10. (15 points) Given that $a$, $b$, and $c$ are the side lengths of $\triangle ABC$, and satisfy
$$
\sum \frac{1}{a} \sqrt{\frac{1}{b}+\frac{1}{c}} \geqslant \frac{3}{2} \sqrt{\prod\left(\frac{1}{a}+\frac{1}{b}\right)},
$$

where “$\sum$” and “$\prod$” denote cyclic sum and cyclic product, respectively, determine the shape of $\triangle ABC$.
10. Let $d=a+b, e=b+c, f=c+a$. Then, with $d, e, f$ as side lengths, a $\triangle D E F$ can be formed. Thus,
$\sin \frac{D}{2}=\sqrt{\frac{1-\cos D}{2}}=\sqrt{\frac{1-\frac{e^{2}+f^{2}-d^{2}}{2 e f}}{2}}$
$=\sqrt{\frac{(d-e+f)(d+e-f)}{4 e f}}=\sqrt{\frac{a b}{(b+c)(c+a)}}$.
Similarly, $\sin \frac{E}{2}=\sqrt{\frac{b c}{(c+a)(a+b)}}$,
$\sin \frac{F}{2}=\sqrt{\frac{c a}{(a+b)(b+c)}}$.
In $\triangle D E F$, it always holds that
$\sin \frac{D}{2}+\sin \frac{E}{2}+\sin \frac{F}{2} \leqslant \frac{3}{2}$,

with equality if and only if $\angle D=\angle E=\angle F=\frac{\pi}{3}$. 
Thus, $\sum \sqrt{\frac{a b}{(b+c)(c+a)}} \leqslant \frac{3}{2}$.
Multiplying both sides of the above inequality by $\sum \frac{1}{a} \sqrt{a+b}$, we get
$$
\sum \frac{1}{c} \sqrt{\frac{1}{a}+\frac{1}{b}} \leqslant \frac{3}{2} \sqrt{\prod\left(\frac{1}{a}+\frac{1}{b}\right)} .
$$

From inequality (2) and the given inequality, we have
$$
\sum \frac{1}{c} \sqrt{\frac{1}{a}+\frac{1}{b}}=\frac{3}{2} \sqrt{\prod\left(\frac{1}{a}+\frac{1}{b}\right)},
$$

which is the equality case of inequality (1).
$$
\begin{array}{l}
\text { Therefore, } \angle D=\angle E=\angle F=\frac{\pi}{3} \\
\Rightarrow d=e=f \Rightarrow a+b=b+c=c+a \\
\Rightarrow a=b=c \Rightarrow \triangle A B C \text { is an equilateral triangle. }
\end{array}
$$