powers fun

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 + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})}

Find the minimum value of the expression above such that x x is positive.


The answer is 6.

View wiki
Alex Wang by Alex Wang , 25, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Alex Wang
Apr 15, 2015

f ( x ) = ( x + 1 x ) 6 − ( x 6 + 1 x 6 ) − 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) f(x) = \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} = ( x + 1 x ) 6 − ( x 6 + 2 + 1 x 6 ) ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) = \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + 2 + \frac{1}{x^6})}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} = ( x + 1 x ) 6 − ( x 3 + 1 x 3 ) 2 ( x + 1 x ) 3 + ( x 3 + 1 x 3 ) = \frac{\left( x + \frac{1}{x} \right)^6 - (x^3 + \frac{1}{x^3})^2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} = ( x + 1 x ) 3 − ( x 3 + 1 x 3 ) = ( x + \frac{1}{x} )^3 - (x^3 + \frac{1}{x^3}) = 3 ( x + 1 x ) = 3 (x + \frac{1}{x})

By the AM-GM inequality, x + 1 x ≥ 2 x + \frac{1}{x} \ge 2 .

Therefore, our answer is 3 ∗ 2 = 6 3 * 2 = \boxed{6}

4 Helpful 0 Interesting 1 Brilliant 0 Confused
Utkarsh Singh
Apr 15, 2015

Used AM-GM inequality. Is that correct in this case?

2 Helpful 0 Interesting 0 Brilliant 0 Confused

yes, my solution does use the AM-GM inequality.

Alex Wang - 6 years, 1 month ago

Log in to reply

I used it without actually simplifying the expression. This way by using AM-GM inequality, I found the minimum value of both numerator and the denominator, which was actually incorrect because to find the minimum of f(x) I should be considering the maximum value of the denominator. Hence the doubt.

Utkarsh Singh - 6 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...