Given,

A + B = $${\pi \over 6}$$

$$ \therefore $$   tan(A + B) = tan$$\left( {{\pi \over 6}} \right)$$ = $${1 \over {\sqrt 3 }}$$

We know,

tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$

$$ \Rightarrow $$  $${1 \over {\sqrt 3 }}$$ = $${y \over {1 - \tan A\tan B}}$$

where y = tan A + tan B

$$ \Rightarrow $$    tanA tanB = 1 $$-$$ $$\sqrt 3 $$ y

Also AM   $$ \ge $$  GM

$$ \Rightarrow $$    $${{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B} $$

$$ \Rightarrow $$   y $$ \ge $$ 2$$\sqrt {1 - \sqrt 3 y} $$

$$ \Rightarrow $$   y² $$ \ge $$ 4 $$-$$ 4$${\sqrt 3 y}$$

$$ \Rightarrow $$    y² + 4$${\sqrt 3 y}$$ $$-$$ 4 $$ \ge $$ 0

$$ \Rightarrow $$   y  $$ \le $$ $$-$$ 2$$\sqrt 3 $$ $$-$$ 4   

or   y $$ \ge $$ $$-$$ 2$$\sqrt 3 $$ + 4

(y $$ \le $$ $$-$$ 2$$\sqrt 3 $$ $$-$$ 4 is not possible as tan B > 0)