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Re: which one is greater than or equal?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg41325] Re: which one is greater than or equal?
  • From: Bill Rowe <listuser at earthlink.net>
  • Date: Wed, 14 May 2003 08:19:18 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 5/13/03 at 4:18 AM, shbui at yahoo.com (Son Bui) wrote:

>can any one tell me which one is greater than? Please give your proof.

>sqrt((a1+a2+...+an)/n)  ?   (sqrt(a1)+sqrt(a2)+...+sqrt(an))/n

In general, this is indeterminate. But if the problem is restricted so that all of the a_n are real positive numbers, it looks like the square root of the mean will be equal to or greater than the mean of the square roots. To see this simplify the problem to consider only two real positive numbers a,b.

Choose a third real number c = a/b

Then

Sqrt[(a +b)/2] = Sqrt[b/2]Sqrt[1+c]  and

(Sqrt[a] + Sqrt[b])/2 = Sqrt[b]/2 (1 + Sqrt[c])

So, the ratio of the two things becomes

(Sqrt[c]+1)/Sqrt[2(c +1)]

The limit of this last is 1/Sqrt[2] has c tends to infinity and equals 1/Sqrt[2] when c = 0.

Plotting this expression shows it is every where less than or equal to one, equaling 1 when c = 1

Going back to the more general case and letting a_n be any real value, then clearly the square root of the mean can be zero. But that means the mean of the square roots is a  non zero complex number. So, it follows for the general case you can get essentially any answer you like by appropriate choice of values for the a_n.


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