Key Concepts and Formulas

• Trigonometric Identity for Tangent of Sum: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. This identity relates the sum of tangents to the tangent of the sum of angles.
• Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For non-negative real numbers $x$ and $y$, $\frac{x+y}{2} \ge \sqrt{xy}$, with equality if and only if $x=y$. This inequality is crucial for finding minimum values of sums when the product is constant or can be related to the sum.
• Quadratic Inequalities: Understanding how to solve inequalities of the form $ax^2+bx+c \ge 0$ or $\le 0$ is necessary to determine the range of the variable.

Step-by-Step Solution

Step 1: Utilize the Tangent Sum Identity

• Action: Apply the tangent addition formula to the given sum $A+B = \frac{\pi}{6}$.
• Why: This will connect the given constant sum of angles to the expression $\tan A + \tan B$ that we need to minimize, and the product $\tan A \tan B$. $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
• Action: Substitute the value of $A+B$ and $\tan(A+B)$.
• Why: We know $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. $\frac{1}{\sqrt{3}} = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
• Action: Let $S = \tan A + \tan B$ and $P = \tan A \tan B$. Rearrange the equation to express $P$ in terms of $S$.
• Why: This will allow us to substitute $P$ in the AM-GM inequality later. $\frac{1}{\sqrt{3}} (1 - P) = S$ $1 - P = S\sqrt{3}$ $P = 1 - S\sqrt{3} \quad (*)$

Step 2: Apply the AM-GM Inequality

• Action: Consider the terms $\tan A$ and $\tan B$. Apply the AM-GM inequality to these terms.
• Why: We are given $A > 0$, $B > 0$, and $A+B = \frac{\pi}{6}$. This implies $0 < A < \frac{\pi}{6}$ and $0 < B < \frac{\pi}{6}$. In this interval, $\tan A > 0$ and $\tan B > 0$, so AM-GM is applicable. $\frac{\tan A + \tan B}{2} \ge \sqrt{\tan A \tan B}$
• Action: Substitute $S$ and $P$ into the inequality and square both sides.
• Why: Squaring is permissible because both sides are non-negative ($\frac{S}{2} > 0$ and $\sqrt{P} > 0$). This gives a relationship between $S^2$ and $P$. $\frac{S}{2} \ge \sqrt{P}$ $\frac{S^2}{4} \ge P$ $S^2 \ge 4P \quad (**)$

Step 3: Combine the Equations and Solve the Inequality

• Action: Substitute the expression for $P$ from equation $(*)$ into inequality $(**)$.
• Why: This eliminates $P$ and gives an inequality solely in terms of $S$. $S^2 \ge 4(1 - S\sqrt{3})$
• Action: Rearrange the inequality into a standard quadratic form.
• Why: Solving quadratic inequalities will give us the possible range for $S$. $S^2 \ge 4 - 4\sqrt{3}S$ $S^2 + 4\sqrt{3}S - 4 \ge 0$
• Action: Find the roots of the quadratic equation $S^2 + 4\sqrt{3}S - 4 = 0$.
• Why: The roots of the quadratic determine the intervals where the inequality holds. Using the quadratic formula $S = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $S = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(-4)}}{2(1)}$ $S = \frac{-4\sqrt{3} \pm \sqrt{48 + 16}}{2}$ $S = \frac{-4\sqrt{3} \pm \sqrt{64}}{2}$ $S = \frac{-4\sqrt{3} \pm 8}{2}$ The roots are $S_1 = \frac{-4\sqrt{3} - 8}{2} = -2\sqrt{3} - 4$ and $S_2 = \frac{-4\sqrt{3} + 8}{2} = 4 - 2\sqrt{3}$.
• Action: Determine the valid range for $S$.
• Why: Since the quadratic $S^2 + 4\sqrt{3}S - 4$ has a positive leading coefficient (1), it opens upwards. The inequality $S^2 + 4\sqrt{3}S - 4 \ge 0$ holds for $S \le S_1$ or $S \ge S_2$. We know $S = \tan A + \tan B$. Since $A, B \in (0, \frac{\pi}{6})$, $\tan A > 0$ and $\tan B > 0$, so $S$ must be positive. $S_1 = -2\sqrt{3} - 4$ is negative. $S_2 = 4 - 2\sqrt{3} \approx 4 - 2(1.732) = 0.536$, which is positive. Therefore, for $S$ to be positive and satisfy the inequality, we must have $S \ge 4 - 2\sqrt{3}$.

Step 4: Check for Attainability of the Minimum

• Action: Determine the condition for equality in the AM-GM inequality.
• Why: The minimum value is achieved when the equality condition is met. Equality in AM-GM holds when $\tan A = \tan B$.
• Action: Use the equality condition $\tan A = \tan B$ along with $A+B = \frac{\pi}{6}$.
• Why: Since $A$ and $B$ are in the interval $(0, \frac{\pi}{6})$, where the tangent function is one-to-one, $\tan A = \tan B$ implies $A=B$. Substituting $A=B$ into $A+B = \frac{\pi}{6}$ gives $2A = \frac{\pi}{6}$, so $A = \frac{\pi}{12}$. Thus, $A=B=\frac{\pi}{12}$.
• Action: Calculate $\tan(\frac{\pi}{12})$.
• Why: To verify that the minimum value obtained is indeed $2\tan(\frac{\pi}{12})$. $\tan\left(\frac{\pi}{12}\right) = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$ $= \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3+1-2\sqrt{3}}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$
• Verification: The minimum value of $\tan A + \tan B$ is $2\tan(\frac{\pi}{12}) = 2(2-\sqrt{3}) = 4-2\sqrt{3}$. This confirms that the minimum value derived from the inequality is attainable.

Common Mistakes & Tips

• Sign of Tangent: Ensure that the terms for AM-GM are positive. Here, $A, B \in (0, \pi/6)$, so $\tan A, \tan B > 0$.
• Equality Condition: Always check if the equality condition for AM-GM can be satisfied within the problem's constraints. If $A=B=\pi/12$, then $A+B=\pi/6$, and $A, B > 0$.
• Quadratic Inequality Interpretation: For a quadratic inequality $ax^2+bx+c \ge 0$ with $a>0$, the solution lies outside the roots. For $a<0$, it lies between the roots.

Summary

We used the tangent addition formula to establish a relationship between the sum and product of $\tan A$ and $\tan B$. Then, we applied the AM-GM inequality to find a lower bound for the sum. By substituting the relationship between the sum and product into the AM-GM inequality, we obtained a quadratic inequality in terms of the sum $S = \tan A + \tan B$. Solving this inequality, and considering that $S$ must be positive, we found that $S \ge 4 - 2\sqrt{3}$. The equality condition for AM-GM was shown to be achievable when $A=B=\frac{\pi}{12}$, confirming that $4 - 2\sqrt{3}$ is indeed the minimum value.

The final answer is $\boxed{4 - 2\sqrt 3}$.