Absolute value Inequality: $a+b+c=2$; $abc=4.$ What's the minimum value of $|a|+|b|+|c|?$

Question

Given $a,b,c$ real numbers, with $a+b+c=2$ and $abc=4.$ What's the minimum value of $|a|+|b|+|c|?$

Answer 1

Using AM-GM one can show that $a,b,c$ cannot all be positive

$abc>0$, then there are two negative and one positive number in $a,b,c$

WLOG let $a>0$, then $b<0$, $c<0$. Let $b=-d$, $c=-e$, $d,e>0$

Then problem becomes following: $a-d-e=2$, $ade=4$, $a+d+e={\rm min}$

Let $d+e=f$, then using AM-GM one can show $de\leq \frac{f^2}{4}$

$a=2+d+e=f+2$

$4=ade=(f+2)de\leq \frac{f^2(f+2)}{4}$

$f^3+2f^2-16\geq 0$, $(f-2)(f^2+4f+8)\geq 0$, $f\geq 2$

$a+d+e=2f+2\geq 6$

Minimum is obtained at $d=e=1$, $a=4$