How to prove the follwing inequality $(a+b)(b+c)(c+a) \ge 8$

Question

If $$a+b+c=3abc$$ 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

Since $${a+b+c\over 3}\geq \sqrt[3]{abc}$$

we get $$abc\geq \sqrt[3]{abc} \implies a^3b^3c^3\geq abc \implies a^2b^2c^2 \geq 1$$

Since $${x+y\over 2}\geq \sqrt{xy} \implies x+y\geq 2\sqrt{xy}$$

so we have $$(a+b)(b+c)(c+a)\geq 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8\sqrt{a^2b^2c^2}\geq 8$$

Answer 2

By AM-GM $$a+b+c=3abc\leq3\left(\frac{a+b+c}{3}\right)^3,$$ which gives $$a+b+c\geq3.$$ Thus, $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)\geq$$ $$\geq\frac{8}{9}(a+b+c)\sqrt{3abc(a+b+c)}=\frac{8}{9}(a+b+c)^2\geq8.$$