<s>[INST] 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 $-3/2$
 [/INST] 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.
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