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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,182

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.**

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,182

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.**

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi

But a, b, c all > 0

Similarly

Bob

Children are not defined by school ...........The Fonz

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: 7,273

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

Children are not defined by school ...........The Fonz

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: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

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.

Children are not defined by school ...........The Fonz

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: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

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

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,182

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.**

**If it ain't broke, fix it until it is.**

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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: 98,182

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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