Prove $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}\leq \frac {1}{8}$, $\alpha, \gamma\, \beta$ being angles of a triangle

Question

Prove $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}\leq \frac {1}{8}$

I defined $f(x,y,z)=\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}$, and wanted to find max/min points under the constraint $\alpha+\beta+\gamma=\pi$.

What I reached, when using the Lagrange multipliers method is as follows:

$\alpha+\beta+\gamma=\pi$, and $\sin \frac{\alpha}{2}\sin \frac{\beta}{2}\cos \frac{\gamma}{2}=\sin \frac{\alpha}{2}\cos \frac{\beta}{2}\sin \frac{\gamma}{2}=\cos \frac{\alpha}{2}\sin \frac{\beta}{2}\sin \frac{\gamma}{2}$

So obviously all points of the sort $(0,0,\pi), (\pi,0,0), (0,\pi,0)$ are fine, but I couldn't find the criticial points and extracting them from the Lagrange function.

Thanks in advance for any assistance!

Answer 1

Using Algebra only,

$$2\sin\frac\alpha2\sin\frac\beta2=\cos\frac{\alpha-\beta}2-\cos\frac{\alpha+\beta}2$$

Now $\displaystyle\cos\frac{\alpha+\beta}2=\cdots=\sin\frac\gamma2$

Let $\displaystyle y=2\sin\frac\alpha2\sin\frac\beta2\sin\frac\gamma2$

$$\implies y=\left(\cos\frac{\alpha-\beta}2- \sin\frac\gamma2\right)\sin\frac\gamma2\iff2\sin^2\frac\gamma2-\cos\frac{\alpha-\beta}2\sin\frac\gamma2+y=0$$ which is a Quadratic Equation in $\sin\dfrac\gamma2$

As $\gamma$ is real, so will be $\sin\dfrac\gamma2$

So, the discriminant $\displaystyle\cos^2\frac{\alpha-\beta}2-4\cdot2\cdot y$ must be $\ge0$

Answer 2

Min of 0 is obvious. All of the sin values are non-negative (since restricted to $(0,\pi/2)$), so the product is non-negative. Clearly 0 can be achieved.

Max of 1 follows immediately from the convexity of $\log \sin \theta$ in the range $(0,\pi/2)$, and applying Jensens

Answer 3

Easy proof :

$a+b+c=\pi/2$

Note that $a,b,c\in(0,\pi/2)$

$(\sin a\sin b\sin c)^{1/3} \leq\frac{\sin a+\sin b+\sin c}{3}\le \frac 1 2 $

First is AM-GM and second is Jensen