9. (16 points) Given that $a, b, c$ are the three sides of $\triangle ABC$, and satisfy
$$
\sum \frac{1}{a} \sqrt{\frac{1}{b}+\frac{1}{c}} \geqslant \frac{3}{2} \sqrt{\Pi\left(\frac{1}{a}+\frac{1}{b}\right)},
$$

where $\sum, \prod$ represent cyclic sum and cyclic product, respectively. Determine the shape of $\triangle ABC$.
9. $\triangle A B C$ is an equilateral triangle.
Let $\sqrt{\frac{1}{b}+\frac{1}{c}}=x, \sqrt{\frac{1}{c}+\frac{1}{a}}=y, \sqrt{\frac{1}{a}+\frac{1}{b}}=z$.
Then $\frac{1}{a}=\frac{y^{2}+z^{2}-x^{2}}{2}, \frac{1}{b}=\frac{z^{2}+x^{2}-y^{2}}{2}$,
$\frac{1}{c}=\frac{x^{2}+y^{2}-z^{2}}{2}$.
Thus, the original inequality becomes
$$
\sum x\left(y^{2}+z^{2}-x^{2}\right) \geqslant 3 x y z \text {. }
$$

By Schur's inequality, we have
$$
\sum x^{3}+3 x y z \geqslant \sum x\left(y^{2}+z^{2}\right) \text {. }
$$

Combining the equality condition of Schur's inequality and $x, y, z > 0$, we know that $x=y=z$, at which point we get $a=b=c$, i.e., $\triangle A B C$ is an equilateral triangle.