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

$\begin{array}{c}f\left(x\right)=x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)\\ f\left(x\right)={\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^3-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)-4\left[{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2-2\right]\\ f\left(x\right)={\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^3-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)-4\left[{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2-2\right]\\ \text{ Let }g\left(t\right)=t^3-3t-4t^2+8\\ \Rightarrow g\left(t\right)=t^3-4t^2-3t+8\\ \Rightarrow g^{\text{'}}\left(t\right)=3t^2-8t-3=(t-3)(3t+1)\\ g^{\text{'}\text{'}}\left(t\right)=6t-8\\ {\text{g}}^{\text{'}\text{'}}\left(3\right)=10>0\Rightarrow \text{g}\left(3\right)\text{ is minimum }\\ \text{g(3)=27-9-36+8=-10}\end{array}$