7. Given positive real numbers $a, b$ satisfy $a b(a+b)=4$. Then the minimum value of $2 a+b$ is $\qquad$ .

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7. $2 \sqrt{3}$.

From $a b(a+b)=4$, we get $a^{2}+a b=\frac{4}{b}$. Then
$$
\begin{array}{l}
(2 a+b)^{2}=4 a^{2}+4 a b+b^{2} \\
=\frac{16}{b}+b^{2}=\frac{8}{b}+\frac{8}{b}+b^{2} \\
\geqslant 3 \times \sqrt{\frac{8}{b} \times \frac{8}{b} \times b^{2}}=12 .
\end{array}
$$

When $b^{2}=\frac{8}{b}$, i.e., $b=2$, the equality holds. At this time,
$$
a=\sqrt{3}-1, b=2 \text {. }
$$

Therefore, the minimum value of $2 a+b$ is $2 \sqrt{3}$.