0 \leq \frac{|\frac{x^2y^2z^2}{x^4+y^4+z^4} |}{\sqrt{x^2+y^2+z^2}} \leq \frac{1}{27} \frac{(x^2+y^2+z^2)^2\sqrt{x^2+y^2+z^2}}{x^4+y^4+z^4} \leq \frac{1}{9} \frac{(x^4+y^4+z^4)\sqrt{x^2+y^2+z^2}}{x^4+y^4+z^4}=\frac{1}{9}\sqrt{x^2+y^2+z^2