5. If $a b c=4, a+b+c=2$, and $a \geqslant b \geqslant c$, then the minimum value of $|a|+|b|+|c|$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
5. C.

If $a>0$.
Also, from the problem, we know $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$

Then, $\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \Rightarrow a \geqslant 4$.
Therefore, $b, c$ are both negative.
$$
\begin{array}{l}
\text { Hence }|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2 a-2 \geqslant 6 .
\end{array}
$$

When $a=4, b=c=-1$, $|a|+|b|+|c|$ takes the minimum value 6.