<s>[INST] 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.
 [/INST] 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,$$\sin(x) \sin(y) \sin(z) \leq \left( \dfrac{3\sqrt{3}}{2 \pi} \right)^3 xyz$$

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