Unsolved Challenge: Σ(cosAcosB)>√(cosAcosBcosC)

[anemone]

Let $ABC$ be a non-obtuse triangle. Prove that $\cos A \cos B+\cos B \cos C+\cos C \cos A>2\sqrt{\cos A \cos B \cos C}$.

 

[jvicens]

Thank you for your interesting post. I would like to provide a proof for the inequality stated in your post.

Let's begin by recalling the cosine rule for a triangle: $c^2 = a^2 + b^2 - 2ab\cos C$, where $a$ and $b$ are the lengths of the sides opposite angles $A$ and $B$ respectively. Since we are given that $ABC$ is a non-obtuse triangle, we know that $C$ is an acute angle, meaning that $\cos C>0$.

Now, let's rewrite the inequality as follows:
$$\cos A \cos B+\cos B \cos C+\cos C \cos A>2\sqrt{\cos A \cos B \cos C}$$
$$\Leftrightarrow \cos A \cos B+\cos B \cos C+\cos C \cos A-2\sqrt{\cos A \cos B \cos C}>0$$
$$\Leftrightarrow (\cos A \cos B+\cos C \cos A)^2-4\cos A \cos B \cos C>0$$
$$\Leftrightarrow (\cos A + \cos C)^2>4\cos A \cos C.$$

Now, using the cosine rule, we can rewrite $(\cos A + \cos C)^2$ as:
$$(\cos A + \cos C)^2 = (\cos^2 A + \cos^2 C + 2\cos A \cos C) = 1 + 2\cos A \cos C.$$

Substituting this into our inequality, we get:
$$1 + 2\cos A \cos C > 4\cos A \cos C.$$
$$\Leftrightarrow 1 > 2\cos A \cos C.$$

Since $\cos C>0$, we can divide both sides by $\cos C$ without changing the direction of the inequality:
$$\Leftrightarrow \frac{1}{\cos C} > 2\cos A.$$

But we know that $\frac{1}{\cos C} = \sec C$, and since $C$ is an acute angle, $\cos C < 1$, meaning that $\sec C > 1$. Therefore, we have:
$$\sec C > 2\cos A.$