Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks

This inequality seems to be somehow too weak, so I'll prove the much stronger inequality:
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$

$$1 + \frac ab + \frac ac \ge 3\sqrt[3]\frac{a^2}{bc} = \frac{3a}{\sqrt[3]{abc}}$$
$$1 + \frac ba + \frac bc \ge 3\sqrt[3]\frac{b^2}{ac} = \frac{3b}{\sqrt[3]{abc}}$$
$$1 + \frac ca + \frac cb \ge 3\sqrt[3]\frac{c^2}{ab} = \frac{3c}{\sqrt[3]{abc}}$$
Sum all the inequalities and we have:
$$3 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb \ge 3\frac{a+b+c}{\sqrt[3]{abc}}$$
$$1 + \frac ac + \frac ab + \frac ba + \frac bc + \frac ca + \frac cb + 1 \ge 2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right) + \left(\frac{a+b+c}{\sqrt[3]{abc}} - 3\right)$$
The last term is obviously positive, since $a+b+c \ge 3\sqrt[3]{abc}$ from AM-GM, hence the proof.
We have equality when $a=b=c$
