由 qeypour 於 星期五 八月 19, 2005 10:51 am
原式相當於證明x^2-x(2ycosC+2zcosA)+y^2+z^2-2yzcosA>=0
等價於(2ycosC+2zcosA)^2-4(y^2+z^2-2yzcosA)<=0
等價於-4(sinB)^2*z^2+(8ycosBcosC+8ycosA)*z-4(sinC)^2*y^2<=0
等價於(8ycosBcosC+8ycosA)^2-64y^2(sinB)^2*(sinC)^2<=0.......(1)
只要證(1)式成立即可,證明如下
64y^2(cosBcosC+cosA)^2-64y^2(sinB*sinC)^2
=64y^2(cosBcosC+cosA+sinBsinC)(cosBcosC+cosA-sinBsinC)
=64y^2[cos(B-C)+cosA]*[cos(B+C)+cosA]
其中cos(B+C)=-cosA代入上式=0
(1)式為0>=0,得證