神马品牌网元素爆破师远古套选择:急求答案 数学的题目
来源：百度文库 编辑：神马品牌网 时间：2024/05/01 01:20:22
(注：“^2”指2次方 )
a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)-a^3-b^3-c^3
=a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)
=cosA*2abc+cosB*2abc+cosC*2abc（余弦定理）
=2abc(cosA+cosB+cosC)
只需证明cosA+cosB+cosC>1即可。
cosA+cosB+cosC = 2cos((A+B)/2)cos((A-B)/2)+1-2sin^2(C/2)
= 2sin(C/2)cos((A-B)/2)+1-2sin^2(C/2)
= 1 + 2sin(C/2)*(cos((A-B)/2)-sin(C/2))
= 1 + 2sin(C/2)*(cos((A-B)/2)-cos((A+B)/2))
= 1 + 4sin(C/2)sin(A/2)sin(B/2)
> 1
a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)-a^3-b^3-c^3
=a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2+c^2)
=cosAabc+cosBabc+cosCabc (A,B,C为三边所对应的角)
=abc(cosA+cosB+cos(180-(A+B)))
=abc(cosA+cosB-cos(A+B))
=abc(cos(A+B/2-B/2)-cos(A+B/2+B/2)+cosB)
=abc(2sin(A+B/2)sin(B/2)+cosB)
=abc(2sin(A+B/2)sin(B/2)+1-2sinB/2sinB/2)
=abc(2sinB/2(sin(A+B/2)-sinB/2)+1)
=abc(2sinB/2*2cos(A+B)/2*sinA/2+1)
=abc+4abc*sinB/2*cos(A+B)/2*sinA/2
=abc+