9. (16 points) Given $x, y, z > 0$. Find
$$
f(x, y, z)=\frac{\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+4 z^{2}}+\sqrt{z^{2}+16 x^{2}}}{9 x+3 y+5 z}
$$

the minimum value.
9. Introduce the parameter $\alpha$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}} \sqrt{1+\alpha^{2}} \geqslant x+\alpha y \\
\sqrt{y^{2}+4 z^{2}} \sqrt{1+\alpha^{2}} \geqslant y+2 \alpha z \\
\sqrt{z^{2}+16 x^{2}} \sqrt{1+\alpha^{2}} \geqslant z+4 \alpha x .
\end{array}
$$

Then $f(x, y, z)$
$$
\begin{array}{l}
=\frac{\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+4 z^{2}}+\sqrt{z^{2}+16 x^{2}}}{9 x+3 y+5 z} \\
\geqslant \frac{(1+4 \alpha) x+(1+\alpha) y+(1+2 \alpha) z}{(9 x+3 y+5 z) \sqrt{1^{2}+\alpha^{2}}} . \\
\text { Let }(1+4 \alpha):(1+\alpha):(1+2 \alpha)=9: 3: 5 .
\end{array}
$$

Then $\alpha=2$.
And by $y=\alpha x, 2 z=\alpha y, 4 x=\alpha z$, we get
$$
\alpha^{3}=8 \Rightarrow \alpha=2 \text {. }
$$

Thus, when $y=z=2 x$, $f(x, y, z)$ achieves its minimum value $\frac{\sqrt{5}}{5}$.