三角形ABC中,求证：sin(A/2)sin(B/2)sin(C/2)小于等于1／8

证明：sin(A/2)*sin(B/2)*sin(C/2)
=sin(A/2)*sin(B/2)*sin[(π-A-B)/2]
=sin(A/2)*sin(B/2)*cos[(A+B)/2]
=-0.5{cos[(A+B)/2]-cos[(A-B)/2]}*cos[(A+B)/2]
=-0.5{cos[(A+B)/2]}^2+0.5cos[(A-B)/2]cos[(A+B)/2]
可以看成是关于cos[(A+B)/2]的二次函数,显然当cos[(A+B)/2]=0.5cos[(A-B)/2]=0.5（此时cos[(A-B)/2]=1)时,sin(A/2)*sin(B/2)*sin(C/2)有最大值
-1/8+1/4=1/8
所以sin(A/2)*sin(B/2)*sin(C/2)