在△ABC中，求证
[sin(B/2)+sin(C/2)]^2+[sin(C/2)+sin(A/2)]^2+[sin(A/2)+sin(B/2)]^2≥3

在△ABC中，求证
[sin(B/2)+sin(C/2)]^2+[sin(C/2)+sin(A/2)]^2+[sin(A/2)+sin(B/2)]^2≤3
证明 根据三角形半角公式:
[sin(A/2)]^2=(s-b)(s-c)/(bc)
[sin(B/2)]^2=(s-c)(s-a)/(ca)
[sin(C/2)]^2=(s-a)(s-b)/(ab)
由均值平均不等式得:
2sin(B/2)+sin(C/2)=2(s-a)√[(s-b)(s-c)]/√(a^2*bc)
≤[(s-a)/a]*[(s-b)/c+(s-c)/b]
所以欲证上述命题只需证:
[sin(B/2)+sin(C/2)]^2+[sin(C/2)+sin(A/2)]^2+[sin(A/2)+sin(B/2)]^2
≤2Σ(s-b)(s-c)/(bc)+Σ[(s-a)/a]*[(s-b)/c+(s-c)/b]
=[1/(abc)]*Σ{2a(s-b)(s-c)+(s-a)*[b(s-b)+c(s-c)]}
=[1/(abc)]*Σ[2as^2+2abc-2s(ab+ac)+(b+c)s^2+a(b^2+c^2)-(ab+ac+b^2+c^2)s]
=[1/(abc)]*{(a+b+c)(bc+ca+ab)+(b+c)(c+a)(a+b)+4abc}
=[1/(abc)]*(3abc)=3.
证毕。