$\tan{A} \cdot \tan{B} \cdot \tan{C}=9$, find $\tan^2{A}+\tan^2{B}+ \tan^2{C}$ In  $\triangle{ABC}$,
$$\tan{A}\cdot \tan{B}\cdot \tan{C}=9$$
$$\tan^2{A}+\tan^2{B}+ \tan^2{C}=\lambda$$
then,

$\lambda$ lies in the interval?


We have the inequality $\displaystyle (ab+bc+ca)^2 \ge 3abc(a+b+c)$ 
Also in $\displaystyle \triangle ABC, \ \tan A+\tan B + \tan C = \tan A \tan B \tan C$, 
Hence, $\displaystyle \left(\sum \tan A \tan B \right)^2 \ge 3 \times 9^2 \Rightarrow \sum \tan A \tan B \notin \left(- 9 \sqrt 3 , 9 \sqrt 3 \right)$
So $\displaystyle \sum \tan^2 A = \left(\sum \tan A \right)^2 - 2 \sum \tan A \tan B \notin (81-18 \sqrt 3, 81+18 \sqrt 3)$
