Dear Gerald Brown and Alex,

I have found one another proof using some well known interesting facts:
Suppose in acute triangle ABC:
AD = diameter of circumcircle of ABC
M = midpoint of BC

1. Prove that ABDC is convex by the fact: a triangle is acute if and only if circumcenter is inside the triangle.
From that we can show: angle(BDC) > 90 (1)
(This result requires that triangle ABC must be acute)

2. Prove that: in a triangle the median from obtuse angle is less than half of its side.
Combine with (1), we can show: MD < BC/2 (2)

3. Prove that: in any triangle any median is less than average of two adjacent sides.
Therefore: AM < (AB + AC)/2 (3)

From (2) and (3) we have:
2*R = AD < AM + MD < (AB + AC + BC)/2 = s what we must prove.

Best regards,
Bui Quang Tuan