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
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![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](/uploads/image/z/13396696-16-6.jpg?t=%E4%B8%80%E9%81%93%E5%88%9D%E7%AD%89%E4%B8%8D%E7%AD%89%E5%BC%8F%2C%E4%BB%8E%E5%B7%A6%E5%90%91%E5%8F%B3%E6%8E%A8%E5%AF%BC%E8%BF%87%E7%A8%8B%E2%88%91yz%282x%2By%2Bz%29%2F%28x%2By%29%28x%2Bz%29%3D%E2%88%91x%E6%AD%A4%E5%A4%84%E2%88%91%E6%98%AF%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C)
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
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
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
∑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.