Euler is Everywhere! #2

1+r⁄R = cosA + cosB + cosC. We know that maximum value of cosA + cosB + cosC = 3/2. So cosA + cosB + cosC ≤ 3/2. Therefore, r/R ≤ 1/2, hence, R≥2r. This is the proof of this inequality! And yes, the equality holds true for an equilateral triangle, as cos(60)+cos(60)+cos(60)=3/2. So, R≥2(2015) i.e., 4030.

For this problem, we use Euler's Inequality, which states that in any triangle with circumradius $R$ and inradius $r,$

$R \geq 2r.$

This can be proven by using the area formulae for a triangle. Thus, $R \geq 2(2015) = \boxed{4030}.$