How to prove the follwing inequality

Question

If $$(a+b+c)abc=3$$ and $$a,b,c > 0$$ prove that $$(a+b)(b+c)(c+a)\geq 8$$

I can fairly easily prove that $(a+b)(b+c)(c+a)\geq8abc$, but then I get stuck.....since then I cannot move forward

If I was to prove that $abc\geq1$ this would have been easy but I am stuck, please help me.

Any help is appreciated, thanks.

edit:I am incredibly sorry that I remembered the question incorrectly

Answer 1

Note: $3=abc(a+b+c)\ge 3(abc)^{4/3}$ (by AM-GM) hence $ abc\le 1$.

Using this we have, using AM-GM again: $$(a+b)(a+c)(b+c) = (a+b+c)(ab+ac+bc) - abc\\ \ge 3 (a+b+c)\sqrt[3]{(abc)^2} - 1\\ = 9 (abc)^{-1/3} - 1\\ \ge 9-1 = 8$$