Có: $x^{4}+y^{4}\geq xy(x^{2}+y^{2})\Leftrightarrow (x-y)^{2}(x^{2}+xy+y^{2})\geq 0$
$\Rightarrow \sum \frac{1}{x^{4}+y^{4}+z}\leq \sum \frac{1}{xy(x^{2}+y^{2})+z}=\sum \frac{z}{x^{2}+y^{2}+z^{2}}=\frac{x+y+z}{x^{2}+y^{2}+z^{2}}\leq \frac{3}{x+y+z}\leq 1$