# 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, 9 months ago

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

Staff - 2 years, 9 months ago

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

- 2 years, 9 months 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, 9 months 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, 9 months 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, 9 months ago