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 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! 
 A: 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 $\ge0$ 
A: 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
A: 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
