Will Factoring solve the problem?

Algebra Level 3

( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) \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)}

If x x is a positive real number , then find the minimum value of the expression above.


The answer is 6.

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Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Chew-Seong Cheong
Sep 17, 2016

This problem has been posted twice:

X = ( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) = ( x + 1 x ) 6 − ( ( x 3 + 1 x 3 ) 2 − 2 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) = ( x + 1 x ) 6 − ( x 3 + 1 x 3 ) 2 ( 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 ) ) ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) = ( x + 1 x ) 3 − ( x 3 + 1 x 3 ) = ( x 3 + 1 x 3 ) + 3 ( x + 1 x ) − ( x 3 + 1 x 3 ) = 3 ( x + 1 x ) By AM-GM inequality x + 1 x ≥ 2 ≥ 3 ( 2 ) = 6 \begin{aligned} 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{{\left(x+\frac{1}{x}\right)}^6-\left(\left(x^3+\frac{1}{x^3}\right)^2 - 2\right)-2}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)} \\ & = \frac{{\left(x+\frac{1}{x}\right)}^6-\left(x^3+\frac{1}{x^3}\right)^2}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)} \\ & = \frac{\cancel{\left(\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right)}\left(\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right)}{\cancel{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}} \\ & = \left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right) \\ & =\cancel{\left(x^3+\frac{1}{x^3}\right)} + 3\left(x+\frac{1}{x}\right) - \cancel{\left(x^3+\frac{1}{x^3}\right)} \\ & = 3\left(\color{#3D99F6}{x+\frac{1}{x}}\right) & \small \color{#3D99F6}{\text{By AM-GM inequality } x + \frac 1x \ge 2} \\ & \ge 3(\color{#3D99F6}{2}) = \boxed{6} \end{aligned}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

What about A.M > G.M inequality. Can't we apply it here.

Archit Tripathi - 4 years, 9 months ago

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Yes, if you can make all terms to be positive.

Chew-Seong Cheong - 4 years, 9 months ago

Oh crap, i forgot if that equation should be "-" i calculated and i forgot to write on "-"

Daniel Sugihantoro - 4 years, 9 months ago

yes you are right !! it seemed like i have seen this somewhere else too.

A Former Brilliant Member - 4 years, 9 months ago

thanks for mentioning same other problems

Razing Thunder - 11 months, 1 week ago

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