$\tan{A} \cdot \tan{B} \cdot \tan{C}=9$, 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?

 A: As lab bhattacharjee noted in comments, for triangle angles we have 
$$\tan A\cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C.\tag{1}$$
Denote $\;\;x=\tan A,\quad y=\tan B, \quad z=\tan C$.
We have
$$x+y+z=9;\\
xyz=9;\tag{2}$$
and we try to estimate expression $$x^2+y^2+z^2.$$
Let use $z$ as parameter ($0<z<9$). Then
$$x+y=9-z;\\
xy=9/z;$$
and we try to estimate expression
$$(x^2+y^2)+z^2=(x+y)^2-2xy+z^2=(9-z)^2-\dfrac{18}{z}+z^2.\tag{3}$$
Now, $x$ and $y$ (according to Vieta's theorem) are solutions of quadratic equation
$$a^2-(9-z)a+\dfrac{9}{z}=0.\tag{4}$$
This equation has  (real) positive solutions if $(9-z)^2z\ge36$.
So, bounds for $z$ are (approximately):
$$z\in I, \mbox{ where } I=[0.498040,6.678221].\tag{5}$$
Considering expression $(3)$ on segment $(5)$, we reach minimal value when $z\approx 4.25098$: $$\min_{z\in I}\{x^2+y^2+z^2\}\approx 36.3897;\tag{min}$$
and we reach maximal value when $z\approx 6.678221$:
$$\max_{z\in I}\{x^2+y^2+z^2\}\approx 47.29396.\tag{max}$$
Minimizing parameters: $x \approx 0.4980402$, $y=z\approx 4.250979.$
Maximizing parameters: $x =y\approx 1.160889$, $z\approx 6.678221.$

This result can be obtained using Lagrange multipliers as well.
A: 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 $\displaystyle \sum \tan^2 A = \left(\sum \tan A \right)^2 - 2 \sum \tan A \tan B \notin (81-18 \sqrt 3, 81+18 \sqrt 3)$
