Suppose z>y>x>0 and S=x2+y2+z2x+y+z, then
S>x2z
S<z2x
Since z>y>x>0, we have z2>y2>x2.
Therefore,
3x2<x2+y2+z2<3z2 (1)and 3x<x+y+z<3z (2)From(2), we get13z<1x+y+z<13x∴3x23z<x2+y2+z2x+y+z<3z23x⇒x2z<x2+y2+z2x+y+z<z2x