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

$f\left(x\right)=x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)$

$\Rightarrow f\left(x\right)={\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^3-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)-4\left\{{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2-2\right\}$

$\Rightarrow f\left(x\right)={\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^3-4{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+8$

Clearly, $f\left(x\right)$ is defined for all $x>0$, 

Let$\sqrt{x}+\frac{1}{\sqrt{x}}=t$, Then, $t\ge 2$ for all $x>0.$

Also, let$g\left(t\right)=t^3-4t^2-3t+8$, Then,

$g^{\mathrm{\prime }}\left(t\right)=3t^2-8t-3$ and $g^{\prime \prime }\left(t\right)=6t-8$

$\Rightarrow g^{\mathrm{\prime }}\left(t\right)=(3t+1)(t-3)$ and $g^{\prime \prime }\left(t\right)=6t-8$

Clearly, $g^{\mathrm{\prime }}\left(t\right)=0$ for $t=3$ and $g^{\prime \prime }\left(t\right)=10>0$

Thus, g is minimum when$t=3$ and the minimum value of g is$g\left(3\right)=-10$.