in the triangle ABC

By Titu's lemma , we have:

$\begin{aligned} \frac 1{\sin \frac A2} + \frac 1{\sin \frac B2} + \frac 1{\sin \frac C2} & \ge \frac {(1+1+1)^2}{\sin \frac A2 + \sin \frac B2 + \sin \frac C2} & \small \blue{\text{Equality occurs when }A=B=C=60^\circ} \\ & = \frac 9{\frac 32} = \boxed 6 \end{aligned}$

Let the lengths of the sides opposite to the angles $A, B, C$ of $\triangle {ABC}$ be $a, b, c$ respectively. Then the given expression equals

$\sqrt {\frac{bc}{(s-b) (s-c) }}+\sqrt {\frac{ca}{(s-c) (s-a) }}+\sqrt {\frac{ab}{(s-a) (s-b) }}\geq 3\sqrt[3] {\frac{abc}{(s-a) (s-b) (s-c) }}$ , where

$s=\frac{a+b+c}{2}$ is the semi-perimeter of the triangle.

Equality holds when $a=b=c$ , so that the required minimum is $3\sqrt[3] {\frac{a^3}{\frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}}}=\boxed 6$ .