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By Thermodude (Thermodude) on Thursday, September 09, 2004 - 06:48 pm: Edit |

I remember making an earlier post for a math problem to be solved algebraically, and I think it went unsolved. So here's one hint to the problem:

a + b + c = 1

Prove that 1/3^a + 1/3^b + 1/3^c >= 3a/3^a + 3b/3^b + 3c/3^c

Hint:

Since 1 = a + b + c....this is the same as

(a + b + c)/3^a + (a + b + c)/3^b + (a + b + c)/3^c >= 3a/3^a + 3b/3^b + 3c/3^c....

subtract the right side from the left to get...

(c -2a + b)/3^a + (a - 2b + c)/3^b + (b + a - 2c)/3^c >= 0

...which becomes...

c/3^a - a/3^a + b/3^a - a/3^a + c/3^b - b/3^b + a/3^b - b/3^b + a/3^c - c/3^c + b/3^c - c/3^c >= 0

....which you can then.....

By Thermodude (Thermodude) on Friday, September 10, 2004 - 01:39 am: Edit |

...perhaps I'll post the next step...I'll wait a little bit more to see if anyone can solve it from here.

By Tongos (Tongos) on Friday, September 10, 2004 - 07:49 pm: Edit |

i believe one can use calculus to figure that one out also. Hey, thermodude, didnt you make the physics olympics? good job! Im hoping on doing the math ones this year.

By Thermodude (Thermodude) on Friday, September 10, 2004 - 11:01 pm: Edit |

Well, I made it as a semifinalist this year for Physics Olympiad, but I didn't get to the training camp. I'm definately going to try out again for it next year...and hopefully I'll get in when I do. Yeah, Tongos, you should definately do the math olympiad next year. Ur excellent at math, and most importantly you love it...which definately makes an excellent combination.

By Thermodude (Thermodude) on Sunday, September 12, 2004 - 09:20 pm: Edit |

k....back to the problem....just group it so that

it becomes...

(c - a)(1/3^a - 1/3^c) + (b - a)(1/3^a - 1/3^b) + (b - c)(1/3^c - 1/3^b) >= 0

Now think....a + b + c = 1

By Thermodude (Thermodude) on Tuesday, September 14, 2004 - 01:18 am: Edit |

Apparently, it looks like interest in this math problem is rather low. Perhaps I should post the answer.

By Thermodude (Thermodude) on Tuesday, September 14, 2004 - 09:40 pm: Edit |

Ah shucks...no wonder no one could solve it....the final step has nothing to do with a + b + c = 1.

Think....what if a > b > c?

(c - a)(1/3^a - 1/3^c) + (b - a)(1/3^a - 1/3^b) + (b - c)(1/3^c - 1/3^b) >= 0

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