I have a geometric proof if anyone could try and prove. Keep in mind for the following:
1. Sides a,b,c form a triangle.
2. Sides (a+x),(b+y),(c+z) also form a triangle.
3. x+y+z=0.
4. Assume without loss of generality that a>b>c, and x>0.
Thanks!
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Comments
I don't think condition 4 is a "without loss of generality".
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Could you explain that? I don't really understand why not. Thanks.
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Say that a=11,b=10,c=9 and x=−1,y=0,z=1. How do you want to reorder the terms (IE WLOG) such that your condition 4 holds?
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x=1 ? That satisfies the fact the x>0. I'm just trying to say that the greatest's (of sides a,b,c) corresponding x,y,z should be assumed to be positive, which would help with the proof. :D
Maybe we could letLog in to reply
x=1? a,b,c,x,,yz are "constants" that are given, and we cannot simply let them be what we want them to be.
Why can we letHowever, "help with the proof" could be elaborated on better. If we can show that when a,b,c and x,y,,z are similarly ordered, then the RHS decreases, then yes we could make the assumption that a>b>c and x>y>z, which gives us x>0.