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 you don't like to my first solution, here is a more simple solution.Assume without loss of generality that A,B,C belongs to the first quadrant.And so, it's easy to see that:
\begin{align*}
     x+y\geq 2\sqrt{xy} \tag{1}
 \end{align*}
\begin{align*}
      x+z\geq 2\sqrt{xz} \tag{2}
 \end{align*}
\begin{align*}
     y+z\geq 2\sqrt{yz} \tag{3}
 \end{align*}
Multiplying (1),(2),(3), we get:
$$ (x+y)(x+z)(y+z) \geq 8xyz$$
$$ \frac{1}{8} \geq \frac{xyz}{(x+y)(x+z)(y+z)}$$
$$ \displaystyle \frac{1}{8} \geq \sqrt{\frac{xy}{(x+z)(y+z)}}\sqrt{\frac{xz}{(x+y)(y+z)}}\sqrt{\frac{yz}{(x+y)(x+z)}} $$
By Ravi transformation and Law of cosines, we get
$$  \frac{1}{8} \geq sen\frac{\alpha}{2} sen\frac{\beta}{2} sen\frac{\gamma}{2}$$
And this is equivalent to:
$$  \frac{1}{8} \geq cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right) cos \left(\frac{\pi}{2}-\frac{\beta}{2}\right) cos \left(\frac{\pi}{2}-\frac{\gamma}{2}\right)$$
Take the substitution $\displaystyle \frac{\pi}{2}-\frac{\alpha}{2}=A$,$\displaystyle \frac{\pi}{2}-\frac{\beta}{2}=B$,$\displaystyle \frac{\pi}{2}-\frac{\gamma}{2}=C$, we get $\displaystyle A+B+C=\pi$ and:
$$  \frac{1}{8} \geq cosAcosBcosC$$
