10. 若对任意 $x>0, \frac{x}{x^{2}+3 x+1} \leqslant a$ 恒成立, 则 $a$ 的取值范围为
because x>0, \therefore \frac{x}{x^{2}+3 x+1}>0$, 易知 $a>0$.

$\therefore \frac{x^{2}+3 x+1}{x} \geqslant \frac{1}{a}$,

$\therefore \frac{1}{a} \leqslant x+\frac{1}{x}+3$.

$\because x>0, x+\frac{1}{x}+3 \geqslant 2 \sqrt{x \cdot \frac{1}{x}}+3=5(x=1$ 时取等号 $)$,

$\therefore \frac{1}{a} \leqslant 5 . \therefore a \geqslant \frac{1}{5}$.