It's really a matter of simplification

Algebra Level 3

( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) \large \frac{\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)}

Find the minimum value of the expression above for all x > 0 x>0 .

Note: This particular problem appeared in the 1998 Putnam Math Competition.


The answer is 6.

Rishav Koirala by Rishav Koirala , 22, Singapore

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1 solution

Sabhrant Sachan
Sep 14, 2016

The question is already posted here

f ( x ) = ( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) ⟹ f ( x ) = ( x + 1 x ) 6 − ( ( x 3 ) 2 + 1 ( x 3 ) 2 + 2 ⋅ x 3 ⋅ 1 x 3 ) ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) f ( x ) = ( ( x + 1 x ) 3 ) 2 − ( x 3 + 1 x 3 ) 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) f ( x ) = ( ( x + 1 x ) 3 − ( x 3 + 1 x 3 ) ) ( ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) ) ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) f ( x ) = 3 ( x + 1 x ) Using A.M - G.M x + 1 x ≥ 2 f ( x ) Min. = 6 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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