Without loss of generality, suppose $a \ge b\ge c$. Then $b+c=2-a$ and $bc=\frac{4}{a}$. Hence, $b$ and $c$ are two real roots of equation $$x^2 -(2-a)x+\frac{4}{a}=0$$ Therefore, the discriminant of this equation must be non-negative:$$(2-a)^2 - 4\times\frac{4}{a}\ge 0\implies (a^2+4)(a-4)\ge 0\implies a\ge \boxed{4}$$ It follows that$$\mid a\mid +\mid b \mid + \mid c \mid \ge \mid a \mid + \mid b+c\mid = \mid a\mid +\mid 2- a\mid=2a-2\ge \boxed{6}$$