A cyclic, non-symmetric inequality If $x,y,z\geq 0$, prove 
$$
(x^2+y^2+z^2)^2\geq 3(x^3y+y^3z+z^3x)
$$
With these nonsymmetric inequalities the usual tools like Muirhaed or Schur do not apply, and also AM-GM doesn't seem to be of any help. Also, it's non-factorizable.

Because
$$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum\limits_{cyc}(x^2-z^2-2xy+xz+yz)^2\geq0$$
Also we have
$$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{6}\sum\limits_{cyc}(x^2+y^2-2z^2-3xy+3xz)^2\geq0$$
