设A、B、C为三角形的三个内角，求证sin（A／2）×sin（B／2）×sin（C／2）≤1/8 !!!

经典的三角变换题目
不妨设α=A/2,β=B/2,γ=C/2
α+β+γ=90,并且sinα,cosα,sinβ,cosβ,sinγ,cosγ均为正
2sinαsinβsinγ
=(cos(α-β)-cos(α+β))sinγ
≤(1-cos(α+β)sinγ
=(1-sinγ)sinγ
=-(sinγ-1/2)^2+1/4
≤1/4
即sinαsinβsinγ≤1/8
等号成立当且仅当α=β,sinγ=1/2,即α=β=γ=30,A=B=C=60
原命题得证

sin（A／2），sin（B／2），sin（C／2）均大于0。

由均值不等式xyz≤((x+y+z)/3)^3，当且仅当xyz相等时等号成立。将x=A/2，y=B/2，z=C/2带入该式子即可。（A=B=C=60度）

sin(A/2)*sin(B/2)*sin(C/2)=sin(A/2)*sin(B/2)*sin(90-(A+B)/2)
=sin(A/2)*sin(B/2)*cos(A/2+B/2)
=sin(A/2)*sin(B/2)*(cos(A/2)*cos(B/2)-sin(A/2)*sin(B/2))
=sin(A/2)*sin(B/2)*cos(A/2)*cos(B/2)-(sin(A/2)*sin(B/2))^2
=1/2*[sin(A)]*1/2*[sin(B)]-1/4*[cos(A/2+B/2)-cos(A/2-B/2)]^2
=1/4*sin(A)*sin(B)-1/4*[cos(A/2+B/2)-cos(A/2-B/2)]^2
=1/4*[sin(A)*sin(B)-[cos(A/2+B/2)-cos(A/2-B/2)]^2]
<=1/8