$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $

Question

In an acute triangle with angles $ A, B $ and $ C $, show that

$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $

I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.

How can I proceed?

Answer 1

$$y=2\cos A\cos B\cos C=[\cos(A-B)+\cos(A+B)]\cos C=[\cos(A-B)-\cos C]\cos C$$

$$\implies\cos^2C-\cos(A-B)\cos C+y=0$$ which is Quadratic Equation in $\cos C$

As $C$ is real $\implies\cos C$ is real,

the discriminant $\cos^2(A-B)-4y\ge0\iff y\le\dfrac{\cos^2(A-B)}4\le\dfrac14$

Answer 2

Homage a lab

$2\cos A\cos B\cos C=(\cos(A+B)+\cos(A-B))\cos C=(\cos(A-B)-\cos C)\cos C=\frac{1}{4}\cos^2(A-B)-(\cos C-\frac{1}{2}\cos(A-B))^2\le\frac{1}{4}$

Answer 3

NTS $\cos A\cos B\cos C \leq 1/8$ Now, since $ABC$ is a triangle, only one of the angles can possibly be equal or greater than $\pi/2$. So, without loss of generality, let $B, C<\pi/2$. Then $\cos B>0,\cos C>0.$

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Case 1: $A\ge\pi/2$ $\Rightarrow \cos A<0 \Rightarrow \cos A\cos B\cos C<0<1/8$
Case 2: $A<\pi/2$ $\Rightarrow \cos A>0$
Then, by the AM-GM inequality: $$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)$$ Now, consider the function $f(x)=\cos x$ where $x \in(0,\pi/2)$. $f$ is a concave function because $f''(x)=-\cos x>0$ for $x\in(0,\pi/2)$. Then, by the Jenson's inequality: $$f(\frac1 3A +\frac 1 3B+\frac 13 C)\ge\frac 13f(A) +\frac13f(B)+\frac13f(C) \\$$ Then:$$\cos(\frac \pi3)\ge \frac13(\cos A+\cos B+\cos C)$$ Which is: $$\frac12\ge\frac13(\cos A+\cos B+\cos C)$$ Combining that result with the result from the AM-GM inequality, we get: $$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)\le\frac12$$ Cube each portion:$$\cos A\cos B\cos C\leq {\frac 1 {27}}(\cos A+\cos B+\cos C)^3\le\frac18$$

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$$\therefore \cos A\cos B\cos C\leq \frac18$$

Answer 4

We need to prove that $$\prod_{cyc}\frac{a^2+b^2-c^2}{2ab}\leq\frac{1}{8}$$ or $$a^2b^2c^2\geq\prod_{cyc}(a^2+b^2-c^2)$$ or $$\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)\geq0,$$ which is Schur.

Done!