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this might be a stupid question, but I wanted to know that if a,b,c are positive reals and $$ab+bc+ca\geq 3$$ Can we say that $$3 \sqrt[3]{(abc)^2} \geq 3$$ By applying am-gm? I'm confused about whether we can do this or not. I'd really appreciate it if someone could explain.

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  • $\begingroup$ Well, you know that the arithmetic mean is bigger than 1, and that the geometric mean is less than the arithmetic mean, so I think there is a step missing to also conclude that the geometric mean is bigger than 1. $\endgroup$ – user546996 Mar 29 '18 at 18:27
  • $\begingroup$ Please don't downvote, I really want to know in which cases it's possible $\endgroup$ – K. Chopra Mar 29 '18 at 18:27
  • $\begingroup$ That wasn't me, I don't know who did that. =( $\endgroup$ – user546996 Mar 29 '18 at 18:28
  • $\begingroup$ @user546996 Thanks a lot $\endgroup$ – K. Chopra Mar 29 '18 at 18:29
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    $\begingroup$ counterexample is: Consider $a = 10^{-10}$, $b=c=2$, then the first inequality holds and the second doesn't. $\endgroup$ – Andreas Mar 29 '18 at 18:32
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I believe we cannot arrive at this conclusion, because of the following counterexample:

Let $a=3$, $b=1-\epsilon$, and $c=\frac{1}{3}$. Then $ab+bc+ca\geq 4-3\epsilon\geq 3$, but $3(abc)^{2/3}<3$.

So for any sufficiently small $\epsilon$ our conclusion doesn't hold.

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  • $\begingroup$ I see. Thanks a lot $\endgroup$ – K. Chopra Mar 29 '18 at 18:35

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