✔ 最佳答案
cos(B/2)cos(C/2)sin(A/2)
=(1/2)sin(A/2){ cos[(B+C)/2]+cos[(B-C)/2] }
=(1/2)sin(A/2){ sin(A/2)+cos[(B-C)/2] }
=(1/4){ 2sin²(A/2)+ 2cos[(B+C)/2]cos[(B-C)/2] }
=(1/4)( 1-cosA+cosB+ cosC )
so,
cos(C/2)cos(B/2)sin(A/2)+cos(A/2)cos(C/2)sin(B/2)+cos(A/2)cos(B/2)sin(C/2)
=(1/4)( 3+cosA+cosB+cosC)
=(1/4){ 3+2cos[(A+B)/2]cos[(A-B)/2]+ 1-2sin²(C/2) }
=(1/4){ 4+2sin(C/2) cos[(A-B)/2]- 2sin(C/2)cos[(A+B)/2] }
=(1/4){ 4+ 4sin(C/2)sin(A/2)sin(B/2) }
= 1+sin(A/2)sin(B/2)sin(C/2)
<= 1+ { [sin(A/2)+sin(B/2)+sin(C/2)]/ 3 }³ (By GP <= AP)
<= 1+ { sin[ (A+B+C)/6 ] }³ ( By Jensen's ineq. and sin"(x/2) <= 0 )
= 1+ 1/8 = 9/8
