Example 11 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 $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition)
Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$

Therefore,
$$
\begin{array}{l}
\Delta=[-(2-a)]^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Rightarrow a^{3}-4 a^{2}+4 a-16 \geqslant 0 \\
\Rightarrow\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$

Solving this, we get $a \geqslant 4$.
When $a=4, b=c=-1$, the minimum value of the maximum among $a, b, c$ is 4.
(2) From $abc>0$, we know that $a, b, c$ are all greater than 0 or one is positive and two are negative.
(i) If $a, b, c$ are all greater than 0, then from (1) we know that the minimum value of the maximum among $a, b, c$ is 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0$, $b<0$, $c<0$. Then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2 a-2 .
\end{array}
$$

From (1), we know $a \geqslant 4$, hence $2 a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions of the problem are satisfied and the equality holds.
Therefore, the minimum value of $|a|+|b|+|c|$ is 6.