My Math Forum 1998 Putnam Mathematical Competition

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 July 21st, 2009, 02:30 AM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 1998 Putnam Mathematical Competition Okay, since hardly no one is posting on Math Olympiads section anymore, I will post some interesting ones here more often. Here's an interesting Putnam problem: B-1. Find the minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$ for $x>0$. For a challenge, try not to look at the solutions on the Putnam site or anything similar (unless if you completely give up).
 July 21st, 2009, 02:32 AM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: 1998 Putnam Mathematical Competition Okay. First thing that comes to my mind is to substitute $u= x+\frac{1}{x}$, simply. Does that help you guys?
 July 21st, 2009, 07:56 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: 1998 Putnam Mathematical Competition u^3 = x^3 + 3x + 3/x + 1/x^3, so x^3 + 1/x^3 = u^3 - 3u. u^6 = x^6 + 6x^4 + 15x^2 + 20 + 15/x^2 + 6/x^4 + 1/x^6 u^4 = x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4 u^2 = x^2 + 2 + 1/x^2 u^6 - 6u^4 = x^6 - 9x^2 - 16 - 9/x^2 + 1/x^6 u^6 - 6u^4 + 9u^2 - 2 = x^6 + 1/x^6 So the numerator is u^6 - u^6 + 6u^4 - 9u^2 + 2 - 2 = 6u^4 - 9u^2 and the denominator is u^3 + u^3 - 3u = 2u^3 - 3u so the fraction simplifies to $3\frac{2u^3-3u}{2u-3}$ Now you need to take the derivative wrt x...
 July 21st, 2009, 02:48 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: 1998 Putnam Mathematical Competition Great job on the solution, CRGreathouse! Since now we have the equation, all we need to do is take the derivative wrt $x$, and obey the inequality of $x>0$. Putnam isn't too hard compared to IMO, as long we take a good time to look at the problem and look for different situations and good solutions, and also using past mathematical experiences and knowledge.

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