## Wednesday, May 18, 2016

### The relation of $2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$.

Prove that if in a triangle $ABC$ we have the following equality that holds

$2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$

then the triangle will be an equilateral triangle.

In any triangle $ABC$, we have the following equality that holds:

$1-2\cos A \cos B \cos C=\cos^2 A+\cos^2 B+\cos^2 C$

This turns the given equality to become

$\cos^2 A+\cos^2 B+\cos^2 C=1-2\cos A \cos B \cos C=\cos A \cos B + \cos B \cos C + \cos C \cos A$

For any angle where $0\lt A,\,B,\,C\lt 180^\circ$, the relation $\cos A \gt \cos B \gt \cos C$ must hold, Now, by applying the rearrangement inequality on $\cos^2 A+\cos^2 B+\cos^2 C$ leads to:

$\cos^2 A+\cos^2 B+\cos^2 C\ge \cos A \cos B + \cos B \cos C + \cos C \cos A$, equality occurs iff $\cos A=\cos B=\cos C$, i.e. $A=B=C$, when that triangle is equilateral, and we're hence done with the proof.