10. When $x$, $y$, $z$ are positive numbers, the maximum value of $\frac{4 x z+y z}{x^{2}+y^{2}+z^{2}}$ is $\qquad$ .
10. $\frac{\sqrt{17}}{2}$.

Notice,
(1) $x^{2}+\frac{16}{17} z^{2} \geqslant 2 \sqrt{\frac{16}{17}} x z$, equality holds if and only if $x=\frac{4}{\sqrt{17}} z$;
(2) $y^{2}+\frac{1}{17} z^{2} \geqslant 2 \sqrt{\frac{1}{17}} y z$, equality holds if and only if $y=\frac{1}{\sqrt{17}} z$.
$$
\begin{array}{l}
\text { Then } x^{2}+y^{2}+z^{2} \\
=\left(x^{2}+\frac{16}{17} z^{2}\right)+\left(y^{2}+\frac{1}{16} z^{2}\right) \\
\geqslant 2 \sqrt{\frac{16}{17}} x z+2 \sqrt{\frac{1}{17}} y z \\
=\frac{2}{\sqrt{17}}(4 x z+y z) \\
\Rightarrow \frac{4 x z+y z}{x^{2}+y^{2}+z^{2}} \leqslant \frac{\sqrt{17}}{2} .
\end{array}
$$

Equality holds if and only if $x: y: z=4: 1: \sqrt{17}$. 
Thus, the maximum value of $\frac{4 x z+y z}{x^{2}+y^{2}+z^{2}}$ is $\frac{\sqrt{17}}{2}$.