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.

From $abc(a+b+c)=3$ we have $$a^2+a(b+c)=\frac {3}{bc}.$$ Therefore we have $$(a+b)(b+c)(c+a)\geq 8\iff$$ $$(a+b)(a+c)\geq \frac {8}{b+c}\iff$$ $$a^2+a(b+c)\geq -bc+ \frac {8}{b+c)}\iff$$ $$(\bullet ) \quad   \frac {3}{bc}\geq -bc+\frac {8}{b+c}.$$ Now $b+c\geq 2\sqrt {bc},$ so $$\frac {8}{b+c}\leq \frac {4}{\sqrt {bc}}.$$ So with $x=\sqrt {bc} $ we see that ($\bullet $ ) is satisfied if $\frac {3}{x^2}\geq -x^2+\frac {4}{x} $, or, equivalently, $\frac {3} {x^2}+x^2-\frac {4}{x}\geq 0 ,$ for all $x>0.$
We have $\frac {3}{x^2}+x^2-\frac {4}{x}=x^{-2}(x-1)(x^3+x^2+x-3). $ The terms $(x-1)$ and $(x^3+x^2+x-3)$ always have the same sign when $x\geq 0.$ (Case 1:$\;x\geq 1.$ Case 2: $\;0<x<1.$)
Inelegant compared to the answer by Booldy, but it works.
