Question 98: As shown in the figure, in the tetrahedron $\mathrm{D}-\mathrm{ABC}$, it is known that $\mathrm{AB}=2, \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BD}}=-3$, let $\mathrm{AD}=\mathrm{a}, \mathrm{BC}=b$, $\mathrm{CD}=\mathrm{c}$, try to find the minimum value of $\frac{\mathrm{c}^{2}}{\mathrm{ab}+1}$.
Question 98, Answer: Let $\vec{\alpha}=\overrightarrow{\mathrm{AB}}, \vec{\beta}=\overrightarrow{\mathrm{BC}}, \vec{\gamma}=\overrightarrow{\mathrm{AD}}$, then $|\vec{\alpha}|=2, \overrightarrow{\mathrm{AC}}=\vec{\alpha}+\vec{\beta}, \overrightarrow{\mathrm{BD}}=\vec{\gamma}-\vec{\alpha}$, and $c=|C D|=|\vec{\gamma}-\vec{\alpha}-\vec{\beta}|$. According to the conditions:
$(\vec{\alpha}+\vec{\beta}) \cdot(\vec{\gamma}-\vec{\alpha})=-3$
$$
\Rightarrow \vec{\alpha} \cdot \vec{\gamma}+\vec{\beta} \cdot \vec{\gamma}-\vec{\alpha} \cdot \vec{\beta}=-3+|\vec{\alpha}|^{2}=1
$$

Thus, we know:

Therefore, the minimum value of $\frac{c^{2}}{a b+1}$ is 2.