Paul
Posts:
769
Registered:
7/12/10


Re: proof of an inequality
Posted:
Apr 24, 2013 8:58 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.pledgerC04AB8.09445924042013@news.eternalseptember.org>, > > >> Ken Pledger <ken.pledger@vuw.ac.nz> wrote: > > >> > > >> > In article <ee54f5fc33a54c578b8ecfcf32f7f585@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 CauchySchwarz > > for the 'inner product' of (x1, ... , xn) and (1, ... , 1). > > > > Leon
No, the CauchySchwarz result is stronger. The CauchySchwarz 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

