Bounds of an Expression 
if $(a,b,c)>0$
and  $abc(a+b+c)=3$
Then what can you say about the bounds of 
$(a+b)(b+c)(c+a)$ ?

Hint:

 think Geometrically!

My Approach
1.assumed $a\geq b \geq c$ 
2.Tried to think geometrically
3.used the Hadwiger-Finsler inequality
and modified it to get $(a+b+c) \geq \sqrt{4A\sqrt{3}+2(a^2+b^2+c^2)}$ (where A is the area of a triangle and a,b,c are it's sides)
4. I suspect that as we're to find bounds of  $\Pi (a+b)$,we could try to find some vaues of (a+b)'s.
Could you please help me with this problem?
 A: The following inequality is true.
$$\sqrt[3]{\frac{(a+b)(a+c)(b+c)}{8}}\geq\sqrt[4]{\frac{abc(a+b+c)}{3}}.$$
For the proof use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc).$$
It gives a lower bound.
The upper bound is $+\infty$.
Can you end it now?
We need to prove that
$$\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^4\geq\left(\frac{abc(a+b+c)}{3}\right)^3.$$
Now, by using $(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$ and by  AM-GM  we obtain:
$$\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^4\geq\left(\frac{(a+b+c)(ab+ac+bc)}{9}\right)^4=$$
$$=\left(\frac{a+b+c}{3}\right)^3\cdot\frac{a+b+c}{3}\cdot\left(\frac{ab+ac+bc}{3}\right)^4\geq$$
$$\geq\left(\frac{a+b+c}{3}\right)^3\cdot\sqrt[3]{abc}\cdot\left(\sqrt[3]{a^2b^2c^2}\right)^4=\left(\frac{abc(a+b+c)}{3}\right)^3.$$
A: Let $g=a+b$,$~f=a+c$,$~e=b+c$,
$\implies a=\frac f2+\frac g2-\frac e2$,$~~b=\frac e2+\frac g2-\frac f2$,   $~~c=\frac e2+\frac f2-\frac g2$
It is easy to see that $e,f,g$ can form the sides of a triangle. Let $\Delta$ denote the area of triangle $EFG$ and $R$ denote the circumradius of the triangle $EFG$.
We have,
$$(s-e)(s-f)(s-g)s=3$$
$$\implies\Delta=\sqrt3$$
$$(a+b)(b+c)(c+a)=efg=8R^3(\sin E\cdot\sin F\cdot \sin G)=\frac{(efg)^3}{8\Delta^3}(\sin E\cdot\sin F\cdot \sin G)$$
$$\implies(efg)^2=\frac{24\sqrt3}{\sin E\cdot\sin F\cdot \sin G}$$
$$0<\sin E\cdot\sin F\cdot \sin G\le\frac{3\sqrt3}{8}$$
$$\implies efg\in\bigg[8,\infty\bigg)$$
$$\implies (a+b)(b+c)(c+a)\in\bigg[8,\infty\bigg)$$
