# 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)

• There is equality when $A=B=C = \frac{\pi}{3}$. Apr 25, 2017 at 12:24
• yeah!,maybe this inequality can solve it? Apr 25, 2017 at 12:27
• What is the source of this problem? Apr 25, 2017 at 12:31
• olypmiad problem last problem Apr 25, 2017 at 12:32
• Which year and country? The last comment isn't really helpful. Apr 25, 2017 at 12:34

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.