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I see now. Thank you.
You were on the right track when you posted that.
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.
It's okay. For I second there I thought we finally had proof!
Arhh! Once more you have spotted my error. Curses. (not aimed at you of course!)
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...
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
Exactly what I had in mind. I will try to do the problem.
One property is
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.
Supposing a was large and b,c were small? I do not know if that step is rigorous enough or requires more.
The equation factorises to these three:
Can you prove that mathematically?