In a triangle, find the minimum and maximum of $\cos(A-B)\cos(B-C)\cos(C-A)$ In a triangle, with $A, B, C$ are three angles, find the minimum and maximum of $$\cos(A-B)\cos(B-C)\cos(C-A)$$
 A: They are multiplication of cosinus of three angles which have sum of zero.
So:
$$\min(\cos(A-B)\cos(B-C)\cos(C-A))$$
Is the same as 
$$\min(\cos(x)\cos(y)\cos(-x-y))$$
or in another form
$$\min(\cos(x)\cos(y)\cos(x+y))$$
Due to symmetric position of $x$ and $y$, the candidates for minimum must be at the boundaries ($\cos . =\pm1$) or where $x=y$
We can check them.
For when $x=y$:
$$A=\cos(x)\cos(y)\cos(x+y)=\cos^2(x)\cos(2x)$$
$$\frac{dA}{dx}=-\sin(2x)-\sin(4x)=0$$
$$\sin(2x)=-2\sin(2x)\cos(2x)$$
$$\sin(2x)(1+2\cos(2x))=0$$
if $\sin(2x)=0$ then $A=0$
if $\cos(2x)=\frac{-1}{2}$ then $A=\frac{-1}{8}$
So, the absolute mimimum is $\frac{-1}{8}$
Update
$\cos(2x)=\frac{-1}{2}$ means that 
$$x=k \pi \pm \frac{\pi}{3}$$
if we set $x$ as $(A-B)$ and $y$ as $(B-C)$ then since $x=y$ then:
$$A-B=B-C=k \pi \pm \frac{\pi}{3}$$
And there is always a degree of a freedom.
If you add any equal value to all A, B, and C the result of the multiplication remains the same. So A, B and C can be any number as far as they satisfy the equation above.
Just an example can be $B=0$, $A=\frac{\pi}{3}$ and $C=\frac{-\pi}{3}$
$$cos(\frac{\pi}{3})\times \cos(\frac{\pi}{3}) \times \cos(\frac{-2\pi}{3})=\frac{-1}{8}$$
Fix
For angles in a triangle, we need to add additional condition:
$$A+B+C=\pi$$
After, I did all my bests, I found these angles cannot be placed in any triangle. It means in a trangle, due to the following contrains:
$$0<A<\pi$$
$$0<B<\pi$$
$$0<C<\pi$$
we would have $A>\frac{-1}{8}$ instead of $A\ge\frac{-1}{8}$. So we can approach very close to $\frac{-1}{8}$ but we can never achieve it in a real triangle.
According to suggestion of JimmyK4542:
$$A=\varepsilon, B=\frac{\pi}{3}, C=\frac{2\pi}{3}-\varepsilon$$
Where $\varepsilon$ is very close to zero causes the multiplication reach very close to $\frac{-1}{8}$ but $\varepsilon$ never touches zero in a real triangle. It just approaches it.
Maximum
The maximum is when all $\cos(.)$ are at their maximum which means all angles are equivalent.
Symmetry

Also, this answer still relies on the fact that the minimum of
    cosxcosycos(x+y) "must be at the boundaries (cos.=±1) or where x=y",
    which needs to be justified. The fact that cosxcosycos(x+y) is
   symmetric with respect to x and y isn't enough to justify this
JimmyK4542 

I agree. It looks like an open discussion: link
A: Let $A-B=x$ etc. $\implies x+y+z=0$
and $S=\cos x\cos y\cos z\iff2S=2\cos x\cos y\cos z=\cos x[\cos(y+z)+\cos(y-z)]$
$\implies2S=\cos x[\cos(-x)+\cos(y-z)]\iff \cos^2x+\cos x\cos(y-z)-2S=0$
which is a Quadratic Equation in $\cos x$
So, the discriminant $\cos^2(y-z)-4\cdot1(-2S)\ge0$
$\iff8S\ge-\cos^2(y-z)=\sin^2(y-z)-1\ge-1\implies S\ge?$ 
The equality occurs if $\sin^2(y-z)=0\iff\sin(y-z)=0$
