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!

## Comments

Sort by:

TopNewestI don't think condition 4 is a "without loss of generality". – Calvin Lin Staff · 1 year, 8 months ago

Log in to reply

– Pickle Lamborghini · 1 year, 8 months ago

Could you explain that? I don't really understand why not. Thanks.Log in to reply

– Calvin Lin Staff · 1 year, 8 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?Log in to reply

– Pickle Lamborghini · 1 year, 8 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. :DLog in to reply

However, "help with the proof" could be elaborated on better.

If we can showthat 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 \( x > 0 \). – Calvin Lin Staff · 1 year, 8 months agoLog in to reply