(6) Given $O$ is the circumcenter of $\triangle A B C$, $A B=2 a, A C=\frac{2}{a}, \angle B A C$
$120^{\circ}$. If $\overrightarrow{A O}=\alpha \overrightarrow{A B}+\beta \overrightarrow{A C}$, then the minimum value of $\alpha+\beta$ is $\qquad$.
$6(\alpha+\beta)_{\text {min }}=2$ Hint: Let the radius of the circumcircle of $\triangle A B C$ be $R$. Then
$$
\begin{aligned}
\overrightarrow{A B} \cdot \overrightarrow{A C} & =|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cos 120^{\circ} \\
& =2 a \cdot \frac{2}{a}\left(-\frac{1}{2}\right)=-2 .
\end{aligned}
$$

Also,
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{A B}=\alpha \overrightarrow{A B^{2}}+\beta \overrightarrow{A C} \cdot \overrightarrow{A B}, \\
\overrightarrow{A O} \cdot \overrightarrow{A C}=\alpha \overrightarrow{A B} \cdot \overrightarrow{A C}+\beta \overrightarrow{A C^{2}}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
R \cdot 2 a \cdot \frac{2}{a}=\alpha \cdot 4 a^{2}-2 \beta, \\
R \cdot \frac{2}{a} \cdot \frac{\frac{1}{a}}{R}=-2 \alpha+\beta \frac{4}{a^{2}}
\end{array}\right. \\
\Rightarrow \alpha=\frac{2 a^{2}+1}{3 a^{2}}, \beta=\frac{a^{2}+2}{3}
\end{array}\right\}
$$

When and only when $a=1$, the equality holds. Therefore, $(\alpha+\beta)_{\min }=2$. At this time, $\triangle A B C$ is an isosceles triangle.