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Problem Find the minimum value of $\displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( x^6 + \frac{1}{x^6}) - 2}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$ and $x \in \mathbf{R}$ and $x > 0$

Discussion:

$\displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( x^6 + \frac{1}{x^6}) - 2}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( (x^3)^2 + (\frac{1}{x^3})^2) - 2}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( (x^3)^2 + (\frac{1}{x^3})^2 + 2)}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( (x^3)^2 + (\frac{1}{x^3})^2 + 2 \times (x^3) \times \frac{1}{x^3})}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { \frac{ {( x + \frac{1}{x} )^3}^2 - (x^3 + \frac{1}{x^3})^2}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { \frac{ {( x + \frac{1}{x} )^3 + (x^3 + \frac{1}{x^3})} {( x + \frac{1}{x} )^3 - (x^3 + \frac{1}{x^3})}}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$

$= \displaystyle { ( x + \frac{1}{x} )^3 - (x^3 + \frac{1}{x^3})}$

$= \displaystyle { 3( x + \frac{1}{x} )}$

Applying Arithmetic Mean – Geometric Mean Inequality (since x is positive), we have:

$\displaystyle { \frac {( x + \frac{1}{x} )}{2} \ge \sqrt {x\times \frac{1}{x}} = 1 }$

$\displaystyle { \Rightarrow ( x + \frac{1}{x} ) \ge 2 }$

$\displaystyle { \Rightarrow 3( x + \frac{1}{x} ) \ge 3\times 2 = 6}$

Hence the minimum value is 6.