Math Problem!!!!

Discus: College Confidential Café: 2004 Archive: Math Problem!!!!
 By Thermodude (Thermodude) on Wednesday, July 14, 2004 - 12:45 am: Edit

Since it appears that many of us on CC enjoy doing math...i have a fun problem here:

Given that a + b + c = 1......

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

I would write the problem on a sheet of paper so it looks cleaning and it will be easier to solve.

 By Noodleman (Noodleman) on Wednesday, July 14, 2004 - 12:32 pm: Edit

Nice try, homework troller.

 By Lame (Lame) on Wednesday, July 14, 2004 - 02:27 pm: Edit

Wow, how easy.

 By Thermodude (Thermodude) on Wednesday, July 14, 2004 - 05:24 pm: Edit

What's a homework troller?....If it is what I think it is, I really just posting this problem for fun. I'm not currently taking any math classes right now, as I just finished BC Calc (got a 5) during the school year. I just thought it would be fun to post a fun math problem. Anyways...if you've got a proof...just post it here.

 By Ubercollegeman (Ubercollegeman) on Wednesday, July 14, 2004 - 07:36 pm: Edit

First of all, this problem is far too difficult and not-oriented to any normal math high school subject to be homework. Inequalities are tough.

Offhand, the inequality looks like it should be >= instead of >, since equality clearly holds at a=b=c=1/3.

 By Ubercollegeman (Ubercollegeman) on Wednesday, July 14, 2004 - 08:06 pm: Edit

Result that I got unintentionally:

Prove that 3^a + 3^b + 3^c has a minimum value of 3^(4/3).

I seem to be getting lost in circles, but I'll give it some more thought.

 By Socalnick (Socalnick) on Thursday, July 15, 2004 - 02:01 am: Edit

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

i think that it should be 1/(3^a) + 1/(3^b) + 1/(3^c) = 3a/(3^a) + 3b/(3^b) + 3c/(3^c)

i may have done some bad math from my first step, i really dont kno its summer

 By Ubercollegeman (Ubercollegeman) on Thursday, July 15, 2004 - 02:32 am: Edit

Socalnick, please don't major in math.

As for me, I'm tired, but I'll give it another shot tomorrow.

 By Lame (Lame) on Thursday, July 15, 2004 - 03:51 am: Edit

Alright, this will be more written than anything but here you go.
So we reduce the original down to this
1/(3^a) + 1/(3^b) + 1/(3^c) > a/(3^(a-1)) + b/(3^(b-1)) + c/(3^(c-1))
k?
Now, since the a-1, b-1, and c-1 will result in a negative number if we use a, b, and c values between 0 and 1 but excluding 1, those things on the right will have their denominators moved to the top. So
1/(3^a) + 1/(3^b) + 1/(3^c) > a3^(1-a) + b3^(1-b) + c3^(1-c)
Now remember the assumption about a b and c? Alright so take for example (of the left side) 1/3^a Now, since the 3^a is on the bottom, we are dividing 1 by it, and since a < 1, this is going to decrease the size of the 3 cause we're doing like a root operation of sorts. Therefore, the 1 is divided by a less "powerful" number than the 3 would have been. All the left side will experience this. The right side, however, will not get such a bonus, even though it is being multiplied as the root operation works against it instead of in favor of it. (Just try using an 'a' value as described). It's easiest to see it working with these fractional values, but it will work for any 3 values which sum to 1, unless a=b=c=1/3, in which case as Ubercollegeman said they will be = to eachother.

 By Apocalypse_Now (Apocalypse_Now) on Thursday, July 15, 2004 - 11:24 am: Edit

Interesting idea Lame, but i don't think that proves it. I like your approach though...

 By Ubercollegeman (Ubercollegeman) on Thursday, July 15, 2004 - 12:50 pm: Edit

Well, Lame, I don't think your method works for two reasons.

1) The problem does not state that a<1, b<1, or c<1. You're assuming that the three are nonnegative.

2) Using your terminology, the right side will not get the bonus, but given your assumption that a<1, b<1, and c<1, the right side will get a different "bonus."

 By Deferreddude (Deferreddude) on Thursday, July 15, 2004 - 02:04 pm: Edit

If 1 = a + b + c, then 3 = 3a + 3b + 3c. Now some weird stuff will happen. Theoretically, the quantity "3a" on the right side of the equation can be written as "a(3a + 3b + 3c)" which will be equal to "3a^2 + 3ab + 3ac." Now that new expression can be written as (3a + 3b + 3c)(a^2) + (3a + 3b + 3c)(3ab) + (3a +3b + 3c)(3ac). This can go on forever and ever and the resulting INFINITELY complex expression will be equal to little old "3a." Reminds me of a fractal.

Yeah I didn't solve anything at all. I'm just bored.

 By Lame (Lame) on Thursday, July 15, 2004 - 08:29 pm: Edit

Actually, the non-negative thing does not matter. If you look at it from a logic stand point, rather than attempting to do it out on paper, you'll realize that. The size of the numbers (excluding possibility they are all 1/3) will have no effect on what I stated with my reduction. Let's use a whole number but larger than 1 or <0 example for our 3 new numbers (for the ease of it). Say a=5, b=-1, and c=-3. So, automatically, 2 of the numbers on the right will be smaller. Here's how...so the left side, take 1/3^b, since b is negative, will become 1*3^|b|, so 3 to in this case the first. So second value on the left is now = to 3. 1/3^c becomes 1*3^|c| which is 3^3, so 27. 1/3^a will not have a reversal of num/denom with the 3^a part since a is now positive, but it will get the "bonus" similar to what I was talking about because it will be divided by a way small number (in this case 3^5 whatever that is). The right hand side, however, does not get such luck. Right off the bat, 2 of the numbers become negative. The other one, a3^(1-a), will result in 5*3^(1-5), which is 5/3^4, which is a rather small fraction. If you demand it, I'll do this out again with 2 positive numbers and 1 big negative number. Really don't see why the logic of that isn't pretty obvious...

 By Thermodude (Thermodude) on Thursday, July 15, 2004 - 09:23 pm: Edit

Lame...I could make exactly the same argument with the initial inequality posted...really..all you are doing is plugging in numbers into an inequality which is already supposed to be true. It would be similar to just plugging in some Pythagorean tripples into a re-arranged form Pythagorean Theorem...and stating that doing so would be a proof when it really would not be. Still, I did like your approach to the problem made in an earlier post. :-)

 By Ubercollegeman (Ubercollegeman) on Thursday, July 15, 2004 - 10:24 pm: Edit

Thermo, does the solution use a relatively obscure inequality (not AM-GM or Cauchy-Schwarz)?

 By Lame (Lame) on Thursday, July 15, 2004 - 10:40 pm: Edit

Considering the OP hasn't even solved it ...

 By Thermodude (Thermodude) on Sunday, July 18, 2004 - 02:14 am: Edit

actually...the proof to this problem can be completed with pre-calc/pre-newtonian mathematics...if i'm correct