6. Let $a_{1}, a_{2}, \cdots, a_{2007}$ be positive real numbers, and
$$
\frac{1}{2+a_{1}}+\frac{1}{2+a_{2}}+\cdots+\frac{1}{2+a_{200}}=\frac{1}{2} \text {. }
$$

Then the minimum value of $a_{1} a_{2} \cdots a_{200}$ is $\qquad$
$6.4012^{2007}$.
Let $x_{i}=\frac{2}{2+a_{i}}$, then
$a_{i}=2 \cdot \frac{1-x_{i}}{x_{i}}$, and $\sum_{i=1}^{200} x_{i}=1$.
Thus, $a_{1} a_{2} \cdots a_{200}$
$$
\begin{array}{l}
=2^{2000} \cdot \frac{1}{x_{1} x_{2} \cdots x_{200}} \prod_{i=1}^{2007}\left(\sum_{j \neq i} x_{j}\right) \\
\geqslant 2^{2000} \cdot \frac{1}{x_{1} x_{2} \cdots x_{200}} \prod_{i=1}^{2000}\left(2006 \prod_{j \neq i} x_{j}^{\frac{1}{2000}}\right) \\
=2^{2007} \times 2006^{2000}=4012^{200} .
\end{array}
$$