Find maximum and minimum of : $\Omega=\sin x\sin y\sin z$ Question : 
Let :  $x,y,z>0$ and $x+y+z=\frac{π}{2}$
Compute the maximum and minimum of the 
following expansion : 
$\Omega=\sin x\sin y\sin z$ 
My attempt  : 
About maximum : 
$\Omega=\sin x\sin y\cos (x+y)=\frac{1}{2}\sin x(\sin (2x+y)-\sin y)$ 
$≤\sin y(1-\sin y)$ as $\sin (2x+y)≤1$
Now : 
Let define the function : $f(x)=x(1-x)$ then 
$f(x)≤f(\frac{1}{2})=\frac{1}{4}$ 
So we get maximum $=\frac{1}{8}$ 
Now is my work correct ? Please if any one have
another method tell me 
Thanks!

For  $\Omega= (\sin x \sin y \sin z)$ by GM-AM
$$(\sin x \sin y \sin z)^{1/3} \le \frac{\sin x+\sin y+ \sin z}{3}\le \sin \frac{(x+y+z)}{3}=\frac{1}{2}.$$ The second inequality is due to Jensens Inequality for $x \in (0,\pi/2).$ as $f''(x)<0.$ So finally
$$0<\sin x \sin y \sin z \le \frac{1}{8},~ x,y,z >0, ~ x+y+z=\pi/2$$
So max of $\Omega$ is $1/8$ which is attained when $x=y=z=\pi/6$ but its min does not exist, however $\Omega >0$.
