# Elementary Inequality question

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.

• 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. – user546996 Mar 29 '18 at 18:27
• Please don't downvote, I really want to know in which cases it's possible – K. Chopra Mar 29 '18 at 18:27
• That wasn't me, I don't know who did that. =( – user546996 Mar 29 '18 at 18:28
• @user546996 Thanks a lot – K. Chopra Mar 29 '18 at 18:29
• counterexample is: Consider $a = 10^{-10}$, $b=c=2$, then the first inequality holds and the second doesn't. – Andreas Mar 29 '18 at 18:32

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.