Prove that $(a+b)(b+c)(c+a) \ge8$ 
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
  Prove that $(a+b)(b+c)(c+a)\ge8$.

My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)$, $(2)$ and $(3)$ together we reach
$$(a+b)(b+c)(c+a) \ge8abc.$$
Now, I need to show that $abc = 1$.
Again by AM-GM inequality we have
$$\frac{a+b+c}{3}\ge\sqrt[3]{abc}
\implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc}
\implies abc\le1$$

Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?
Other solutions to the above question are also welcomed.

If the numbers are positive here is a solution
$$3=abc(a+b+c)\ge 3abc(abc)^{\dfrac 1 3} \Rightarrow abc\le 1 \\$$
$$3=abc(a+b+c)\le \dfrac {(a+b+c)^3} {27} \cdot (a+b+c) \Rightarrow a+b+c \ge 3 \\ $$
$$ (ab+bc+ca)^2 \ge 3abc(a+b+c)=9 \Rightarrow ab+bc+ca \ge 3$$
$$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \ge 9-1=8$$
equality holds if and only if $a=b=c=1$ 
