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.
Best Answer
(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 $(\pi/12,\,4\pi/12)$.