If $A,B$ and $C$ are the interior angles of a triangle, then what is the maximum value for $\sin(A)\cdot\sin(B)\cdot\sin(C)$?

Question

If $A,B$ and $C$ are the interior angles of a triangle, then what is the maximum value for $\sin(A)\cdot\sin(B)\cdot\sin(C)$?

I couldn't think of a specific direction to start the solution with.

Answer 1

Based on comments, you seem to be interested in multivariate calculus application, especially Lagrange multipliers. Here is a sketch of such an approach.

Writting down what we want to maximize and under which condition, we arrive at Langrangian $$ L(A,B,C,\lambda)=\sin A\sin B\sin C-\lambda (A+B+C-\pi). $$ Now you need to evaluate stationary points of this function. You can restrict yourself for a moment to closed region $(A,B,C)\in[0,\pi]^3$, since in closed region the function attains both its minima and maxima (in $(0, \pi)$ you would find it has only one stationary point, it might be tricky to argue it is actually a maximum).

If you calculate these stationary points correctly, you should obtain four points $(A,B,C)$, namely $(0,0,\pi)$, $(0,\pi,0)$, $(\pi,0,0)$ and $(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3})$. Simple evaluation shows that $(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3})$ gives maximum $\frac{3\sqrt{3}}{8}$, and the other points give same minimum $0$.

Answer 2

HINT

By AM-GM

$$\sin A \cdot \sin B \cdot \sin C \le \left(\frac{\sin A+\sin B+\sin C}{3}\right)^3$$

with equality for $\sin A=\sin B=\sin C$.

Answer 3

Another approach:

We know that $\ln\circ\sin$ is concave in $(0,\pi)$, therefore

$$\ln (\sin A\sin B\sin C)=\ln\sin A+\ln\sin B+\ln\sin C$$ $$\le 3\ln\sin\left( \frac{A+B+C}{3}\right)=3\ln \sin\frac\pi3=3\ln\frac{\sqrt 3}{2}$$ So $$\sin A\sin B\sin C\le \frac{3\sqrt 3}{8}$$