Irska

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

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

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

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