right) \\\\
& = R \left(a^2 + b^2 + c^2\right) \\\\
& = RHS
\end{aligned}$
Hence proved.