sin a + sin b + sin c ???

sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2] ≦ 2sin[(A+B)/2]
similarly
sinC + sinD ≦ 2sin[(C+D)/2]
sinA + sinB + sinC + sinD ≦ 2sin[(A+B)/2] + 2sin[(C+D)/2]
2sin[(A+B)/2] + 2sin[(C+D)/2] = 4sin[(A+B+C+D)/4]cos[A+B-C-D)/4] ≦ 4sin[(A+B+C+D)/4]
∴ sinA + sinB + sinC + sinD ≦ 4sin[(A+B+C+D)/4]
let D = (A+B+C)/3
sinA + sinB + sinC + sinD = sinA + sinB + sinC + sin[(A+B+C)/3] ≦
4sin[(A+B+C+((A+B+C)/3)/4] = 4sin[(A+B+C)/3]
∴ sinA + sinB + sinC ≦ 3sin[(A+B+C)/3]
= 3sin[(pi)/3] = 3sqrt(3)/2

2009-12-04 21:14:47 補充：
http://www.enotes.com/math/q-and-a/by-considering-arithmetic-mean-geometric-mean-119089

我晤識用AM>=GM。

2009-12-05 11:18:20 補充：
sinA + sinB + sinC = 4(cosA/2)(cosB/2)(cosC/2)
4(cosA/2)(cosB/2)(cosC/2) ≦ 4{(1/3)[(cosA/2) + (cosB/2) + (cosC/2)]}^3

2009-12-05 11:18:25 補充：
when A/2 = B/2 =　C/2 ,
4{(1/3)[(cosA/2) + (cosB/2) + (cosC/2)]}^3 attains max value, and GM = AM
∴ 4(cosA/2)(cosB/2)(cosC/2) ≦ 4{[cos(pi)/6 + (cos(pi)/6 + (cos(pi)/6]/3}^3
= 4(sqrt(3)/2)^3 = 3sqrt3/2

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