The inequality $\frac{3\sqrt{3}}{2}\frac{27(A+B)(B+C)(C+A)}{8(A+B+C)^3}\ge \sin{A}+\sin{B}+\sin{C}$ 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)

Not a full solution just yet, but we can go like this:
$$\sin A + \sin B + \sin C = 4 \sin\frac{A+B}2\sin\frac{B+C}2\sin\frac{C+A}2.$$
So if we let $x= \frac{B+C}2,y=\frac{C+A}2,z=\frac{A+B}2$, we need to show that
$$xyz\ge \lambda \sin x\sin y\sin z$$
for $x,y,z\in(0,\pi/2)$, and $ x+y+z=\pi$ and some constant $\lambda$ that make it an equality when $x=y=z$.
Now using Wolfram Alpha, we see that (this is where the proof is not complete, but I'm too lazy)
$$f(x) = \ln \frac{\sin x}{x}=\ln (\sin x) -\ln x$$
has negative second derivative in $(0,\pi/2)$. The inequality follows.
