If A > 0, B > 0 and A + B = (π)/(6) , then the minimum positive...

$\tan \left(A + B\right) = \dfrac{tan A + tan B}{1 - tan Atan B}$

$⇒$ $\dfrac{1}{\sqrt{3}} = \dfrac{y}{1 - tan Atan B}$ where $y = tan A + tan B$

$⇒$ $tan A . tan B = 1 - \sqrt{3}y$

and, also $AM \ge GM$

$⇒$ $\dfrac{tan A + tan B}{2} \ge \sqrt{tan Atan B}$

$⇒$ $y \ge 2\sqrt{1 - \sqrt{3}y}$

$⇒$ ${y}^{2} \ge 4 - 4\sqrt{3}y \ge 0$

$⇒$ $y \le - 2\sqrt{3} - 4$ or $y \ge - 2\sqrt{3} + 4$

$y \le - 2\sqrt{3} - 4$, is not possible as $tan A + tan B > 0$

Therefore, $y \ge - 2\sqrt{3} + 4$

Hence, minimum positive value is $4 - 2\sqrt{3}$.