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?

I use the Lagrange's multipliers theorem.
Let us define the functions
$$g(A,B,C)=A+B+C-\pi,$$
$$f(A,B,C)=\cos A\cos B\cos C.$$
Then we have
$$\mathrm{d}g(A,B,C)=\mathrm{d}A+\mathrm{d}B+\mathrm{d}C,$$
$$\mathrm{d}f(A,B,C)=-\sin A\cos B\cos C\mathrm{d}A-\cos A\sin B\cos C\mathrm{d}B-\cos A\cos B\sin C\mathrm{d}C$$
where $\left(\mathrm{d}A,\mathrm{d}B,\mathrm{d}C\right)$ is the coordinate forms on $\mathbb{R}^3$ (for example, $\mathrm{d}A[(1,2,3)]=1$).
Then we apply the Lagrange's multipliers theorem : we must cancel every determinants of the matrix
$$\left(\begin{array}{cccccccc}
1 & 1 & 1 \\
-\sin A\cos B\cos C & -\cos A\sin B\cos C & -\cos A\cos B\sin C
\end{array}\right).$$
This yields
\begin{cases} -\cos A\sin B\cos C + \sin A\cos B\cos C = 0 \\
-\cos A\cos B\sin C + \sin A\cos B\cos C = 0 \\
-\cos A\cos B\sin C + \cos A\sin B\cos C=0
\end{cases}
The first line gives us $\cos C = 0$ (and then $C=\pi/2$) or
$$-\cos A\sin B + \sin A\cos B =0$$
that is
$$\sin(A-B)=0$$
and then $A=B$.
Do the same for the two other lines to get the condition $A=B=C$ (the other conditions are impossible, check that). Because in a triangle, we have $g(A,B,C)=0$, we finally find that $A=B=C=\pi/3$.
