7. Let $A, B$ be two points on the curve $x y=1(x, y>0)$ in the Cartesian coordinate system $x O y$, and let the vector $\vec{m}=(1,|O A|)$. Then the minimum value of the dot product $\vec{m} \cdot \overrightarrow{O B}$ is $\qquad$ .
Let $A\left(a, \frac{1}{a}\right), B\left(b, \frac{1}{b}\right)(a, b>0)$, then $\vec{m}=\left(1, \sqrt{a^{2}+\frac{1}{a^{2}}}\right), \overrightarrow{O B}=\left(b, \frac{1}{b}\right)$. Thus, $\vec{m} \cdot \overrightarrow{O B}=b+\frac{1}{b} \sqrt{a^{2}+\frac{1}{a^{2}}} \geqslant b+\frac{\sqrt{2}}{b} \geqslant 2 \sqrt[4]{2}$, equality holds when $(a, b)=(1, \sqrt[4]{2})$. Therefore, the minimum value of $\vec{m} \cdot \overrightarrow{O B}$ is $2 \sqrt[4]{2}$.