Find maximum of $\cos(x)\cos(y)\cos(z)$ when $x + y + z = \frac{\pi}{2}$ and $x, y, z > 0$

Question

Find maximum value of P = $\cos(x)\cos(y)\cos(z)$, given that $x + y + z = \frac{\pi}{2}$ and $x, y, z > 0$.

Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.

Answer 1

Let $f(x)=\ln\cos{x}.$

Thus, $$f''(x)=-\left(\tan{x}\right)'=-\frac{1}{\cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain: $$\cos{x}\cos{y}\cos{z}=e^{\sum\limits_{cyc}\ln\cos{x}}\leq e^{3\ln\cos\frac{x+y+z}{3}}=\frac{3\sqrt3}{8}.$$ The equality occurs for $x=y=z=\frac{\pi}{6},$ which says that we got a maximal value.

Answer 2

A geometric approach:

Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=\frac{\pi}{2}-A$, $y=\frac{\pi}{2}-B$, $z=\frac{\pi}{2}-C$. Since $A+B+C=\pi$, $x+y+z=\frac{\pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $\sin A\sin B\sin C$.

Now, we get clever. The area of triangle $ABC$ is equal to $2R^2\sin A\sin B\sin C$, where $R$ is the circumradius of the triangle. Why? Because $2R\sin A=a$ and $2R\sin B=b$, we can reduce to the more familiar area formula $\frac12ab\sin C$. Maximizing $\sin A\sin B\sin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $\Gamma$.

And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $\Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $\Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^\circ$ angles $A,B,C$, for $x=y=z=\frac{\pi}{6}$ and $\cos x\cos y\cos z = \sin A\sin B\sin C=\left(\frac{\sqrt{3}}{2}\right)^3=\frac{3\sqrt{3}}{8}$.

Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $\Gamma^3$ (the positions of the three points) to $\mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.