For which values of $\gamma$ is the following statement true: If one of the angles of a triangle is $\gamma$, then

$$
\frac{T}{\sqrt{a^{2} b^{2}-4 T^{2}}}+\frac{T}{\sqrt{b^{2} c^{2}-4 T^{2}}}+\frac{T}{\sqrt{c^{2} a^{2}-4 T^{2}}} \geq \frac{3 \sqrt{3}}{2}
$$

where $a, b, c$ are the sides of the triangle, and $T$ is the area of the triangle.
Using the notation and thought process of problem F. 3008, we need to determine those obtuse angles $\gamma$ for which, given $\alpha, \beta > 0$ and $\alpha + \beta = \pi - \gamma$,

$$
|\operatorname{tg} \alpha| + |\operatorname{tg} \beta| + |\operatorname{tg} \gamma| \geq 3 \sqrt{3}
$$

always holds. On the left side, for a fixed $\gamma$, the minimum value of $|\operatorname{tg} \alpha| + |\operatorname{tg} \beta|$ is $2 \operatorname{tg} \frac{\pi - \gamma}{2}$, since $\alpha$ and $\beta$ are acute angles, and the tangent function is positive, increasing, and convex in the domain of acute angles. Therefore, $\gamma$ is good if and only if it satisfies the inequality

$$
2 \operatorname{tg} \frac{\pi - \gamma}{2} - \operatorname{tg} \gamma \geq 3 \sqrt{3}
$$

Let $x = \operatorname{tg}(\gamma / 2)$. We are looking for those $x$ for which

$$
x > 1 \quad \text{and} \quad \frac{2}{x} + \frac{2 x}{x^{2} - 1} \geq 3 \sqrt{3}
$$

Solving the cubic equation using the known Cardano formula, we get for $x$ that $1,3090 \cdots = x_{0} \geq x > 1$, where $x_{0}$ denotes the largest root of the equation

$$
\frac{2}{x} + \frac{2 x}{x^{2} - 1} = 3 \sqrt{3}
$$

Combining everything, the sought values of the obtuse angle $\gamma$ are

$$
105.248 \ldots^{\circ} = 2 \operatorname{arctg} x_{0} \geq \gamma > 90^{\circ}
$$