Since abc = 2>0, there must be two of them are negative and one is positive.

Let a<0, b<0 and c>0. The absolute symbols could be removed

|a|+|b|+|c|

=-a-b+c

=-(a+b)+c

Since a+b+c=0, then

On the other hand,

a+b = -c

and

ab = \dfrac{2}{c}

Apply the Vieta's formula, a,b are the two roots of the quadratic equation

x^2+cx+ \dfrac{2}{c}=0

The equaiton has real roots if and only if its discriminant is not less than 0. That is,

c^2-4\cdot \dfrac{2}{c}\geq 0

Solving the inequality gives

c\geq 2

Therefore, the minimum value of |a|+|b|+|c| is 4