Minimize it!

$f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left( x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+\left( x^3+\dfrac{1}{x^3}\right)} \implies f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left( (x^3)^2+\dfrac{1}{(x^3)^2}+2\cdot x^3\cdot\dfrac{1}{x^3}\right)}{\left(x+\dfrac{1}{x}\right)^3+\left( x^3+\dfrac{1}{x^3}\right)} \\ f(x)=\dfrac{\left( \left( x+\dfrac{1}{x} \right)^3 \right)^2-\left( x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left( x^3+\dfrac{1}{x^3}\right)} \\ f(x)=\dfrac{\left( \left(x+\dfrac{1}{x}\right)^3-\left( x^3+\dfrac{1}{x^3}\right) \right) \left( \cancel{\left(x+\dfrac{1}{x}\right)^3+\left( x^3+\dfrac{1}{x^3}\right)} \right)}{\cancel{\left(x+\dfrac{1}{x}\right)^3+\left( x^3+\dfrac{1}{x^3}\right)}} \\ f(x)=3\left( x+\dfrac{1}{x} \right) \\ \text{Using A.M - G.M} \quad x+\dfrac{1}{x} \ge 2 \\ \boxed{f(x)_{\text{Min.}} = 6}$

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$\text{Please mention that } x>0$

Let $y = x + \dfrac 1x$ . Then we have:

$\begin{aligned} y^3 & = \left(x + \dfrac 1x \right)^3 = x^3 + 3x + \frac 3x + \frac 1{x^3} = x^3 + \frac 1{x^3} + 3y \end{aligned}$

$\begin{aligned} \implies x^3 + \frac 1{x^3} & = y^3-3y \\ \left(x^3 + \dfrac 1{x^3} \right)^2 & = (y^3-3y)^2 \\ x^6 + 2 + \dfrac 1{x^6} & = y^6 - 6y^4 + 9y^2 \\ \implies x^6 + \dfrac 1{x^6} & = y^6 - 6y^4 + 9y^2 - 2 \end{aligned}$

Now we have:

$\begin{aligned} f(x) & = \frac { {\left( x + \frac{1}{x} \right)}^{6} - \left( x^6 + \frac{1}{x^6} \right) - 2}{ {\left( x + \frac{1}{x} \right)}^{3} + \left( x^3 + \frac{1}{x^3} \right)} \\ & = \frac {y^6 - y^6 + 6y^4 - 9y^2 + 2-2}{y^3 + y^3 - 3y} \\ & = \frac {6y^4 - 9y^2}{2y^3 - 3y} \\ & = 3y \\ & = 3 \left(\color{#3D99F6}{x + \frac 1x} \right) & \small \color{#3D99F6}{\text{By AM-GM inequality}} \\ & \ge 3 \left(\color{#3D99F6}{2} \right) = \boxed{6} \end{aligned}$

simple any number plus its inverse is always greater than or equal to 2, x+1/x>=2