ABC is an arbitrary triangle. Show that sin(A/2) sin(B/2) sin(C/2) < 1/4.
Solution
wlog A ≤ B ≤ C. Then A/2 ≤ 30o, so sin A/2 ≤ 1/2. Also B/2 is acute and B/2 = (90o-C/2)-A/2 < 90o-C/2. Hence sin B/2 sin C/2 < sin(90o-C/2) sin C/2 = cos C/2 sin C/2 = 1/2 sin C ≤ 1/2.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03