<s>[INST] Finding the minimum value of a function. Find the minimum value of the function:
$$f(x) = \frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2 - 2}{\left(x + \frac{1}{x}\right)^3 +\left(x^3 + \frac{1}{x^3}\right)}$$
for $x>0$.
I know that this function simplifies into something a little 'nicer' than what we have above. However, I have hit a brick wall after that. Please refrain from using derivatives to find the minimum value. 
 [/INST] The numerator of $f(x)$ is $$
\begin{align}
\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2=&\Bigg(\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\Bigg)\times
\\&\Bigg(\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\Bigg)\,.
\end{align}$$
Hence, $f(x)=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)=3\left(x+\frac{1}{x}\right)$.  The rest is easy, and the answer is that the minimum value of $f(x)$ is $6$, which is obtained only at $x=1$.
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