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 $-3/2$

If $\displaystyle2x+2y+2z=n\pi,$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
As $\displaystyle x+y=n\pi-y, \cos(x+y)=(-1)^n\cos z$
$$\implies2\cos^2z\pm2\cos z\cos(x-y)-(1+F)=0$$ which is a Quadratic Equation in $\cos z$
So, the discriminant must be $\ge0$ 
$$\implies (2\cos(x-y))^2\ge 4.\cdot2(-1-F)\iff 2F\ge-2-2\cos^2(x-y)\ge-3 $$
The equality occurs if $\cos^2(x-y)=1\implies \cos2(x-y)=2\cos^2(x-y)-1=1\iff 2(x-y)=2m\pi$ where $m$ is an integer
