Prove Equilateral Triangle from Cosine Equality

[anemone]

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.

 

[Albert1]

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

$S=2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1---(1)$

then the triangle will be an equilateral triangle.

my solution:

Spoiler

if $(1)$ is true,it is easy to see that triangle $ABC$ must be an acute triangle
(if one of those 3 angles A,or B, or C $\geq 90^o$ then (1) will fail)
that is $0<cos A,cos B,cos C<1$
using $AP\geq GP$
$S\geq 4\sqrt[4]{2cos^3Acos^3Bcos^3C}=1$
(equality occurs at $A=B=C=60^o$)
so it is an equilateral triangle

 

[anemone]

Thanks for participating, Albert!

My solution:

Spoiler

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.