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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.

Chatuspathi:

December 27, 2015

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