# Proof of a trigonometric inequality

Does anyone know the proof of the following inequality

$$\sin(A)\sin(B)\sin(C)\le\left(\frac{3\sqrt{3}}{2\pi}\right)^3ABC$$

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 $\log \left(\dfrac{x}{\sin(x)}\right)$.
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,$$\sin(x) \sin(y) \sin(z) \leq \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3 xyz$$
• @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 $$\dfrac{f(x)+f(y)+f(z)}3 \leq f((x+y+z)/3)$$ – user17762 Jan 9 '13 at 1:02
• the inequality can be obtained if $t=1/3$ and $y=\frac{y+z}{2}$ right? – user10444 Jan 9 '13 at 1:23
• @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 $t_j's \geq 0$. – user17762 Jan 9 '13 at 1:25
• isn't the inequality you wrote in the comment wrong? On the right shouldn't it be $tf(x)+(1-t)f(y)$? – user10444 Jan 9 '13 at 1:26