(In reply to

re: Solution by Charlie)

After the step:

BE + EC < BZ + ZY + YC = (1/2)*(AB + BC + AC).......(*)

But, BE = EC = R, where R is the length of circumradius of the triangle ABC, while E is the circumcenter of the triangle.

Hence, BE+ EC = 2R........(i)

In terms of the law of sines,

(Reference: http://en.wikipedia.org/wiki/Law_of_sines), we have:

BC/sin A = CA/sin B = AB/sin C = 2R, giving:

(1/2)*(AB + BC + AC) = R(sin C + sin A + sin B).......(ii)

Substituting (i) and (ii) in (*), we have:

2R< R(sin A + sin B + sin C)

Or, sin A + sin B + sin C > 2

*Edited on ***October 14, 2007, 3:13 pm**