nice problem! make the substitution $a=\sqrt[3]x, b=\sqrt[3]y, c=\sqrt[3]z$, and we have that $abc=1$, so the problem is equivalent to $(abc+a^3)(abc+b^3)(abc+c^3)\geq 2+2(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}) \\\Leftrightarrow (bc+a^2)(ac+b^2)(ab+c^2)\geq 2+2(a^2b+b^2c+ac^2) \\\Leftrightarrow a^3+b^3+c^3+(ab)^3+(ac)^3+(bc)^3\geq2(a^2b+b^2c+ac^2)$ but from am-gm we have that $a^3+(ab)^3+1\geq3a^2b$, taking the cyclic sum we obtain $a^3+b^3+c^3+(ab)^3+(ac)^3+(bc)^3+3\geq2(a^2b+b^2c+ac^2)+a^2b+b^2c+ac^2\geq2(a^2b+b^2c+ac^2)+3\sqrt[3]{(a^2b)(b^2c)(ac^2)}=2(a^2b+b^2c+ac^2)+3 \\\Rightarrow a^3+b^3+c^3+(ab)^3+(ac)^3+(bc)^3\geq2(a^2b+b^2c+ac^2)$ and we're done!!! maybe the substitution is useless but it is easier, at least for me, to work without any kind of roots...

Source : Baltic Way 2003 , problem 3 .

Amir.S wrote: Let $x$, $y$ and $z$ be positive real numbers such that $xyz = 1$. Prove that $$\left(1+x\right)\left(1+y\right)\left(1+z\right)\geq 2\left(1+\sqrt[3]{\frac{x}{z}}+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}\right).$$ I thought this is APMO 1998/3 for a second. Anyway, here's cleaner writeup than @2above but mostly the same. Note that the inequality is equivalent to proving $$x+y+z+xy+yz+zx\geq 2\left(\sqrt[3]{\frac{x}{z}}+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}\right).$$Of course, the RHS term look very sus... Indeed, we'll be using AM-GM. $$x+xy+1\geq 3\sqrt[3]{x\cdot xy\cdot 1}=3\sqrt[3]{\frac{x}{z}}$$and vice versa. Sum it up to get $$x+y+z+xy+yz+zx+3\geq 3\left(\sqrt[3]{\frac{x}{z}}+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}\right).$$Now, we'll use AM-GM one last time. We have $$\sqrt[3]{\frac{x}{z}}+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}\geq 3$$and it's easy to see that we're done.