Find the maximum and minimum of $\cos x\sin y\cos z$. 
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.
 A: (Edit 2017-01-14. This answer is incorrect. When I gave the answer, I misread the first requirement as $x,y,z\ge\pi/12$ --- it should be $x\ge y\ge z\ge\pi/12$. But I will leave it unmodified because the solution for the problem with the relaxed constraint is still interesting.)
You may rewrite the problem as finding the extrema of $f(x,z)=\cos(x)\cos(z)\cos(x+z)$ with $x, z \ge \pi/12$ and $x+z \le 5\pi/12$. Note that
$$\cos(x)\cos(z)\cos(x+z)=\frac12 [\cos(x+z)+\cos(x-z)] \cos(x+z)$$
and so
\begin{align*}
& [\cos(5\pi/12)+\cos(3\pi/12)] \cos(5\pi/12)\\
\le& [\cos(5\pi/12)+\cos(x-z)] \cos(5\pi/12)\\
\le&[\cos(x+z)+\cos(x-z)] \cos(x+z)\\
\le& [\cos(\pi/6)+\cos(x-z)] \cos(\pi/6)\\
\le& [\cos(\pi/6)+\cos(0)] \cos(\pi/6).
\end{align*}
Hence the maximum of $f$ occurs at $x=z=\pi/12$ and the minima occur at $(x,z)=(4\pi/12,\,\pi/12)$ or $(\pi/12,\,4\pi/12)$.
A: 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 $\displaystyle x=y = \frac{5\pi}{24}$
