ta có $1+x^{3}+y^{3}\geq 3\sqrt[3]{x^{3}y^{3}}=3xy$$ \Rightarrow \frac{\sqrt{1+x^{3}+y^{3}}}{xy} \geq \frac{\sqrt{3xy}}{xy}=\frac{\sqrt{3}}{\sqrt{xy}}$
TT $\Rightarrow VT \geq \sqrt{3}(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}})$
$ \geq \sqrt{3} 3\sqrt[3]{\frac{1}{\sqrt{xyz}^{2}}}=3\sqrt{3}$ do $xyz=1$
dấu "="$\Leftrightarrow x=y=z=1$