6. Given $a, b, c \in [0,1]$. Then
$$
\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}
$$

the range of values is
6. $[0,2]$.

Obviously, the minimum value of the required algebraic expression is 0, which is achieved if and only if $a=b=c=0$.

When $a+b+c>0$, $a+b+c \leqslant 1+b+c \leqslant 2(1+b c)$.
Thus, $\frac{a}{b c+1} \leqslant \frac{2 a}{a+b+c}$.
Similarly, $\frac{b}{c a+1} \leqslant \frac{2 b}{a+b+c}, \frac{c}{a b+1} \leqslant \frac{2 c}{a+b+c}$.
Adding the above three inequalities yields
$$
\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1} \leqslant 2 \text {. }
$$

When one of $a, b, c$ is 0 and the other two are 1, the equality holds.