Paul
Posts:
500
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.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
The OP probably means the average of x ^ 2. Under this interpretation, yours is the best way of seeing it.
Paul

