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!
 A: Since $\ln\sin{x}$ is a concave function on $\left(0,\frac{\pi}{2}\right)$ and $e^x$ increases, by Jensen we obtain:
$$\prod_{cyc}\sin{x}=e^{\sum\limits_{cyc}\ln\sin{x}}\leq e^{3\ln\sin\frac{x+y+z}{3}}=\frac{1}{8}.$$
The equality occurs for $x=y=z=\frac{\pi}{6},$ which says that we got a maximal value.
The minimum does not exist.
But since $\Omega>0$ and $\Omega\rightarrow0^+$ for $x=y=z\rightarrow0^+,$ we see that $$\inf{\Omega}=0.$$
A: 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 $\Omega >0$.
A: $$2\Omega=2\sin x \sin y\sin z=[\cos(x-y)-\cos(x+y)]\cos(x+y)$$
$$\cos^2(x+y)-\cos(x-y)\cos(x+y)+2\Omega=0$$
which is a quadratic equation in $\cos(x+y)$
The discriminant must be $\ge0$
i.e., $$\cos^2(x-y)-8\Omega\ge0$$
$$\implies8\Omega\le\cos^2(x-y)\le1$$
Can you take it from here?
