6. Given positive real numbers $x, y$ satisfy
$\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y$, then the minimum value of $x+y$ is $\qquad$.
$$
\begin{array}{l}
2 x+\sqrt{4 x^{2}+1} \geqslant \\
\frac{y}{\sqrt{y^{2}+4}-2}=\frac{\sqrt{y^{2}+4}+2}{y}=\frac{2}{y}+\sqrt{\frac{4}{y^{2}}+1},
\end{array}
$$

Notice that the function $u=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing on $(0,+\infty)$, so $x \geqslant \frac{1}{y}$.
Therefore, $x+y \geqslant \frac{1}{y}+y \geqslant 2$, with equality holding when $x=y=1$.