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.
 A: 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
$$f(x)=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)=3\left(x+\dfrac{1}{x}\right)$$
A: 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 $6$.
