Prove that $(a+b)(b+c)(c+a) \ge8$

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 $a=b=c=1$

A geometric approach is to notice that $a,b,c>0$ and $abc(a+b+c)=3$ grant that there is a triangle with side lengths $a+b,a+c,b+c$ and area $\sqrt{3}$. Since the area of a triangle is given by the product of its side lengths divided by four times the length of the circumradius, the problem boils down to understanding what is the minimum circumradius for a given area, and it is pretty clear that the minimum is achieved by the equilateral triangle, since $$ 2R = \frac{BC}{\sin\widehat{BAC}} $$ and if a variable $A$ point travels on a line parallel to a fixed $BC$ segment (so that the area of $ABC$ is constant), the maximum $\widehat{BAC}$ angle is achieved when $BA=CA$.