3. Given $\triangle A B C, O$ is a point inside $\triangle A B C$, $\angle A O B=\angle B O C=\angle C O A=\frac{2 \pi}{3}$. Then the largest number $m$ such that
$$
A B+B C+C A \geqslant m(A O+B O+C O)
$$
holds is $(\quad$.
(A) 2
(B) $\frac{5}{3}$
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}{2}$
3. (C).

As shown in Figure 2, draw $OD \perp BC$ intersecting $BC$ at point $D$, and $OE$ as the angle bisector of $\angle BOC$. Let $OB=x$, $OC=y$, $OA=z$. Then, by $\angle BOC=\frac{2\pi}{3}$ and the cosine rule, we have
$$
\begin{array}{l}
BC=\sqrt{x^{2}+xy+y^{2}}. \\
\text{And } S_{\triangle BOC}=\frac{1}{2} BC \cdot OD = S_{\triangle BOE} + S_{\triangle OOE} \\
=\frac{1}{2} OE \cdot x \cdot \sin 60^{\circ} + \frac{1}{2} OE \cdot y \cdot \sin 60^{\circ} \\
=\frac{1}{2} OE \cdot (x+y) \cdot \frac{\sqrt{3}}{2},
\end{array}
$$

Then, by $OE \geqslant OD$, we get
$$
\sqrt{x^{2}+xy+y^{2}} \geqslant \frac{\sqrt{3}}{2}(x+y).
$$

Similarly, we have
$$
\begin{array}{l}
\sqrt{y^{2}+yz+z^{2}} \geqslant \frac{\sqrt{3}}{2}(y+z), \\
\sqrt{x^{2}+xz+z^{2}} \geqslant \frac{\sqrt{3}}{2}(x+z).
\end{array}
$$

Therefore, $\sqrt{x^{2}+xy+y^{2}} + \sqrt{y^{2}+yz+z^{2}}$
$$
+ \sqrt{x^{2}+xz+z^{2}} \geqslant \sqrt{3}(x+y+z),
$$

which means $AB + BC + CA \geqslant \sqrt{3}(BO + CO + AO)$. The equality holds if and only if $x=y=z$. Hence, $m=\sqrt{3}$.