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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: B1. Find the minimum value of for . 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 , 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 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 , and obey the inequality of . 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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