Prove that the product of sines of angles of a triangle is min if it is equilateral

Question

I was solving problems about extreme values and conditional extrema , I encounter this problem :

Prove that the product of the sines of the three angles in a triangle is minimum if the triangle is equilateral .

I started by writing the objective function and differentiating it to find the critical points but I do not know how to complete the solution.

$$F(x,y,z)=sin(x)sin(y)sin(z)$$ then $$F_x= cos(x)sin(y)sin(z)=0$$ hence $$x=(2n+1)\pi/2 \ \ ,\ \ y=n\pi \ \ ,\ \ z=n\pi$$ where $$n=0,\pm1,\pm2,...$$ Similarly $$F_y=0$$ gives $$x=n\pi\ \ ,\ \ y=(2n+1)\pi/2 \ \ ,\ \ z=n\pi$$ Finally $$F_z=0$$ gives $$x=n\pi \ \ ,\ \ y=n\pi \ \ ,\ \ z=(2n+1)\pi/2 $$

Answer 1

We wish to minimise $\ln\sin x +\cdots$ subject to $x+y+z=\pi$ with nonnegative angles. The function $\ln\sin p$ has first derivative $\cot p$ and second derivative $-\csc^2 p<0$, and the desired result follows from Jensen's theorem for concave functions.

Answer 2

An alternate approach uses the Law of Sines. Minimizing $\sin a \sin b\sin c$ is the same as minimizing $ABC$, since $\frac{A}{\sin a} = \frac{B}{\sin b} = \frac{C}{\sin c}$. Fixing the perimeter at $n$ (since the ratio of the sides is clearly independent of the perimeter for given angles), this is the same as minimizing $$xyz,\ x+y+z=n,\ x \ge 0, y\ge 0, z\ge 0.$$ It's not hard to see that the maximum occurs for $x=y=z$.

Answer 3

First of all I think the exercise would be to maximise the product, because otherwise the statement is not even true (just take $x$ almost $\pi$ and $y=z=\frac{\pi-x}{2}.$)

In that case there's a more direct approach.

Observe that $\sin$ is concave on $[0,\pi].$ So by applying Jensen we get \begin{equation}\frac{\sin(x)+\sin(y)+\sin(z)}{3}\leq \sin\left(\frac{x+y+z}{3}\right)=\frac{\sqrt{3}}{2}. \end{equation} Using this and AM-GM we conclude \begin{equation}\sin(x)\sin(y)\sin(z)\leq \left( \frac{\sin(x)+\sin(y)+\sin(z)}{3}\right)^3\leq \frac{9\sqrt{3}}{8} \end{equation}with equality if and only if $x=y=z$ (due to AM-GM and Jensen).