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.

Then,

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**