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

Thomas Andrews
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Rafael Guimarães
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    What have you tried? – Thomas Andrews May 04 '23 at 12:48
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    Using average inequality: if a,b,c are positive: (a+b+c)/3 >= (abc)^1/3 = 4^1/3; a+b+c>=3*4^1/3>2. so a,b, or c is negative.
    Applying absolute value inequality is not useful |a|+|b|+|c|>= a+b+c=2, but |a|+|b|+|c| can never be 2 as a,b,c can not be all positive Absolute value inequality has to be used at some point     – Rafael Guimarães May 04 '23 at 12:54
  • What have you tried? – Bumblebee May 04 '23 at 12:58
  • Also, average inequality is not very useful as abc=4, so |a|,|b|,|c| are not all free variables – Rafael Guimarães May 04 '23 at 13:00
  • Are you allowed to use Lagrange multipliers? Or maybe I should ask: Have you heard about them? – Arthur May 04 '23 at 13:02
  • No, i dont think so – Rafael Guimarães May 04 '23 at 13:10
  • There is exactly one positive number in $a,b,c$. WLOG, let $a>0$, then you can write $|a|+|b|+|c|=a-b-c$ and solve without absolute values. Or you can take $b=-d$, $c=-e$ and solve in positive numbers $a,d,e$: $a-d-e=2$, $ade=4$, $a+d+e={\rm min}$ – Ivan Kaznacheyeu May 04 '23 at 13:11
  • Basically, if $a>0,$ then $b,c$ have to be the same sign. If $b,c$ are both positive, $|a|+|b|+|c|=a+b+c=2.$ Otherwise, $|a|+|b|+|c|=a-(b+c)=2a-2.$ So you need to rule out all of $a,b,c>0$ and then you need to find the smallest positive $a$ where we can find $b,c.$ – Thomas Andrews May 04 '23 at 13:42
  • Put your work in your question. – Thomas Andrews May 04 '23 at 13:48

1 Answers1

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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 $d=e=1$, $a=4$

Ivan Kaznacheyeu
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  • Try to avoid answering questions that lack of context. – Gary May 04 '23 at 13:25
  • @Gary OP's personal work, a decent start on the problem, was given in the comment section. Should it have been put in the actual question post? Yes. Does it still count? I think it does. – Arthur May 04 '23 at 13:34