一道初等不等式,从左向右推导过程∑yz(2x+y+z)/(x+y)(x+z)=∑x此处∑是循环求和

Answer 1

∑yz(2x+y+z)/(x+y)(x+z)=yz(2x+y+z)/(x+y)(x+z)+xy(2z+x+y)/(z+x)(z+y)+zx(2y+z+x)/(y+z)(y+x)
=[xyz+yz(x+y+z)]/(x+y)(x+z)+[xyz+xy(x+y+z)/(z+x)(z+y)+[xyz+zx(x+y+z)]/(y+z)(y+x)
=xyz/(x+y)(x+z)+yz(x+y+z)/(x+y)(x+z)+xyz/(z+x)(z+y)+xy(z+y+z)/(z+x)(z+y)+xyz/(y+z)(y+x)+zx(x+y+z)/(y+z)(y+x)=xyz[(y+z)/(x+y)(x+z)(y+z)+(x+y)/(x+y)(x+z)(y+z)+(x+z)/(x+y)(x+z)(y+z)]+(x+y+z)[yz(y+z)/(x+y)(x+z)(y+z)+xy(x+y)/(x+y)(x+z)(y+z)+zx(z+x)/(x+y)(x+z)(y+z)]
=xyz[2(x+y+z)/(x+y)(x+z)(y+z)]+(x+y+z)[y^2z+yz^2+x^2y+xy^2+z^2x+zx^2]/(x+y)(x+z)(y+z)
=(x+y+z)(2xyz+y^2z+yz^2+x^2y+xy^2+z^2x+zx^2)/(x+y)(x+z)(y+z)=(x+y+z)((x+y)(x+z)(y+z)/(x+y)(x+z)(y+z)=x+y+z=∑x.证毕.