The product of the sines of a triangle

Concavity. On $(0,\pi)$ the function $g=\ln\sin$ has $g'=\cot$ and $g''=-\dfrac1{\sin^2}<0$, so $g$ is strictly concave.

Jensen. For the three angles with equal weights,
$$\frac{g(A)+g(B)+g(C)}3\le g\!\left(\frac{A+B+C}3\right)=g\!\left(\frac\pi3\right)=\ln\frac{\sqrt3}2.$$
Hence $\ln(\sin A\sin B\sin C)\le3\ln\frac{\sqrt3}2$, i.e.
$$\sin A\sin B\sin C\le\left(\frac{\sqrt3}2\right)^3=\frac{3\sqrt3}8.$$

Equality. By strict concavity, iff $A=B=C=\frac\pi3$.

Remark. The same method gives $\sin\frac A2\sin\frac B2\sin\frac C2\le\frac18$.

Lesson. A product under a fixed-sum constraint is a sum after taking logarithms: check the concavity of $\ln f$ and use Jensen.