Example 4 If real numbers $a, b, c$ satisfy $a+b+c=0$, $abc=2, c>0$, then ( ).
(A) $ab<0$
(B) $|a|+|b| \geqslant 2$
(C) $|a|+|b| \geqslant 4$
(D) $0<|a|+|b| \leqslant 1$
Solution: It is easy to know that $a+b=-c$.
Since $abc=2, c>0$, therefore, $a, b$ are both negative, i.e., $|a|+|b|=|a+b|=|c|$, and $|ab|=\frac{2}{|c|}$.
From $|a|+|b| \geqslant 2 \sqrt{|a| \cdot|b|}$, we get $|c| \geqslant 2 \sqrt{\frac{2}{|c|}}$
Solving this, we get $|c| \geqslant 2$.
Therefore, $|a|+|b|=|c| \geqslant 2$.
Hence, the answer is (B).