2. (2003 Shanghai Competition Problem) Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
2. (1) Without loss of generality, let $a=\max \{a, b, c\}$. From the given conditions, we have
$$
a>0, b+c=2-a, bc=\frac{4}{a} \text {. }
$$

Therefore, $b$ and $c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$. Thus,
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0 \Leftrightarrow a^{3}-4 a^{2}+4 a-16 \geqslant 0 \\
\Leftrightarrow\left(a^{2}+4\right)(a-4) \geqslant 0 \Leftrightarrow a \geqslant 4 .
\end{array}
$$

When $a=4, b=c=-1$, the conditions are satisfied, so the minimum value is 4.
(2) From $abc=4>0$, we know that $a, b, c$ are all positive or one positive and two negative.

If $a, b, c$ are all positive, then $a, b, c$ all belong to the interval $(0,2)$, which contradicts the conclusion of (1). Therefore, $a, b, c$ must be one positive and two negative. By symmetry, without loss of generality, let $a>0, b<0, c<0$. Thus,
$$
|a|+|b|+|c|=a-(b+c)=a-(2-a)=2a-2 \geqslant 2 \times 4-2=6 \text {. }
$$

When $a=4, b=c=-1$, the equality holds.
Therefore, the minimum value of $|a|+|b|+|c|$ is 6.