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# Geometric/Algebraic Inequality

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!

Note by Pickle Lamborghini
2 years, 1 month ago

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I don't think condition 4 is a "without loss of generality".

Staff - 2 years, 1 month ago

Could you explain that? I don't really understand why not. Thanks.

- 2 years, 1 month ago

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?

Staff - 2 years, 1 month ago

Maybe we could let $$x = 1$$ ? 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

- 2 years, 1 month ago

Why can we let $$x = 1$$? $$a, b, c, x, ,y z$$ are "constants" that are given, and we cannot simply let them be what we want them to be.

However, "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 $$x > 0$$.

Staff - 2 years, 1 month ago