trigonometry - Find the maximum and minimum of $cos xsin ycos z$.

Given $x\geq y\geq z\geq\pi/12$, $x+y+z=\pi/2$, find the maximum and minimum of $\cos x\sin y\cos z$.

I tried using turn $\sin y$ to $\cos(x+z)$, and Jensen Inequality, but filed. Please help. Thank you.

*p.s. I'm seeking for a solution without calculus.

Answer

Let $$P=\cos x\sin y\cos z=\frac{\cos z}{2}\bigg[2\cos x \sin y\bigg] = \frac{\cos z}{2}\bigg[\sin(x+y)-\sin(x-y)\bigg]\leq \frac{\cos z}{2}\cdot \sin (x+y)$$

So $$P\leq \frac{\cos z \cdot \cos z}{2}=\frac{1}{4}(1+\cos 2z)\leq \frac{1}{4}\bigg[1+\cos 2\cdot \frac{\pi}{12}\bigg] = \frac{2+\sqrt{3}}{8}$$

Above equality hold when $\sin (x-y) = 0\Rightarrow x=y$ and given $\displaystyle x+y = \frac{\pi}{2}-z$ and $\displaystyle x\geq y \geq z\geq \frac{\pi}{12}$

And for $\max(P),$ We must have $\displaystyle z = \frac{\pi}{12}$ and $\displaystyle x=y = \frac{5\pi}{24}$