Elementary Inequality question

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.

Answer 1

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.