陈胜利老师schur01软件中的征解问题qq.2.1
qq.2.1 设$x,y,z,m\in R$, $(x-y)(y-z)(z-x)\geq 0$,则
$\sum{\left(\frac{x-y}{y-z}-m\right)^2}\geq 2m^2+2m+5.$
证明：因为$\sum{\left(\frac{x-y}{y-z}-m\right)^2}-(2m^2+2m+5)=\left(\sum{\frac{x-y}{y-z}}-m+1\right)^2\geq 0,$
所以原不等式成立.
qq.2.1 设$x,y,z,m\in R$, $(x-y)(y-z)(z-x)\geq 0$,则
$\sum{\left(\frac{x-y}{y-z}-m\right)^2}\geq 2m^2+2m+5.$
证明：因为$\sum{\left(\frac{x-y}{y-z}-m\right)^2}-(2m^2+2m+5)=\left(\sum{\frac{x-y}{y-z}}-m+1\right)^2\geq 0,$
所以原不等式成立.
