3. Given $0 \leqslant a, b, c \leqslant 1$. Then
$$
\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1}
$$

The maximum value is ( ).
(A) 1
(B) 2
(C) 3
(D) $\frac{3}{2}$
3. B.

When $a+b+c=0$, we have $a=b=c=0$, then the original expression $=0$.
When $a+b+c>0$, by
$$
a+b+c \leqslant 1+b+c \leqslant 2(1+b c),
$$

we know $\frac{a}{b c+1} \leqslant \frac{2 a}{a+b+c}$.
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 three inequalities, we get
$$
\frac{a}{b c+1}+\frac{b}{c a+1}+\frac{c}{a b+1} \leqslant 2 \text {. }
$$

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