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.

Here is one way to go about it. We want to express$\left(x^6+\frac1{x^6}\right)$ in terms of $\left(x+\frac1x\right)$.
$$\left(x+\frac1x\right)^6={x}^{6}+6{x}^{4}+15{x}^{2}+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac1{x^6}$$
We can remove the $x^4$ and $\frac1{x^4}$ terms by subtracting $6\left( x+\frac1x\right)^4$, which gives
$$\left(x+\frac1x\right)^6-6\left( x+\frac1x\right)^4={x}^{6}-9{x}^{2}-16-\frac{9}{x^2}+\frac1{x^6}$$
Continuing this process and using the substitution $u=x+\frac1x$ leads to 
$$x^6+\frac1{x^6}=\left(x+\frac1x\right)^6-6\left(x+\frac1x\right)^4+9\left(x+\frac1x\right)^2-2=u^6-6u^4+9u^2-2$$
and similarly
$$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=u^3-3u$$
Now you have the function
$$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\frac{u^6-(u^6-6u^4+9u^2-2)-2}{u^3+(u^3-3u)}$$ 
$$=\frac{6u^4-9u^2}{2u^3-3u}=3u=3\left(x+\frac1x\right)$$
Since $x+\frac1x\geq2$, the minimum value of $f(x)$ is thus $6$.
