| 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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