$a,b,c$ real, summing to two and multiplying to four; find minimum of $|a|+|b|+|c|$ 
Knowing that $a$, $b$ and $c$ are real numbers with
  $$a+b+c=2\qquad abc=4$$
  find the smallest value of $|a|+|b|+|c|$.

The answer's supposed to be 6 but I don't know how to do it.

Let $a\geq b\geq c$.
If $c>0$ then by AM-GM $$2=a+b+c\geq3\sqrt[3]{abc}=3\sqrt[3]4,$$
which is contradiction.
Thus, $c<0$ and since $abc>0$ we obtain $b<0$.
Now, we can write our problem in the following words.

Let $a$, $b$ and $c$ be positive numbers such that $abc=4$ and $a-b-c=2$.
Find the minimal value of $a+b+c$.

Indeed, the condition gives: $bc=\frac{4}{a}$ and $b+c=a-2$, which says that
$$(a-2)^2\geq\frac{16}{a}$$ or
$$(a-4)(a^2+4)\geq0.$$
Thus, $b+c=a-2\geq2$ and $a+b+c\geq4+2=6$.
The equality occurs in the last case for $a=4$ and $b=c=1$, which is
$a=4$ and $b=c=-1$ for the original problem.
Done!
