| Marcos |
If a + b + c = 1 and a,b,c are positive reals, how do I show that: a3 + b3 + c3 + 3abc > = 2/9 Marcos P.S. Hints only please! At least to begin with... |
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Michael
Doré
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Well the first thing I'd try (although it will result in a bit of algebra) is to write the right hand side as 2/9 * (a+b+c)3 . Then expand, simplify and you're left needed to prove a homogeneous inequality, which hopefully shouldn't be too hard. The nice thing is that once you've reduced it to showing: a3 +b3 +c3 + 3abc > = 2/9 * (a+b+c)3 you _know_ that you will never need to use the condition a+b+c = 1 again. Can you see why? |
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| Marcos |
Hi Michael, I'm feeling stupid (and rather tired after attempting two BMO papers!), but I can't seem to get much out of the homogeneous inequality. Is there some standard strategy to solving them? Marcos P.S. And no, I'm not 100% sure why I won't need to use the condition again. (although I think that since we're introducing the relation as a cube - which is the highest power that was in the inequality originally - it won't be necessary to do so again) |
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| Marcos |
I managed to do it but following a different approach: By doing all sorts of nice (!) factorisations and simplifications (which I won't do again here) I showed that 4[a3 + b3 + c3 + 3abc] - 1 = 3(2a - 1)(2b - 1)(2c - 1) By AM-GM, the RHS > = -1/9 and the result follows. In any case, I'd be really grateful if you could guide me through your method Michael. Marcos |
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| Michael Doré |
Sorry Marcos, I made a mistake. Your proof looks great! |
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| David
Loeffler |
There is a fairly standard approach to symmetric polynomial inequalities. Write [r,s,t] for the sum of the 6 possible permutations xr ys zt + xs yr zt + ... Then our inequality 9(a3 + b3 + c3 ) + 27abc > = 2(a + b + c)3 reduces to 9/2 [3,0,0] + 9/2 [1,1,1] > = 2( 1/2 [3,0,0] + 3[2,1,0] + [1,1,1]) or 7/2 [3,0,0] + 5/2[1,1,1] > = 6[2,1,0] Now it's clear that [3,0,0] > = [1,1,1] so it suffices to establish that [3,0,0] + [1,1,1] > = 2 [2,1,0] But this is precisely x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) > = 0 which is a case of Schur's inequality, easily established by taking WLOG x > = y > = z. David |
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| Marcos |
Thanks Michael although my proof is extremely messy in comparison to the elegant, (almost) 3-line proof provided by David Thanks David, I think I better keep that technique in mind (particularly the neat notation) now that olympiad selection is coming up. (at least here in Cyprus) Marcos |
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| James |
David, would you mind explaining a little further, im a bit lost on how you get those numbers and then how to manipulate them. James |
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| Marcos |
James, if it helps these are what they mean in full: [3,0,0] = 2a3 + 2b3 + 2c3 [1,1,1] = 6abc [2,1,0] = a2 b + a2 c + ab2 + ac2 + bc2 + b2 c Try seeing where they come from, based on David's definition of [r,s,t] now. Marcos |
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| David
Loeffler |
Essentially all you do is expand out the brackets, and collect like terms together; then you rewrite each symmetric set of terms in the [r,s,t] notation. The notation isn't actually telling you anything you didn't already know; it just allows you to organise what would otherwise be breathtakingly messy algebra with relatively little pain. (Marcos: You can dispatch any homogenous symmetric polynomial inequality quite quickly using this notation, Muirhead's theorem (see here ) and Schur's inequality - they're definitely worth knowing.) David |
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| Marcos |
Muirhead's theorem is brilliant! You can even prove AM-GM really quickly using it! It'll probably take a few more examples though before I'm comfortable using it... Do you think it's okay to use inequalities such as these in olympiad problems without proof? (As I'm guessing the proof is probably quite hard) Marcos |
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| David
Loeffler |
The proof of Muirhead's theorem relies on using AM-GM, so don't get too excited... You can probably get away with using these in olympiads, but don't quote me on that. David |
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| Kerwin
Hui |
AFAIK, the official policy is that you can quote (and use) any theorem provided they are correctly stated. Kerwin |
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| Marcos |
David, oh well. In any case, how do you prove Muirhead's theorem? Marcos P.S. If you can quote any theorem I'm surprised nobody's been sneaky and quoted some theorem that doesn't exist (but which if true solves a problem) and then proved it after the olympiad and published it. I'm guessing it could be quite hard to find out whether the theorem was published after the competition or not, if the person is clever enough. |
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| David
Loeffler |
But Olympiad papers are marked on the spot, so it wouldn't help you. I do know of a case where a contestant, absolutely stuck, wrote down '... and the result follows by Muller's theorem', leaving his country's team leader frantically running around trying to find out what Muller's theorem actually was until he got the contestant to admit that it was fictitious! This has become something of an in-joke in UK IMO circles (see here , for example). The proof of Muirhead's theorem is quite involved; essentially you show that you can take combinations of terms from [r,s,t, ...] and AM-GM them to get each term of [r',s',t',...] in such a way that when you add this over the permutations of [r',s', t'...] you have used each term from [r,s,t,...] exactly once in total. The details are in Hardy, Littlewood and Polya's monograph 'Inequalities' (CUP); unfortunately I've left my copy thereof at home this term, so I can't be any more explicit than that. David |