Given a1, a2, ..., an positive real numbers, and defined
pk = ak \prod { i = 1 to n, i k} [(ai)2 - (ak)2]
how to prove that \sum \frac{1}{pi} is positive?
I had an idea of proof, but not sure it would work....
I have the idea written in the attached .png file.
EDIT: See the .png file here.
http://imgur.com/a/RrLRo
Dear Kelvin John, I like this question. Your solution looks nice.
However, I have doubts concerning (1.2). For instance, if cj=1, ci=9, cl=16, we have
ci - cj = 8 > ci - cl = -7, but the absolute values are in a reverse order.
Hint:
Without loss of generality, consider a1< a2
The idea, after expansion of some details, leads to a correct proof correct for even n. How about the odd values on n?
Dear Kelvin John, I like this question. Your solution looks nice.
However, I have doubts concerning (1.2). For instance, if cj=1, ci=9, cl=16, we have
ci - cj = 8 > ci - cl = -7, but the absolute values are in a reverse order.
Hint:
Without loss of generality, consider a1< a2
Thanks Viera answer I've got that the proof for even n is quite sufficient. Thank you Viera for keeping calm while giving interesting answers to interesting questions:)
Best regards, Joachim
PS. Meanwhile I have realized, that replacing ak2 by ck and and assuming increasing ordering (without losses, as Viera has noticed) , we are getting for the sum S of the question the following expression with the use of divided difference of order n-1:
(n-1)! (-1)n-1 S = (n-1)! [ c1, c2, . . . , cn : f(c) ],
which equals the value f(n-1)(b) of the derivative of f of order n-1 at some point b from [c1 , cn ], where in this case f(c) = 1/\sqrt{c}.
For the calculus of divided differences and their reprepresentation see e.g.
https://pdfs.semanticscholar.org/49e1/031ea202bb1a8f37ddcc0b5a7ef51a01152a.pdf
Having this and the sign changes of the derivative of the inverse square root one gets positive value of S, for any choice of positive ck -s .
Further conclusion is the this holds for every negative power of c put in place of f(c), and also for every Laplace transform of a positive measure on R+ since then the derivatives of order n have sign equal (-1)n (cf. the William Feller Bible on probability about the completely monotone functions) JoD
Thank you both for helping with the proof.. I am doing corrections and some improvements of my idea based on your considerations. Realized I can construct a square matrix of order n. Which entries are
xij = cj - ci for all i j and
xij = 1 whenever i=j
This way we obviously obtain a skew-symmetric matrix.
I'm reminding myself of its (skew-symmetric matrix) columns properties what are the product for each pi together with the respective ai. Seeing if this maybe helps. Perhaps not.
Noticing the relation to divided differences was helpful, and I'm investigating about.
Thank you both again!
Good luck, Kelvin; take however into account, that for skweness you need zeros on the main diagonal.
Joachim
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