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?

Due to the inequality of means and the law of cosines, we know that:
\begin{equation*}
       4a^4=((a^2+b^2-c^2)+(a^2+c^2-b^2))^2\geq 4(a^2+b^2-c^2)(a^2+c^2-b^2)=\\ 16a^2bc\cos\beta \cos\gamma \Rightarrow a^2\geq 4bc\cos\beta \cos\gamma 
  \end{equation*}
\begin{equation*}
       4b^4=((a^2+b^2-c^2)+(b^2+c^2-a^2))^2\geq 4(a^2+b^2-c^2)(b^2+c^2-a^2)=\\ 16ab^2c\cos\alpha \cos\gamma \Rightarrow b^2\geq 4ac\cos\alpha \cos\gamma 
  \end{equation*}
\begin{equation*}
       4c^4=((a^2+c^2-b^2)+(b^2+c^2-a^2))^2\geq 4(a^2+c^2-b^2)(b^2+c^2-a^2)=\\ 16abc^2\cos\alpha \cos\beta \Rightarrow c^2\geq 4ab\cos\alpha \cos\beta 
 \end{equation*}
Multiplying the last three inequalities and extracting the root (assuming, without loss of generality, that alpha, beta and gamma are in the first quadrant) we will have the required inequality.
