3. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have its major axis endpoints at $A$ and $B$. If there is a point $P$ on the ellipse $C$ such that $\angle A P B=120^{\circ}$, then the minimum value of the eccentricity of the ellipse $C$ is $($.
(A) $\frac{1}{3}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{2}{3}$
(D) $\frac{\sqrt{6}}{3}$
3. D.

Let $P\left(x_{0}, y_{0}\right)$, then $\tan 120^{\circ}=\frac{2 a\left|y_{0}\right|}{x_{0}^{2}+y_{0}^{2}-a^{2}}=-\sqrt{3}$.
Substitute $x_{0}^{2}-a^{2}=-\frac{a^{2}}{b^{2}} y_{0}^{2}$ into the above equation, we get
$\left|y_{0}\right|=\frac{2 a b^{2}}{\sqrt{3}\left(a^{2}-b^{2}\right)}$.
Also, $-b \leqslant y_{0} \leqslant b$, thus,
$$
\frac{2 a b^{2}}{\sqrt{3}\left(a^{2}-b^{2}\right)} \leqslant b \text {, }
$$

which implies $\left(\frac{b}{a}\right)^{2}+\frac{2}{\sqrt{3}}\left(\frac{b}{a}\right)-1 \leqslant 0$.
Solving this, we get $\frac{b}{a} \leqslant \frac{1}{\sqrt{3}}$.
Therefore, $e^{2}=1-\left(\frac{b}{a}\right)^{2} \geqslant 1-\frac{1}{3}=\frac{2}{3}$.
Hence, $\frac{\sqrt{6}}{3} \leqslant e<1$.