9. (16 points) Given a tetrahedron, if there exists a face (lying in the plane $\alpha$) such that when the other three faces are rotated outward around their edges lying on the plane $\alpha$ until they lie on the plane $\alpha$, the four faces together form a triangle on the plane $\alpha$, then the tetrahedron is called a "flat-unfolding tetrahedron".

If there is a flat-unfolding tetrahedron, one of whose faces has side lengths $a$, $b$, and $c$, determine the relationship between $a$, $b$, and $c$, and find the volume of the tetrahedron (expressed in terms of $a$, $b$, and $c$).
9. As shown in Figure 3, if the tetrahedron PDEF is a flat tetrahedron, and the triangle after being flattened along the plane of $\triangle DEF$ is $\triangle ABC$, then $DE$, $EF$, and $FD$ are the midlines of $\triangle ABC$. The four congruent triangles formed by the three midlines of $\triangle ABC$ constitute the four faces of the tetrahedron, thus, the three dihedral angles at a common vertex of the tetrahedron are exactly equal to the three interior angles of $\triangle ABC$. Therefore, $\triangle ABC$ is an acute triangle, and its side lengths are $2a$, $2b$, and $2c$.
Assume $PD = EF = a$,
$$
PE = DF = b, PF = DE = c \text{. }
$$

Then, since $\triangle ABC$ is an acute triangle,
$$
b^{2} + c^{2} > a^{2}, \quad c^{2} + a^{2} > b^{2}, \quad a^{2} + b^{2} > c^{2} \text{. }
$$

Next, we calculate the volume $V$ of the tetrahedron $PDEF$.
Let the circumradius of $\triangle DEF$ be $R$,
$$
\angle FDE = \alpha, \quad \angle DEF = \beta, \quad \angle EFD = \gamma.
$$

Then $\angle PDE = \beta$, $\angle PDF = \gamma$.
Let the dihedral angle $P-DE-F$ be $\varphi$. Then
$$
\begin{array}{l}
\cos \varphi = \frac{\cos \gamma - \cos \alpha \cdot \cos \beta}{\sin \alpha \cdot \sin \beta}. \\
\text{Thus, } \sin \varphi = \sqrt{1 - \left(\frac{\cos \gamma - \cos \alpha \cdot \cos \beta}{\sin \alpha \cdot \sin \beta}\right)^{2}} \\
= \sqrt{1 - \left(1 - \frac{2 \cos \alpha \cdot \cos \beta}{\sin \alpha \cdot \sin \beta}\right)^{2}} \\
= \frac{2 \sqrt{\cos \alpha \cdot \cos \beta \cdot \cos \gamma}}{\sin \alpha \cdot \sin \beta}.
\end{array}
$$

Notice that
$$
S_{\triangle DEF} = S_{\triangle PDE} = \frac{abc}{4R},
$$

and $a = 2R \sin \alpha, b = 2R \sin \beta$, then
$$
\begin{aligned}
V & = \frac{2 S_{\triangle DEF} \cdot S_{\triangle PDE} \sin \varphi}{3 DE} \\
& = \frac{1}{3} abc \sqrt{\cos \alpha \cdot \cos \beta \cdot \cos \gamma} \\
& = \frac{1}{12} \sqrt{2\left(b^{2} + c^{2} - a^{2}\right)\left(c^{2} + a^{2} - b^{2}\right)\left(a^{2} + b^{2} - c^{2}\right)}.
\end{aligned}
$$