6. Let $G$ be the centroid of $\triangle ABC$, $r$ be the inradius of $\triangle ABC$, and the distances from point $G$ to sides $BC$, $CA$, and $AB$ be $GD$, $GE$, and $GF$ respectively. Let $s=\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}$. Then
(A) $s>\frac{3}{r}$
(B) $s=\frac{3}{r}$
(C) $s<\frac{3}{r}$
(D) Cannot be determined
6.B.

As shown in Figure 5, $A M$ is the median on $B C$, and $B G$, $C G$ are connected. Then
$$
\begin{array}{l}
S_{\triangle A C B}=S_{\triangle A C C} \\
=S_{\triangle B C C}=\frac{1}{3} S_{\triangle A B C} . \\
\text { Also, } S_{\triangle B C C}=\frac{1}{2} B C \cdot G D,
\end{array}
$$

Therefore,
$$
\frac{1}{G D}=\frac{B C}{2 S_{\triangle B C C}}=\frac{3 a}{2 S_{\triangle A B C}} .
$$

Similarly, $\frac{1}{G E}=\frac{3 b}{2 S_{\triangle A B C}}, \frac{1}{G F}=\frac{3 c}{2 S_{\triangle A B C}}$.
Adding the above three equations, we get
$$
\frac{1}{G D}+\frac{1}{G E}+\frac{1}{G F}=\frac{3(a+b+c)}{2 S_{\triangle A B C}} \text {. }
$$

Noting that $S_{\triangle A B C}=\frac{1}{2}(a+b+c) r$, we have
$$
\frac{1}{G D}+\frac{1}{G E}+\frac{1}{G F}=\frac{3}{r} \text {. }
$$