Minimise this!

Let's use LaGrange Multipliers here: we are interested in minimizing $f(A,B,C) = \sin^{2}\frac{A}{2} + \sin^{2}\frac{B}{2} + \sin^{2}\frac{C}{2}$ subject to $g(A,B,C) = A + B + C = \pi.$ Computing $grad(f) = \lambda \cdot grad(g)$ gives:

$2\sin \frac{A}{2} \cdot \frac{1}{2} \cos \frac{A}{2} = \frac{1}{2} \sin(A) = \lambda;$

$2\sin \frac{B}{2} \cdot \frac{1}{2} \cos \frac{B}{2} = \frac{1}{2} \sin(B) = \lambda;$

$2\sin \frac{C}{2} \cdot \frac{1}{2} \cos \frac{C}{2} = \frac{1}{2} \sin(C) = \lambda;$

Or $\sin(A) = \sin(B) = \sin(C) \Rightarrow A = B = C = \frac{\pi}{3}$ is our critical point. Taking the Hessian matrix of $f$ now yields:

$\begin{bmatrix} \frac{1}{2} \cos(A) & 0 & 0 \\ 0 & \frac{1}{2} \cos(B) & 0 \\ 0 & 0 & \frac{1}{2} \cos(C) \end{bmatrix}$

and evaluating at the critical point $A = B = C = \frac{\pi}{3}$ gives:

$\begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix} = \frac{1}{4} I_{3x3}$

which is positive-definite, thus a global minimum. Ultimately, $f(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}) = 3 \cdot \sin^{2} (\frac{\pi}{6}) = \boxed{\frac{3}{4}}.$