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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)

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  • $\begingroup$ There is equality when $A=B=C = \frac{\pi}{3}$. $\endgroup$
    – wythagoras
    Apr 25, 2017 at 12:24
  • $\begingroup$ yeah!,maybe this inequality can solve it? $\endgroup$
    – math110
    Apr 25, 2017 at 12:27
  • $\begingroup$ What is the source of this problem? $\endgroup$
    – wythagoras
    Apr 25, 2017 at 12:31
  • $\begingroup$ olypmiad problem last problem $\endgroup$
    – math110
    Apr 25, 2017 at 12:32
  • $\begingroup$ Which year and country? The last comment isn't really helpful. $\endgroup$
    – wythagoras
    Apr 25, 2017 at 12:34

1 Answer 1

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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.

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  • $\begingroup$ Beautiful solution! +1. $\endgroup$
    – Michael Rozenberg
    Apr 25, 2017 at 16:52

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