<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] As Berci suggested ,$x=A-B,y=B-C \to f=\cos {x} +\cos{y}+cos{(x+y)}$
$f=2\cos{\dfrac{x+y}{2}} \cos{\dfrac{x-y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 \ge 2\cos{\dfrac{x+y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 =2\left(\cos{\dfrac{x+y}{2}}+\dfrac{1}{2}\right)^2-\dfrac{3}{2} \ge -\dfrac{3}{2}$
first " $\ge$ " : $\cos{\dfrac{x+y}{2}}<0,x=y$
second " $\ge$ " : $\cos{\dfrac{x+y}{2}}=-\dfrac{1}{2}$
so the "=" will hold  when $x=y= \dfrac{2\pi}{3}(or \dfrac{4\pi}{3})+2k\pi $
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