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:
]]>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?
]]>