Find maximum and minimum of : $\Omega=\sin x\sin y\sin z$

Question

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!

Answer 1

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 $\Omega >0$.

Answer 2

$$2\Omega=2\sin x \sin y\sin z=[\cos(x-y)-\cos(x+y)]\cos(x+y)$$

$$\cos^2(x+y)-\cos(x-y)\cos(x+y)+2\Omega=0$$

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?

Answer 3

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 $$\inf{\Omega}=0.$$