In $\triangle ABC$, show that $$\dfrac{3\sqrt{3}}{2}\dfrac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$$
My attempt: Since $A+B+C=\pi$, it suffices to show that
$$\dfrac{3\sqrt{3}}{2}\cdot \dfrac{27(\pi-A)(\pi-B)(\pi-C)}{8\pi^3}\ge\sin{A}+\sin{B}+\sin{C}$$
then it seem to hard to prove it (because Jensen's inequality can't solve it, and the tangent line also can't solve it)