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July 21st, 2009, 02:30 AM   #1
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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 for .

For a challenge, try not to look at the solutions on the Putnam site or anything similar (unless if you completely give up).
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July 21st, 2009, 02:32 AM   #2
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Re: 1998 Putnam Mathematical Competition

Okay. First thing that comes to my mind is to substitute , simply. Does that help you guys?
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July 21st, 2009, 07:56 AM   #3
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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...
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July 21st, 2009, 02:48 PM   #4
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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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