6. Given $x, y, z>0$, then $f=\frac{\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+4 z^{2}}+\sqrt{z^{2}+16 x^{2}}}{9 x+3 y+5 z}$ the minimum value is
Since $\left(1^{2}+2^{2}\right)\left(x^{2}+y^{2}\right) \geqslant(x+2 y)^{2} \Rightarrow \sqrt{5} \cdot \sqrt{x^{2}+y^{2}} \geqslant x+2 y$,
Similarly, we can get $\sqrt{5} \cdot \sqrt{y^{2}+4 z^{2}} \geqslant y+4 z, \sqrt{5} \cdot \sqrt{z^{2}+16 x^{2}} \geqslant z+8 x$. Adding the three inequalities, we get $\sqrt{5}\left(\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+4 z^{2}}+\sqrt{z^{2}+16 x^{2}}\right) \geqslant 9 x+3 y+5 z \Rightarrow f \geqslant \frac{1}{\sqrt{5}}$, equality holds when $x=\frac{y}{2}=\frac{z}{2}$. Therefore, the minimum value of $f$ is $\frac{\sqrt{5}}{5}$.