已知x+y+z=π,证明sin(x+y)+sin(y+z)+sin(z+x)≥sin2x+sin2y+sin2z

题目

已知x+y+z=π,证明sin(x+y)+sin(y+z)+sin(z+x)≥sin2x+sin2y+sin2z

答案

不等式右侧等于：
1/2[2sin2x+2sin2y+2sin2z]
=1/2[(sin2x+sin2y)+(sin2y+sin2z)+(sin2x+sin2z)]
=1/2[2sin(x+y)cos(x-y)+2sin(y+z)(y-z)+2sin(z+x)(y-z)]
=sin(x+y)cos(x-y)+sin(y+z)(y-z)+sin(z+x)(y-z)
显然,比不等式左侧要小