What is the minimum value of a complex expression?

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The minimum value of the complex expression \[\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}\] for \(x>0\) was discussed in the context of Problem B-1 from the 1998 William Lowell Putnam Mathematical Competition. The solution provided by Opalg demonstrates the application of algebraic manipulation and properties of symmetric functions. The analysis confirms that the expression simplifies to a minimum value of 1.

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Here is this week's POTW:

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Find the minimum value of
\[\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}\]
for $x>0$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 244 - Dec 12, 2016

This was Problem B-1 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows:

In the denominator, $(x+1/x)^3+(x^3+1/x^3) = 2x^3 + 3x + 3/x + 2/x^3 = 2(x+1/x)^3 - 3(x+1/x).$

In the numerator, $(x+1/x)^6-(x^6+1/x^6)-2 = 6x^4 + 15x^2 + 18 + 15/x^2 + 6/x^4 = 6(x+1/x)^4 - 9(x+1/x)^2.$

Put those together and let $y= x+1/x$, getting $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)} = \frac{6y^4 - 9y^2}{2y^3 - 3y} = \frac{3y(2y^3 - 3y)}{2y^3 - 3y} = 3y = 3(x+1/x).$$ The minimum value of $x+1/x$ is $2$, at $x=1$. Therefore the minimum value of $\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$ is $6$.
 

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