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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Given $a b(a+b)=4 \Rightarrow a^{2}+a b=\frac{4}{b}$, then $(2 a+b)^{2}=4 a^{2}+4 a b+b^{2}=\frac{16}{b}+b^{2}$ $=\frac{8}{b}+\frac{8}{b}+b^{2} \geqslant 3 \sqrt[3]{64}=12$, equality holds when $b=2, a=\sqrt{3}-1$.
Therefore, the minimum value of $2 a+b$ is $2 \sqrt{3}$.