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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,735

Supposing a was large and b,c were small? I do not know if that step is rigorous enough or requires more.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

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.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,735

One property is

a + b > c

a + c > b

b + c > a

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

Exactly what I had in mind. I will try to do the problem.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi

But a, b, c all > 0

Similarly

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zetafunc.****Guest**

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****Moderator**- Registered: 2010-06-20
- Posts: 6,395

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

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...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

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.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

It's okay. For I second there I thought we finally had proof!

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,735

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I see now. Thank you.

I suppose the proof would look like this:

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,735

Hi;

Yes, that is what I would do. If it is wrong then at least you have company.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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