Let $\Delta$ stand for the area of $\triangle ABC$. Consider the LHS. Note that
$$\begin{align*} a \sin B + b \sin C + c \sin A &= \frac{ac \sin B}{c} + \frac{ab \sin C}{a} + \frac{bc \sin A}{b} \\ &= 2\Delta \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \\ &= \frac{2\Delta}{a+b+c} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) (a+b+c) \\ &= r \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) (a+b+c) \\ &\geq 9r \quad (\text{with equality iff } a=b=c). \end{align*}$$

Consider the RHS. Since $a = 2R \sin A$, etc, we have that
$$\begin{align*} a \sin B + b \sin C + c \sin A \\ &= 2R(\sin A \sin B + \sin B \sin C + \sin C \sin A) \\ &= 2R(\sin A \sin B + \sin C(\sin A + \sin B)) \\ &\le 2R \left( \sin^2 \left( \frac{A+B}{2} \right) + 2 \sin C \sin \left( \frac{A+B}{2} \right) \right) \quad \text{(with equality iff } A=B) \\ &= 2R \left( \sin^2 \left( \frac{\pi-C}{2} \right) + 2 \sin C \sin \left( \frac{\pi-C}{2} \right) \right) \\ &= 2R \left( \cos^2 \left( \frac{C}{2} \right) + 4 \sin \left( \frac{C}{2} \right) \cos^2 \left( \frac{C}{2} \right) \right) \\ &= 2R \cos^2 \left( \frac{C}{2} \right) \left( 1 + 4 \sin \left( \frac{C}{2} \right) \right) \\ &= 2R(1-x^2)(1+4x) \quad (x = \sin(C/2)) \\ &= 2R \left( \frac{9}{4} - (2x-1)^2 \left( x + \frac{5}{4} \right) \right) \\ &\le \frac{9R}{2} \quad \text{(with equality iff } x=1/2\text{).} \end{align*}$$

Thus the inequality holds, and there is equality throughout iff
$$A = B, \quad \text{and} \quad \sin\left(\frac{C}{2}\right) = \frac{1}{2} \quad \text{i.e., iff} \quad A = B = C = \frac{\pi}{3}.$$

One can obtain the RHS a little differently using the facts that
$$ab + bc + ca \le \frac{1}{3}(a+b+c)^2 \quad \text{and} \quad \sin A + \sin B + \sin C \le \frac{3\sqrt{3}}{2}.$$