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- bobbym
- 2012-10-02 04:06:20
Hi;
Yes, that is what I would do. If it is wrong then at least you have company.
- zetafunc.
- 2012-10-02 01:29:59
I see now. Thank you.
I suppose the proof would look like this:
- bobbym
- 2012-10-01 04:27:20
Hi zetafunc.;
zetafunc wrote:
a(b + c) + bc = 1
b(a + c) + ca = 1
c(a + b) + ab = 1
You were on the right track when you posted that.
You need to prove that a,b,c <1.
Let's assume WLOG that a>1 then
By the triangle inequality
If a>1 then (b+c) > 1 and a(b+c) >1 but bc cannot be less than or equal to 0 ( see equation 2 ) so we have a contradiction. Therefore a,b,c<1
Now put your proof all together and present it.
- anonimnystefy
- 2012-09-30 20:48:56
It's okay. For I second there I thought we finally had proof!
- bob bundy
- 2012-09-30 20:24:08
Arhh! Once more you have spotted my error. Curses. (not aimed at you of course!)
My brain did this.
triangle property
b < a + c and b(a+c) < 2 .... => 2b < 2.
But it was wishful thinking.
I should have written b^2 < 2 which is not any help. Sorry zetafunc. Back to the drawing board.
Bob
ps. Nevertheless, some triangle property seems essential here.
- anonimnystefy
- 2012-09-30 20:05:10
bob bundy wrote:
How'd you get 2b<b(a+c)? It would imply that a+c>2, so one of them has to be greater than 1...
- bob bundy
- 2012-09-30 19:59:27
hi zetafunc
Your post with (a+1)(b+1)(c+1) = 4 + (a-1)(b-1)(c-1) is the way to go with this.
But you just needed to justify a,b,c all < 1
I experimented using Sketchpad with a number of values for a b and c and found
it doesn't hold if a b c are not the sides of a triangle and the expression = 4 when any of a b or c = 1. (and is > 4 if over 1)
So the two constraints (triangle and ab + bc + ca = 1) are necessary.
Therefore you have to use a property of triangles.
My contribution uses b < a + c
Bob
- zetafunc.
- 2012-09-30 19:49:38
bob bundy wrote:
hi
But a, b, c all > 0
Similarly
Bob
Thanks for this, I didn't think about setting c(b+a) > 0... so, would all my reasoning be mathematically sound? I agree with what you have written above -- I'm wondering if a geometric solution is also possible however, since it appears that this solution doesn't take advantage of any triangle properties...
- bob bundy
- 2012-09-30 19:35:12
hi
But a, b, c all > 0
Similarly
Bob
- anonimnystefy
- 2012-09-30 09:00:59
Exactly what I had in mind. I will try to do the problem.
- bobbym
- 2012-09-30 08:56:54
One property is
a + b > c
a + c > b
b + c > a
- anonimnystefy
- 2012-09-30 08:45:40
You are forgeting one thing which might be crucial. a, b and c are sides of a triangle. There is a great chance that has some other purpose than just stating that a, b and c are positive.
- bobbym
- 2012-09-30 08:16:47
Supposing a was large and b,c were small? I do not know if that step is rigorous enough or requires more.
- zetafunc.
- 2012-09-30 08:07:09
The equation factorises to these three:
a(b + c) + bc = 1
b(a + c) + ca = 1
c(a + b) + ab = 1
so surely a, b and c must all be smaller than 1?
- bobbym
- 2012-09-30 08:05:15
Because ab + bc + ca = 1, so a, b and c are all smaller than 1
Can you prove that mathematically?