minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$

How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$

Starting with $F=\cos(a-b)+\cos(b-c)+\cos(c-a)$ expand the cosines via the addition formula for cosine, and get $$\cos a \cos b + \cos b \cos c + \cos c \cos a \\ +\sin a \sin b + \sin b \sin c + \sin c \sin a.$$ Then after applying $(u+v+w)^2-u^2-v^2-w^2=2uv+2vw+2wu,$ the double $2F$ of our objective function may be seen to be $$2F=(\cos a+\cos b +\cos c)^2+(\sin a +\sin b + \sin c)^2-3.$$ Note we have combined the terms e.g. $-\cos^2 a -\sin^2 a=-1$ to obtain the final $-3.$ Thus $2F \ge -3$ i.e. $F \ge -3/2.$ Since there are values of $a,b,c$ which achieve $F=-3/2$ this finishes a proof.