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.

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 $x=1$.
• I can't answer that question. There are so many tricks when dealing with inequalities. The only main idea of this problem is to notice that $\left(x^6+\frac{1}{x^6}\right)+2=\left(x^3+\frac{1}{x^3}\right)^2$. – Batominovski Jul 7 '15 at 18:52