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

The inequality is unchanged if you multiply $a$, $b$, and $c$ by a positive number $k$, so WLOG $abc = 1$. Then the left hand is greater than or equal to 
$$1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
and $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge a + b + c$$
since
\begin{align}&\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\\
&= \frac{1}{3}\left(\frac{a}{b} + \frac{a}{b} + \frac{b}{c}\right) + \frac{1}{3}\left(\frac{b}{c} + \frac{b}{c} + \frac{c}{a}\right) + \frac{1}{3}\left(\frac{c}{a} + \frac{c}{a} + \frac{a}{b}\right)\\
&\ge\sqrt[3]{\frac{a}{b}\frac{a}{b}\frac{b}{c}} + \sqrt[3]{\frac{b}{c}\frac{b}{c}\frac{c}{a}} + \sqrt[3]{\frac{c}{a}\frac{c}{a}\frac{a}{b}}\\
&= \frac{a}{\sqrt[3]{abc}} + \frac{b}{\sqrt[3]{abc}} + \frac{c}{\sqrt[3]{abc}}\\
&= a + b + c.
\end{align}
