<s>[INST] 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.
 [/INST] 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.
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