6. Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\angle F_{1} P F_{2}=$ $60^{\circ}$. Then the minimum value of the product of the eccentricities of the ellipse and the hyperbola is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) 1
(D) $\sqrt{3}$
6. B.

Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n(m>n)$.
The equation of the ellipse is $\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1$, and the equation of the hyperbola is $\frac{x^{2}}{a_{2}^{2}}-\frac{y^{2}}{b_{2}^{2}}=1$,
with the semi-focal distances of the two curves being $c_{1}$ and $c_{2}$, and $c_{1}=c_{2}$.
By the definition of conic sections, we have
$$
m+n=2 a_{1}, m-n=2 a_{2} \text {. }
$$

Thus, $m=a_{1}+a_{2}, n=a_{1}-a_{2}$.
By the cosine rule, we get
$$
\begin{array}{l}
m^{2}+n^{2}-m n=4 c_{1}^{2}=4 c_{2}^{2} \\
\Rightarrow \quad\left(a_{1}+a_{2}\right)^{2}+\left(a_{1}-a_{2}\right)^{2}-\left(a_{1}+a_{2}\right)\left(a_{1}-a_{2}\right) \\
\quad=4 c_{1}^{2}=4 c_{2}^{2} \\
\Rightarrow a_{1}^{2}+3 a_{2}^{2}=4 c_{1}^{2}=4 c_{2}^{2} \\
\Rightarrow \frac{1}{e_{1}^{2}}+\frac{3}{e_{2}^{2}}=4 .
\end{array}
$$

By the AM-GM inequality, we have
$$
\begin{array}{l}
4=\frac{1}{e_{1}^{2}}+\frac{3}{e_{2}^{2}} \geqslant 2 \sqrt{\frac{3}{e_{1}^{2} e_{2}^{2}}} \\
\Rightarrow e_{1} e_{2} \geqslant \frac{\sqrt{3}}{2} .
\end{array}
$$

When $e_{1}=\frac{\sqrt{2}}{2}, e_{2}=\frac{\sqrt{6}}{2}$, the equality holds.
Thus, the minimum value of the product of the eccentricities of the ellipse and the hyperbola is $\frac{\sqrt{3}}{2}$.