I discovered an inequality.. reverse schur

by pythag011, Dec 15, 2008, 11:39 PM

Instead of schur, we have

Prove that if a,b,c are positive reals, then $ b(a - b)(a - c) + c(b - c)(b - a) + a(c - a)(c - b) \le 0.$

is \operatorname working? $ \operatorname{SPAM} x.$ No it isn't. Oh well.

Is reverse schur equivalent to schur? Is it strong or weak?

Hmm...

Should I try to bash all geometry problems using complex numbers?

Oh well...

Can this be generalized? (Reverse schur).

Interesting...

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2 Comments

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post some of the geometry problems you are doing?

hmm

your ineqality doesn't seem to be very happy for $ a = b$...

like if $ a = b = 1$ and $ c = 2$ then it becomes $ 1\le 0$ which is wrong

HOPEFULLY I AM NOT MISREADING STUFF

by The QuattoMaster 6000, Dec 16, 2008, 12:10 AM

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I can't remember the orginial inequalities :(.

by pythag011, Dec 16, 2008, 2:53 AM

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