# Minimal value problem (RMO 2015 Chennai Solution)

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.