<s>[INST] Find the minimum value of $t^3+t^2-2t-2$ given that $t$ is greater than or equal to 2 The original question was to find the range of the function f defined by:
$$f(x)=\frac {(1+x+x^2)(1+x^4)}{x^3}$$ for $x>0$
Evidently, differentiating is not very helpful. So I wrote $f(x)$ as:
$$f(x)=t^3+t^2-2t-2=(t^2-2)(t+1)$$
Where $t=x+\frac 1x$
The maximum value clearly approaches $+\infty$.
How do I find the minimum value, where $t >=2$? Here, too differentiating would not help, unless I want to plot a graph, which would again be somewhat cumbersome.
 [/INST] The following is your solution, and is very nice.  Note that our function is equal to 
$$\frac{1+x+x^2}{x}\cdot \frac{1+x^4}{x^2},$$
which is 
$$\left(x+1+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right).$$
Make the substitution $x+\frac{1}{x}=t$. Our function is equal to 
$$(t+1)(t^2-2).\tag{1}$$
Since $t\ge 2$, the function (1) is increasing, so has a minimum value of $6$ at $x=1$.  
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