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

 Messages: [ Previous | Next ]
 Paul Posts: 780 Registered: 7/12/10
Re: proof of an inequality
Posted: Apr 24, 2013 4:04 AM

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.

Since abs(x + y) <= abs(x) + abs(y), the mean of abs(xi) is >= abs(mean(xi)).

Hence LHS >= (mean(abs(xi))) ^ 2 >= (abs(mean(xi))) ^ 2 = OP's RHS.

Paul Epstein

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