Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Paul

Posts: 407
Registered: 7/12/10
Re: proof of an inequality
Posted: Apr 24, 2013 8:58 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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


No, the Cauchy-Schwarz result is stronger. The Cauchy-Schwarz result gives you a stronger form of the OP's result with the OP's RHS multiplied by n.

Because the OP's inequality is so weak, you can just use properties of max, mean, and symmetry without using properties of squaring -- that's what people don't seem to get.

I don't see a problem with any of my previous posts on this thread -- I don't see what is "strange".

Paul



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.