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.
 A: Hint: $$\frac{a}{b} + \frac{a}{c} + 1 \ge 3 \frac{a}{\sqrt[3]{abc}}$$ by AM-GM.
A: I know it is a very late to answer this, but i came across a very nice method to prove this.
$$\displaystyle \left(1+\frac a b \right) \left(1+\frac b  c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$\displaystyle \iff (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)  \geq 3+2\left(\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$\displaystyle \iff \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)  \geq \frac{3}{a+b+c}+\frac{2}{\sqrt[3]{abc}}$$
Replacing by their inverses
$$\iff 3AM\ge HM+2GM$$ 
Which is evident.
A: We can begin by clearing denominators as follows
$$a^2c+a^2b+b^2a+b^2c+c^2a+c^2b\geq 2a^{5/3}b^{2/3}c^{2/3}+2a^{2/3}b^{5/3}c^{2/3}+2a^{2/3}b^{2/3}c^{5/3}$$
Now by the Arithmetic Mean - Geometric Mean Inequality, 
$$\frac{2a^2c+2a^2b+b^2a+c^2a}{6} \geq a^{5/3}b^{2/3}c^{2/3}$$
That is, 
$$\frac{2}{3}a^2c+\frac{2}{3}a^2b+\frac{1}{3}b^2a+\frac{1}{3}c^2a \geq 2a^{5/3}b^{2/3}c^{2/3}$$
Similarly, we have 
$$\frac{2}{3}b^2a+\frac{2}{3}b^2c+\frac{1}{3}c^2b+\frac{1}{3}a^2b \geq 2a^{2/3}b^{5/3}c^{2/3}$$
$$\frac{2}{3}c^2b+\frac{2}{3}c^2a+\frac{1}{3}a^2c+\frac{1}{3}b^2c \geq 2a^{2/3}b^{2/3}c^{5/3}$$
Summing these three inequalities together, we obtain the desired result.
A: 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.
