Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

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?

Board footer

Powered by FluxBB