In $\triangle ABC$, prove that $\;$ $a \cos A + b \cos B + c \cos C = 4 R \sin A \sin B \sin C = \dfrac{a b c}{2 R^2}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
and $\;$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
Now,
$a \cos A + b \cos B + c \cos C$
$= 2 R \sin A \cos A + 2 R \sin B \cos B + 2 R \sin C \cos C$ $\;\;$ [by sine rule]
$= R \left(2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C\right)$
$= R \left(\sin 2A + \sin 2B + \sin 2C\right)$
$\left[\text{Note: } \sin \left(\alpha + \beta\right) = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right)\right]$
$= R \left[2 \sin \left(\dfrac{2 A + 2 B}{2}\right) \cos \left(\dfrac{2 A - 2 B}{2}\right) + \sin 2C\right]$
$= R \left[2 \sin \left(A + B\right) \cos \left(A - B\right) + 2 \sin C \cos C\right]$
$\begin{aligned}
\bigg[ \text{Note: In } \triangle ABC, \;\; & A + B + C = \pi \\\\
& i.e. \; A + B = \pi - C \\\\
& i.e. \; \sin \left(A + B\right) = \sin \left(\pi - C\right) = \sin C \\\\
\text{Also, } & C = \pi - \left(A + B\right) \\\\
& i.e. \; \cos C = \cos \left[\pi - \left(A + B\right)\right] = - \cos \left(A + B\right) \bigg]
\end{aligned}$
$= 2 R \left[\sin C \cos \left(A - B\right) - \sin C \cos \left(A + B\right)\right]$
$= 2 R \sin C \left[\cos \left(A - B\right) - \cos \left(A + B\right)\right]$
$\left[\text{Note: } \cos \left(\alpha - \beta\right) = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)\right]$
$= 2 R \sin C \times 2 \sin \left(\dfrac{A - B + A + B}{2}\right) \sin \left(\dfrac{A + B - A + B}{2}\right)$
$= 4 R \sin C \sin \left(\dfrac{2 A}{2}\right) \sin \left(\dfrac{2 B}{2}\right)$
$= 4 R \sin A \sin B \sin C$
$= 4 R \times \dfrac{a}{2 R} \times \dfrac{b}{2 R} \times \dfrac{c}{2 R}$
$= \dfrac{a b c}{2 R^2}$
$\therefore \;$ $a \cos A + b \cos B + c \cos C = 4 R \sin A \sin B \sin C = \dfrac{a b c}{2 R^2}$
Hence proved.