... 2ab 4ab sin2 (C /2) (a − b )2 = = , c2 c2 c2 we have (a − b )2 (a − c )2 (b − c )2 + + c2 b2 a2 exp ≤ b2 a2 R2 c2 = 2 2 4r 4ab sin (C /2) 4ac sin (B /2) 4bc sin (A /2) This is the square of the required ... , 2 2 2 x so sin(A /2) sin(B /2) sin(C /2) = r/(4R) Now e ≤ 1/(1 − x) for all x ∈ [0, 1) Since ≤ (a − b )2 /c2 < and r = (s − a) tan 1− a + b2 − 2ab cos C − a − b2 + 2ab 4ab sin2 (C /2) (a − b )2 = ... circumradius and inradius of that triangle, respectively Show that (a − b )2 (b − c )2 (c − a )2 R ≥ exp + + 2r 2c2 2a 2b2 Solution by A K Shafie, IASBS, Zanjan, Iran Write A, B, C for the angles...