(Moved) Inequality Proof
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D
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What are your thoughts?
If you are stuck at the beginning tell us and we'll start with the definitions.
Please share your work with us ... even if you know it is wrong
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Hint:
(a + b)3 = a3 + 3a2b + 3ab2 + b3
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On what basis have you concluded that this is "easy"? Please reply showing all of your work so far. Thank you!
Steven G
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If you do not see how to do this in one step than try it in cases. If a=b then the result immediately follows. What if a<b? b<a?
In my opinion you should have at least gotten the result when a=b.
I see how to do this with calculus. Can it really be proved with algebra? I did not see how to go with Subhotosh's hint
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D
Deleted member 4993
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I don't either! I went as far as
2(a+b)3 ≥ b(b + 3a)2
I was hoping I would get (a+b) as a factor on the RHS.
The reason I threw that hint in to see his response - which has been nil up to now....
Steven G
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So maybe this should be moved to the calculus section?
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Why? The poster put this in algebra, so presumably the poster should be using algebra. Let's try to help the poster where he is, not where we might like to go, okay? Thank you!
Question for the original poster: You posted as follows:
Is the initial 2 only multiplied against the a^3, or is it on the whole expression? In other words, which of the following is the intended inequality?
. . . . .\(\displaystyle \mbox{i) }\, 2\left(a^3\, +\, b^3\right)\, \geq\, 3 a^2 b\, \mbox{ for all }\, a,\, b\, \geq\, 0\)
. . . . .\(\displaystyle \mbox{ii) }\, \left(2 a^3\right)\, +\, b^3\, \geq\, 3 a^2 b\, \mbox{ for all }\, a,\, b\, \geq\, 0\)
Thank you!
Steven G
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I would never have suggested that the post should be move if it were not for the fact that I felt that the post was already moved for being posted in the wrong place by looking at the title of this post--(Moved) Inequality ProofWhy? The poster put this in algebra, so presumably the poster should be using algebra. Let's try to help the poster where he is, not where we might like to go, okay? Thank you!
On what basis have you concluded that this is "easy"? Please reply showing all of your work so far. Thank you!
Well I used to do this kind of exercises but I can't do it anymore; I would have provided you with some more info or ideas but theres the language barrier, i am not that good in english to expose all my thoughts about Maths
D
Deleted member 4993
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My fault - I moved the post from "Arithmetic".
Unreal?
Are "a" and "b", real numbers or, say, non-negative integers so that the proposition could possibly be proved by induction?
Are "a" and "b", real numbers or, say, non-negative integers so that the proposition could possibly be proved by induction?