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By Thermodude (Thermodude) on Tuesday, July 20, 2004 - 09:57 pm: Edit |

Here is the problem:

Given: a + b + c = 1

Prove: 1/(3^a) + 1/(3^b) + 1/(3^c) >= (3a)/(3^a) + (3b)/(3^b) + (3c)/(3^c)

Now, I know Al0 solved this problem using partial derivaties, but his proof did involve calculus...and the wonderful thing about this problem it is solvable using simple algebra. (I think an Arab mathematician of the 10th century AD would have the mathematical knowledge to solve it) SO.....THE CHALLENGE IS ON!!!!!! Provide a proof using algebra!!!!!!!

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 10:26 pm: Edit |

Since a+b+c=1, 3^(a+b+c)=3. Use this property, multiply both sides by 3^(a+b+c) to get

RTP: 3^(b+c) + 3^(a+c) + 3^(a+b) >= a(3^(b+c+1)) + b(3^(a+c+1)) + c(3^(a+b+1))

When a=b=c=1/3, LHS= RHS

If a=(1/3)+x, b=(1/3)-x and c=1/3, after some basic algebra the RTP reduces to

0 >=? x(3^((5/3) -x)) (1-3^(2x))

(1 - 3^(2x)) will always be < 0, so LHS < RHS for this (a,b,c) combo.

I *think* similar logic will hold to show that if LHS <= RHS for some (a0,b0,c0), it will continue to hold for (a0+x, b0-x, c0) , or for (a0,b0+x, c0-x), or for (a0+x, b0, c0-x). Since we know LHS=RHS for (1/3, 1/3, 1/3), we can always work our way towards an arbitrary (a,b,c) by doing simultaneous increases/decreases on any pair of the triplets, and keeping LHS <= RHS while doing so.

By Thermodude (Thermodude) on Wednesday, July 21, 2004 - 12:37 am: Edit |

Nice solution...but isn't it rather a lengthy one? Also, I am somewhat confused about the keeping one variable constant portion. That constant could be any value, which would then force one to look through an infinite amount of combos. The proof I have in mind for this problem took up only 1/2 a page.

BTW, anyone reading this thread who believes he/she has a proof, you can post it.

By Thermodude (Thermodude) on Wednesday, July 21, 2004 - 02:26 pm: Edit |

C'mon...people...give this problem a shot....can be solved in less than a page of algebra(sorry..optimizerdad...but i think ur proof would require one to check an infinite number of combinations of triplets...so its not really valid....if i'm incorrect in this statement...feel free to refute what i just said)

By Tongos (Tongos) on Wednesday, July 21, 2004 - 03:14 pm: Edit |

i will give it a stab thermodude,

'member me, tongos, from that x^x=5, logarithms rock!

By Ubercollegeman (Ubercollegeman) on Wednesday, July 21, 2004 - 03:23 pm: Edit |

I think AM-GM might be a fairly easy way to solve this, but I'm not going to try it much. I have my own math stuff to worry about .

By Apocalypse_Now (Apocalypse_Now) on Wednesday, July 21, 2004 - 03:42 pm: Edit |

Dear God Man!!! While in the midst of working on this problem, I stumbled upon a much greater and astounding fact that will undoubtedly shock the mathematical community! After about two pages of furious writing and mind-bending algebraic transformations (some of which, in retrospect, even I can barely grasp), I came face to face with the horrifying nature of my discovery. Of course I naturally assumed that I must have done something wrong in my derivations, that I must have erred in some manner. However, after checking my work for the seventh time, the incredible, nay, the miraculous truth of it began to sink in. Indeed, my results will shake our notions of reality to the very core. College Confidential is the quickest way for me to get these results out to the public as soon as possible. So, yes, I will post my findings. Here. Now. On this very message. Hold on to your skivvies, ladies and gentlemen...

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WARNING: DO NOT READ ANY FURTHER IF YOU WISH TO CONTINUE LIVING IN YOUR BLISSFUL, IGNORANT EXISTENCE!!!

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And My Final Results Are...

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0=0

Eureka!!! Eureka!!!

By Korey (Korey) on Wednesday, July 21, 2004 - 04:16 pm: Edit |

Sorry, Apocalypse. Everyone KNOWS that 0=1 ;-)

By Thermodude (Thermodude) on Wednesday, July 21, 2004 - 04:39 pm: Edit |

Hey...sup Tongos!!!...nice to see u on my thread...lol...yeah...this problem's pretty fun to do.... :-D

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