9. (8th "Hope Cup" Invitational Competition Question) If $a+b+c=1$, what is the maximum value of $\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1}$?
9. Let $\vec{a}=(\sqrt{3 a+1}, \sqrt{3 b+1}, \sqrt{3 c+1}), \vec{b}=(1,1,1)$, given that $a+b+c=1$, by the theorem we have
$$
\begin{array}{l} 
6=(3 a+1)+(3 b+1)+(3 c+1) \\
=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(1 \cdot \sqrt{3 a+1}+1 \cdot \sqrt{3 b+1}+1 \cdot \sqrt{3 c+1})^{2}}{1^{2}+1^{2}+1^{2}} \\
=\frac{(\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1})^{2}}{3} . \\
(\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1})^{2} \leqslant 18, \\
\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1} \leqslant 3 \sqrt{2} .
\end{array}
$$
The maximum value of $\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1}$ is $3 \sqrt{2}$. This maximum value is achieved when and only when $a=b=c=\frac{1}{3}$.