If  $A\ge 0,B\ge 0,A+B=\frac{\pi }{3}$  and $y=\tan A.\tan B$ then

$\tan A+\tan B=\sqrt{3}\left(1-\tan A\tan B\right)$

Let $\tan A\tan B=y$

$\sqrt{3}\left(1-y\right)=\tan A+\tan B\ge 2\sqrt{y}$

s. o. b. s

$3\left(1-y^2\right)\ge 4y$

$3+3y^2-6y-4y\ge 0$

$3y^2-10y+3\ge 0$

$3y^2-9y-y+3\ge 0$

$3y\left(y-3\right)+\left(y-3\right)\ge 0$

$\left(y-3\right)\left(y-\frac{1}{3}\right)\ge 0$

$y\in \left(-\infty ,\frac{1}{3}\right)\cup \left[3,\infty \right)\to \left(1\right)$

But $0<\tan A<\sqrt{3}$

$0<\tan B<\sqrt{3}$

$0<y<3\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$y\in \left(-\infty ,\frac{1}{3}\right]$

$\therefore \text{ max value of tanA and tanB is }\frac{1}{3}$  

$\therefore \text{ minimum value of tanA and tanB is }0$