In $\triangle ABC$, prove that $4 \left[b c \cos^2 \left(\dfrac{A}{2}\right) + c a \cos^2 \left(\dfrac{B}{2}\right) + a b \cos^2 \left(\dfrac{C}{2}\right)\right] = \left(a + b + c\right)^2$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
LHS & = 4 \left[b c \cos^2 \left(\dfrac{A}{2}\right) + c a \cos^2 \left(\dfrac{B}{2}\right) + a b \cos^2 \left(\dfrac{C}{2}\right)\right] \\\\
& \left[\text{Note: } \cos^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos \theta}{2}\right] \\\\
& = 4 \left[b c \left(\dfrac{1 + \cos A}{2}\right) + c a \left(\dfrac{1 + \cos B}{2}\right) + a b \left(\dfrac{1 + \cos C}{2}\right)\right]\\\\
& = 2 \left[b c \left(1 + \dfrac{b^2 + c^2 - a^2}{2 b c}\right) + c a \left(1 + \dfrac{c^2 + a^2 - b^2}{2 c a}\right) \right. \\\\
& \left. \hspace{3cm} + a b \left(1 + \dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \;\; \left[\text{by cosine rule}\right] \\\\
& = 2 b c + b^2 + c^2 - a^2 + 2 c a + c^2 + a^2 - b^2 + 2 a b + a^2 + b^2 - c^2 \\\\
& = a^2 + b^2 + c^2 + 2 a b + 2 b c + 2 c a \\\\
& = \left(a + b + c\right)^2 = RHS
\end{aligned}$
Hence proved.