Does anyone know the proof of the following inequality


where $A,B,C$ are the vertex angles of a triangle.


First note that $\log(\sin(x)/x)$ is concave for $0 \leq x,y,z < \pi$. Hence, we have that $$\dfrac{\log(\sin(x)/x) + \log(\sin(y)/y) + \log(\sin(z)/z)}3 \leq \log \left( \dfrac{\sin((x+y+z)/3)}{(x+y+z)/3} \right)$$ Now you should be able to finish it off. Essentially same as Jensen's inequality if we consider the function $\log \left(\dfrac{x}{\sin(x)}\right)$.

Move your cursor over the gray area for the complete answer.

Taking $x+y+z = \pi$, we get that $$\dfrac{\log(\sin(x)/x) + \log(\sin(y)/y) + \log(\sin(z)/z)}3 \leq \log \left( \dfrac{3\sqrt{3}}{2 \pi} \right)$$Hence,$$\log \left(\dfrac{\sin(x) \sin(y) \sin(z)}{xyz}\right) \leq \log \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3$$Hence,$$\sin(x) \sin(y) \sin(z) \leq \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3 xyz$$

  • $\begingroup$ Could you please explain how you got to the inequalities in the answer? $\endgroup$ – user10444 Jan 9 '13 at 0:56
  • $\begingroup$ @user10444 For a concave function, you have $f(tx + (1-t)y) \geq t(f(x) + (1-t) f(y))$ for $0\leq t \leq 1$ and under sufficient smoothness this is equivalent to the second derivative being negative, which is easy for you to check on $\log(\sin(x)/x)$. Essentially the above condition amounts to saying that $y$ coordinate of the center of mass is lesser than the functional value of the $x$ coordinate of the center of mass. From this, you should be able to derive that $$\dfrac{f(x)+f(y)+f(z)}3 \leq f((x+y+z)/3)$$ $\endgroup$ – user17762 Jan 9 '13 at 1:02
  • $\begingroup$ the inequality can be obtained if $t=1/3$ and $y=\frac{y+z}{2}$ right? $\endgroup$ – user10444 Jan 9 '13 at 1:23
  • $\begingroup$ @user10444 Yes. In general for a concave function, we have $$f \left(\sum_{k=1}^n t_k x_k\right) \geq \sum_{k=1}^n t_k f(x_k)$$ where $t_1 + t_2 + \cdots +t_n = 1$ and $t_j's \geq 0$. $\endgroup$ – user17762 Jan 9 '13 at 1:25
  • $\begingroup$ isn't the inequality you wrote in the comment wrong? On the right shouldn't it be $tf(x)+(1-t)f(y)$? $\endgroup$ – user10444 Jan 9 '13 at 1:26

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