2016/119)show that if A, B,C are angles of triangle $\tan\, A \tan\, B \tan\, C>=3\sqrt3$

because A B and C are angles of a triangle we have
$A+B= (180^\circ-C)$
taking tan of both sides
$\tan (A + B) = -\tan \, C$
or $\frac{\tan\, A + \tan \, B}{1- \tan\, A \tan\, B} = -\tan \, C$
or $\tan \, A + \tan \, B = - \tan C + \tan\, A \tan\, B \tan \, C$
or  $\tan \, A + \tan \, B + \tan C =  \tan\, A \tan\, B \tan \, C$
so $\tan\, A \tan\, B \tan \, C = \tan \, A + \tan \, B + \tan C$
using AM GM inequalty we have
$\frac{\tan \, A + \tan \, B + \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\frac{\tan \, A  \tan \, B  \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\tan \, A  \tan \, B  \tan C >= 3 \sqrt[3]{\tan \, A  \tan \, B  \tan\, C}$
cube both sides to get $(\tan \, A  \tan \, B  \tan C)^3 >= 27 (\tan \, A  \tan \, B + \tan C)$
or $(\tan \, A  \tan \, B  \tan C)^2 >= 27 $
or $(\tan \, A  \tan \, B  \tan C) >= 3\sqrt{3}$