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I found this post and many others with the same inequality, and I wondered if in the case of a negative RHS, if the absolute would be bigger than $abc$. After a couple of examples I couldn't find any contradictions. Is there any way to prove this?

Memat
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  • I'm curious what type of examples you considered. (Short answer is it's false, by considering $c = a + b + \varepsilon$) – Brian Moehring Dec 11 '22 at 18:20
  • @BrianMoehring Oh, yeah! My bad, I copy pasted it wrong! – Memat Dec 11 '22 at 18:27
  • The edit doesn't change my comment's point. For $\varepsilon =0$ you would have $0$ is greater than a positive number, which is absurd, so by continuity it remains absurd at least for small $\varepsilon$. I am still curious what type of examples you considered. – Brian Moehring Dec 11 '22 at 18:32
  • @BrianMoehring if $c=a+b$ then none of $(a+b-c),(a-b+c),(-a+b+c)$ is negative, so it's not taken into account – Memat Dec 11 '22 at 18:51
  • @BrianMoehring I just wrote a short python program that tries different combinations – Memat Dec 11 '22 at 18:52

1 Answers1

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The task's statement is wrong.

Here is a family of counterexamples:

Let $a=b= xc$ then we have to show $-(2x-1) c^3 \ge x^2 c^3$. If $c >0$ this becomes

$1-2x - x^2\ge 0 $. Let $0 < x <0.5$ and $c >0$ then only the first term is negative, as required. However, then $f(x) = 1-2x - x^2\ge 0 $ does not hold, since $f(x) \ge 0 $ holds only for $0 < x < \sqrt{2} -1 \simeq 0.414$ whereas for all $0.414 \simeq \sqrt{2} -1 < x < 0.5$ we have that $f(x) \le 0 $. This contradicts your claim. $\qquad \Box$

Andreas
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