\(VT=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)\)
\(=3\left(x+\dfrac{1}{x}\right)\ge3.2\sqrt{x.\dfrac{1}{x}}=6\)