Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$. 
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b  c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.

I expanded the LHS, and realized I have to prove $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$, but I don't know how. Please help. Thank you.

The same inequality mentioned by @Sanchez used three times takes us to:
$$\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+3\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}$$
On the other hand, AM-GM gives us:
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
so that:
$$-3 \geq -\frac{a+b+c}{\sqrt[3]{abc}}$$
And adding the first and third inequalities gives the result.
