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Topic: proof of an inequality
Replies: 12   Last Post: Apr 24, 2013 9:02 AM

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 Paul Posts: 780 Registered: 7/12/10
Re: proof of an inequality
Posted: Apr 23, 2013 3:14 PM

On Tuesday, April 23, 2013 5:11:14 PM UTC+1, oercim wrote:
> Okey, I guess I got it. Let n=2,
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> x1^2+x2^2>=((x1+x2)^2)/4
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> ==>4(x1^2+x2^2)>=(x1+x2)^2
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> ==>3(x1^2+x2^2)>=2x1x2
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> if x1=x2, it is obvious it holds,
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> if x1 is not equal x2, then one is greater than other . Let say x1>x2
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> Then x1^2>x1x2. I guess it is easy to apply n variable case. Thanks alot.

I'm not sure whether you're getting there or not.
I'll give two further hints for a cleaner solution. 1) Replace x^2 by a simpler function. 2) Think of the relationship between max and mean.

Paul

Date Subject Author
4/23/13 oercim@yahoo.com
4/23/13 Paul
4/23/13 oercim@yahoo.com
4/23/13 Paul
4/23/13 oercim@yahoo.com
4/23/13 oercim@yahoo.com
4/23/13 Paul
4/23/13 Ken.Pledger@vuw.ac.nz
4/23/13 Virgil
4/24/13 Paul
4/24/13 Leon Aigret
4/24/13 Paul
4/24/13 Paul