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

]]>I suppose the proof would look like this:

]]>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.

]]>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.

]]>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...

]]>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

]]>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...

]]>But a, b, c all > 0

Similarly

Bob

]]>a + b > c

a + c > b

b + c > a

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?

]]>Because ab + bc + ca = 1, so a, b and c are all smaller than 1

Can you prove that mathematically?

]]>