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. Solution: Clearly $a>b, a, b, c$ are the three sides of a triangle $\Leftrightarrow a-b<c<a+b \Leftrightarrow$
$$
\begin{array}{l}
x+y-\sqrt{x^{2}+x y+y^{2}}<m \sqrt{x y}<x+y+ \\
\sqrt{x^{2}+x y+y^{2}} \\
\quad \Leftrightarrow \frac{x+y}{\sqrt{x y}}-\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}<m<\frac{x+y}{\sqrt{x y}}+
\end{array}
$$
$\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}$ holds for any positive numbers $x, y$.
$$
\begin{array}{c}
\text { Since } \frac{x+y}{\sqrt{x y}}+\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}} \geqslant 2+\sqrt{3}, \\
\frac{x+y}{\sqrt{x y}}-\frac{\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}}=\frac{x+y-\sqrt{x^{2}+x y+y^{2}}}{\sqrt{x y}} \\
=\frac{\sqrt{x y}}{x+y+\sqrt{x^{2}+x y+y^{2}}} \leqslant \frac{1}{2+\sqrt{3}}=2-\sqrt{3},
\end{array}
$$

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