In which acute-angled triangle is the value of $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma$ the smallest?
$$
0=\operatorname{tg} 180^{\circ}=\operatorname{tg}(\alpha+\beta+\gamma)=\frac{\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}+\operatorname{tg} \gamma}{1-\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta} \operatorname{tg} \gamma}=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma}{1-(\operatorname{tg} \alpha \operatorname{tg} \beta+\operatorname{tg} \alpha \operatorname{tg} \gamma+\operatorname{tg} \beta \cdot \operatorname{tg} \gamma)}
$$

Since we are dealing with acute angles, the tangents are all finite, so the value of the fraction can only be 0 if the numerator is 0. Therefore,

$$
\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma=\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma
$$

Given that all three tangents are positive due to the acute angles, the well-known relationship between the arithmetic and geometric means can be written for them:

$$
\sqrt[3]{\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma} \leq \frac{\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma}{3}
$$

Multiplying through and cubing,

$$
27 \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma \leq(\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma)^{3}
$$

Using (1) and (2), we get

$$
27 \operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma \leq(\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma)^{3}
$$

Simplifying and taking the square root,

$$
\sqrt{27} \leq \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \operatorname{tg} \gamma
$$

From (3), it is clear that $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma$ cannot take a value less than $\sqrt{27}$. $\sqrt{27}$ can only be taken if the equality sign holds in the inequality between the arithmetic and geometric means. This can only happen if $\operatorname{tg} \alpha=\operatorname{tg} \beta=\operatorname{tg} \gamma$. This means that

$$
\alpha=\beta=\gamma=60^{\circ}
$$

Since $\operatorname{tg} 60^{\circ}=\sqrt{3}$, in the case of an equilateral triangle, $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma=(\sqrt{3})^{3}=\sqrt{27}$, so the tangent product indeed takes the minimum value given by (3).