Each of the angles ABC (or, B) , BCA(or, C) and CAB(or, A) of the triangle ABC is acute (given). Accordingly, the circumcenter (E, say) must lie inside the triangle.
Let X, Y and Z respectively denote the mid points of BC, CA and AB.
Accordingly, it follows that E must be located in the interior of one of the quadrilaterals BCYZ, CAZX, or ABXY.
Without loss of generality, we can assume that E is located in the interior of BCYZ.
Let R be the length of the circumradius of the triangle ABC.
2R = BE + EC
Since triangles AZY and ABC are similar, with BZ = AB/2, and
AY = AC/2, we must have ZY = BC/2.
Now, it can easily be shown that:
BE + EC < BZ + ZY + YC = (1/2)*(AB + BC + AC)
Or, 2R < R*(sin A + sin B + sin C) (in terms of the law of sines)
Or, sin A + sin B + sin C > 2
Edited on October 14, 2007, 6:25 am