8. If for any positive numbers $x, y$, it can be made that $a=x+y$, $b=\sqrt{x^{2}+x y+y^{2}}$, $c=m \sqrt{x y}$ are the three sides of a triangle, then the range of the positive number $m$ is $\qquad$
8. $2-\sqrt{3}b$.  
$a, b, c$ are the three sides of a triangle  
$$
\begin{aligned}
\Leftrightarrow & a-b<c<a+b \\
\Leftrightarrow & x+y-\sqrt{x^{2}+x y+y^{2}}<m \sqrt{x y} \\
& <x+y+\sqrt{x^{2}+x y+y^{2}}
\end{aligned}
$$
$$
\Leftrightarrow \frac{x+y}{\sqrt{x y}}-\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}<m<\frac{x+y}{\sqrt{x y}}+\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}
$$

For any positive numbers $x, y$, it always holds.
$$
\begin{array}{l}
\text { Since } \frac{x+y}{\sqrt{x y}}+\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}} \geqslant 2+\sqrt{3}, \\
\frac{x+y}{\sqrt{x y}}-\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}} \\
=\frac{x+y-\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}} \\
=\frac{\sqrt{x y}}{x+y+\sqrt{x^{2}+x y+y^{2}}} \leqslant \frac{1}{2+\sqrt{3}}=2-\sqrt{3},
\end{array}
$$

Therefore, $2-\sqrt{3}<m<2+\sqrt{3}$ is the solution.