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! 

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
