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 $-3/2$
 A: 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.
A: 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 $
A: 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
A: This result can be used (substituting $(A-B)$ as $X$, $(B-C)$ as $Y$)
