9. Given that $a, b, c$ are the sides of $\triangle \mathrm{ABC}$, and satisfy $\sum_{c y c} \frac{1}{a} \sqrt{\frac{1}{b}+\frac{1}{c}} \geq \frac{3}{2} \sqrt{\prod_{c y c}\left(\frac{1}{a}+\frac{1}{b}\right)}$, determine the shape of the triangle.
9 Analysis: The triangle 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 $\frac{1}{a}=\frac{y^{2}+z^{2}-x^{2}}{2}, \frac{1}{b}=\frac{z^{2}+x^{2}-y^{2}}{2}$,
$\frac{1}{c}=\frac{x^{2}+y^{2}-z^{2}}{2}$, then the original inequality becomes $\sum_{c y c} x\left(y^{2}+z^{2}-x^{2}\right) \geq 3 x y z$, and by Schur's inequality we have $\sum_{c y c} x^{3}+3 x y z \geq \sum_{c y c} x\left(y^{2}+z^{2}\right)$, combining the equality condition of Schur's inequality and $x, y, z>0$ we know $x=y=z$, at this point we can get $a=b=c$, i.e., this triangle is an equilateral triangle.