From - Sun Aug 6 01:13:39 2000 Path: rQdQ!sn-xit-01!supernews.com!newsfeed.stanford.edu!news-spur1.maxwell.syr.edu!news.maxwell.syr.edu!newsfeed.icl.net!colt.net!newsfeeds.belnet.be!news.belnet.be!skynet.be!newsfeed2.news.nl.uu.net!newsfeed1.news.nl.uu.net!sun4nl!surfnet.nl!surfnet.nl!cwi.nl!not-for-mail From: "Peter L. Montgomery" Newsgroups: sci.math Subject: Re: IMO problem 2 Message-ID: Sender: pmontgom@cwi.nl NNTP-Posting-Host: bark.cwi.nl Organization: CWI, Amsterdam References: Date: Sat, 5 Aug 2000 04:31:12 GMT Lines: 43 Xref: rQdQ sci.math:395151 In article pharvey@derwent.co.uk (Paul Harvey) writes: >Does there exist an elegant solution to IMO #2, that is > >Let abc = 1, a,b,c are +ve reals > >(a-1+1/b)(b-1+1/c)(c-1+1/a)<=1 > >Only by expanding fully, and arguing several different cases (a>1, >b<1, c<1 etc. ) can I see a solution. > Since a*b*c = 1, there exist positive real numbers x1, x2, x3 such that a = x1/x2 b = x2/x3 c = x3/x1 We must prove (*) (x1 - x2 + x3) * (x2 - x3 + x1) * (x3 - x1 + x2) <= x1 * x2 * x3 At most one of the parenthesized quantities on the left of (*) can be negative. If, for example, both x1 - x2 + x3 and x2 - x3 + x1 are negative, then their sum 2*x1 would be negative, a contradiction. If exactly one of these is negative, then the left side of (*) is <= 0, and the result follows. Otherwise we must have (x1 + x2 - x3) * (x1 - x2)^2 + (x2 + x3 - x1) * (x2 - x3)^2 + (x3 + x1 - x2) * (x3 - x1)^2 >= 0. Expand this and divide by 2. The result is equivalent to (*). I suspect this proof (found with the assistance of Maple) can be simplified. -- E = m c^2. Einstein = Man of the Century. Why the squaring? Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI