Let  $f\left(x\right)=x^{\frac{3}{2}}+x^{\frac{-3}{2}}-4(x+\frac{1}{x})$, then the minimum value of $f\left(x\right)$  for all permissible real values of  x  is

$$\begin{gathered} f\left(x\right)={(\sqrt{x}+\frac{1}{\sqrt{x}})}^3-3(\sqrt{x}+\frac{1}{\sqrt{x}})-4[{(\sqrt{x}+\frac{1}{\sqrt{x}})}^2-2] \\ Let, \sqrt{x}+\frac{1}{\sqrt{x}}=t \\ g\left(t\right)=t^3-3t-4(t^2-2) \\ {g}^1\left(t\right)=3t^2-8t-3 \\ {g}^1\left(t\right)=0 \Rightarrow t=3 \\ {g}^{11}\left(t\right)=6t-8 \\ {g}^{11}\left(3\right)=10>0 \end{gathered}$$

$\therefore$ minimum value is  $g\left(3\right)=27-36-9+8=-10$