Range of function $ f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ Where $x>0$

Find the range of the function $\displaystyle f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ where $x>0$

$\bf{My\; Try::}$ Let $\sqrt{x}=t\;,$ Then $\displaystyle f(t) = t^3+\frac{1}{t^3}-4\left(t^2+\frac{1}{t^2}\right)\;,$

Now After Simplification, We get $\displaystyle f(t) = \left(t+\frac{1}{t}\right)^3-3\left(t+\frac{1}{t}\right)-4\left[\left(t+\frac{1}{t}\right)^2-2\right]$

Now Put $\displaystyle t+\frac{1}{t} = u\;,$ Then $\displaystyle \sqrt{x}+\frac{1}{\sqrt{x}} = u\;,$ So we get $u\geq 2$ (Using $\bf{A.M\geq G.M}$)

And our function convert into $\displaystyle f(u) = u^3-4u^2-3u+8\;,$ Where $u\geq 2$

Now Using Second Derivative Test, $f'(u) = 3u^2-8u-3$ and $f''(u) = 6u-8$

So for Max. and Min., We put $\displaystyle f'(u)=0\Rightarrow u=3$ and $f''(3)=10>0$

So $u=3$ is a point of Minimum.

So $f(2)=8-4(4)-3(2)+8 = -6$ and $f(3) = -10$

and Graph is Like this

So Range is $$\displaystyle \left[-10,\infty \right)$$

My question is can we solve it any other way, Like using Inequality

If yes, Then plz explain here

Thanks

We have $$\color{blue}{x\sqrt{x} + \dfrac1{x\sqrt{x}}-4\left(x+\dfrac1x\right) = \underbrace{\dfrac{\left(1+\sqrt{x}\right)^2\left(x-3\sqrt{x}+1\right)^2}{x^{3/2}}}_{\text{Is non-negative}}-10}$$ Hence, the minimum is $-10$