Extreme of $\cos A\cos B\cos C$ in a triangle without calculus. 
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.

I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing  helps.
On this topic we learned also about Cauchy inequality, but I have no experience with it.
The answer according to Mathematica is when $A=B=C=60$.
Any ideas?

If $y=\cos A\cos B\cos C,$
$2y=\cos C[2\cos A\cos B]=\cos C\{\cos(A-B)+\cos(A+B)\}$
As $A+B=\pi-C,\cos(A+B)=-\cos C$
On rearrangement we have $$\cos^2C-\cos C\cos(A-B)+2y=0$$
As $C$ is real, so will be $\cos C$
$\implies$ the discriminant  $$\cos^2(A-B)-8y\ge0\iff y\le\dfrac{\cos^2(A-B)}8\le\dfrac18$$
The equality occurs if $\cos^2(A-B)=1\iff\sin^2(A-B)=0$
$\implies A-B=n\pi$ where $n$ is any integer
As $0<A,B<\pi, n=0\iff A=B$ and consequently $$\cos^2C-\cos C+2\cdot\dfrac18=0\implies \cos C=\dfrac12\implies C=\dfrac\pi3$$
$\implies A=B=\dfrac{A+B}2=\dfrac{\pi-C}2=\dfrac\pi3=C$
