Question

已知正数x y z满足x+y+z小于等于xyz求证(1-y2)(1-z2）/yz+(1-x2)(1-z2）/xz+(1-

Answer 1

即证x(1-y2)(1-z2)+y(1-x2)(1-z2)+z(1-x2)(1-y2)=4xyz

左边=x(1-z2-y2-y2z2)+y(1-z2-x2+x2z2)+(1-y2-x2+x2y2)

=(x+y+z)-xz2-xy2+xy2z2-yz2+yx2+yx2z2-zy2-zx2+zx2y2

=xyz-xy(y+x)-xz(x+z)-yz(y+z)+xyz(xy+yz+zx)

=xyz-xy(xyz-z)-xz(xyz-y)-yz(xyz-x)+xyz(xy+yz+zx)

=xyz+xyz+xyz+xyz

=4xyz=右边．

得证