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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 24, 2013 9:02 AM

On Wednesday, April 24, 2013 11:50:46 AM UTC+1, Leon Aigret wrote:
> On Wed, 24 Apr 2013 01:04:18 -0700 (PDT), pepstein5@gmail.com wrote:
>
>
>

> >On Wednesday, April 24, 2013 1:00:13 AM UTC+1, Virgil wrote:
>
> >> In article <ken.pledger-C04AB8.09445924042013@news.eternal-september.org>,
>
> >> Ken Pledger <ken.pledger@vuw.ac.nz> wrote:
>
> >>
>
> >> > In article <ee54f5fc-33a5-4c57-8b8e-cfcf32f7f585@googlegroups.com>,
>
> >> > oercim <oercim@yahoo.com> wrote:
>
> >> >
>
> >> > > ....
>
> >> > > x1^2+x2^2+....+xn^2>=ave(x)^2
>
> >> > >
>
> >> > > does this inequality always hold? ....
>
> >>
>
> >> x^2 + y^2 >= ((x+y)/2)^2
>
> >>
>
> >
>
> >It's simpler than anyone else on this thread seems to realise. You don't need to use properties of squaring -- squaring can be replaced by any symmetric function that increases on the positive x axis.
>
> >
>
> >LHS = (abs(x1))^2 + ... (abs(xn))^2 >= (max(abs(xi))) ^ 2 >= (mean(abs(xi))) ^ 2.
>
> > ...
>
>
>
> That looks a bit strange. An other reason why it is simple is because
>
> the inequalty is just a somewhat disguised version of Cauchy-Schwarz
>
> for the 'inner product' of (x1, ... , xn) and (1, ... , 1).
>
>
>
> Leon

The OP probably means the average of x ^ 2. Under this interpretation, yours is the best way of seeing it.

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