The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ Problem : 
The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$
Can I use this in numerator and denominator : 
The minimum value of $a +\frac{1}{a}$ 
Using A.M and G.M inequality : 
$a +\frac{1}{a} \geq 2\sqrt{a \times \frac{1}{a}}$ 
$\Rightarrow a +\frac{1}{a} \geq 2$ .....(1) 
By putting the minimum value of (1) in $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ 
we get ; $\frac{2^6-2-2}{2^3+2}$ but I think this is wrong especially denominator as we need to find the maximum value of denominator to get the minimum value. 
Please suggest ,thanks.

hint: let 
$$f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+x^{-3}}=3\left(x+\dfrac{1}{x}\right)\ge 6$$
because
$$\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2=\left(x+\dfrac{1}{x}\right)^6-\left(x^3+\dfrac{1}{x^3}\right)^2$$
so
$$f(x)=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)=3\left(x+\dfrac{1}{x}\right)$$
