AO+BO+CO≥6r where r is the radius of the inscribed circle

[solakis1]

From the entrance examinations to Ghana University ,from high school, i got the following problem:

If O is the center of the inscribed circle in an ABC trigon,then prove that: \(\displaystyle AO+BO+CO\geq 6r\) where r is the radius of the inscribed circle.

 

[HallsofIvy]

I first thought that your "ABC trigon" is what I would call a "triangle" but that would make the problem impossible. If ABC is a triangle inscribed in a circle, of radius r, then OA, OB, and OC are equal to r so that OA+ OB+ OC= 3r which is less than 6r.

 

[castor28]

I first thought that your "ABC trigon" is what I would call a "triangle" but that would make the problem impossible. If ABC is a triangle inscribed in a circle, of radius r, then OA, OB, and OC are equal to r so that OA+ OB+ OC= 3r which is less than 6r.

The question is about the circle inscribed in a triangle, not about a triangle inscribed in a circle.