Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


Paul
Posts:
676
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.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.
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



