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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.

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up vote 4 down vote accepted

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

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