6. Given positive real numbers $x, y$ satisfy
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y \text {. }
$$

Then the minimum value of $x+y$ is $\qquad$
6. 2 .

It is easy to know that $\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y$, which means $2 x+\sqrt{4 x^{2}+1} \geqslant \frac{2}{y}+\sqrt{4\left(\frac{1}{y}\right)^{2}+1}$. Let $f(x)=2 x+\sqrt{4 x^{2}+1}$. Then $f(x) \geqslant f\left(\frac{1}{y}\right)$.
Moreover, the function $f(x)$ is monotonically increasing when $x>0$, so, $x \geqslant \frac{1}{y} \Rightarrow x y \geqslant 1$.
Therefore, $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, with equality holding if and only if $x=y=1$.
Thus, $(x+y)_{\min }=2$.