From: Pierre Bornsztein Subject: Re: A difficult inequality Date: Mon, 30 Aug 1999 12:22:21 +0200 Newsgroups: sci.math Keywords: Shapiro's Inequality (for small dimensions) > This is neat and eliminates -- well, not really :-) -- the need for my next > post, which would ask for a generalization in a different direction, namely > > a_1/(a_2+a_3) + a_2/(a_3+a_4) + ... + a_n/(a_1+a_2) >= n/2 > This is a well-known inequality. it is known as Shapiro's Inequality (SI). Unfortunately, SI is true for even n =< 12 and odd n =< 23, and is false otherwise. Denote by f(n) = inf{ a_1/(a_2+a_3) + a_2/(a_3+a_4) + ... + a_n/(a_1+a_2) } where a_1,..a_n are nonnegative. If m = lim f(n)/n as n-> infinity, then it has been proved that : m = 0,4945668... (by Drinfeljd). References : The Finch site about mathematical constants. D.S Mitrinovic, J.E. Pecaric, A.M.Fink : Classicals and new inequalities in Analysis. (Kluwer 1993) Pierre. ============================================================================== From: baloglou@panix.com (George Baloglou) Subject: Re: A difficult inequality Date: 31 Aug 1999 01:25:29 -0400 Newsgroups: sci.math [reply address hidden in signature/URL -- baloglou@panix.com accepts no mail] In article <7qdiru$9lh$1@panix3.panix.com> I wrote: > >This is neat and eliminates -- well, not really :-) -- the need for my next >post, which would ask for a generalization in a different direction, namely > > a_1/(a_2+a_3) + a_2/(a_3+a_4) + ... + a_n/(a_1+a_2) >= n/2 > >Let me illustrate for other readers how your method works for n = 5: Well, Prof. Israel's method does indeed work for n = 5, and the solution offered below is certainly correct, but ... that's about it: I do not think that it can be ("easily") extended to n = 6 or higher -- nor did Bob made such a false claim ... I just made it in a careless, intoxicating moment :-) I indicate the difficulties of extending this approach beyond n = 5 further below. >Setting S = a+b+c+d+e and L = a/(b+c) + b/(c+d) + c/(d+e) + d/(e+a) + e/(a+b), > >L = S*[a/(S*(b+c)) + b/(S*(c+d)) + c/(S*(d+e)) + d/(S*(e+a)) + e/(S*(a+b))], > >so that Jensen's inequality makes the right side greater than or equal to > > 1 >S * ------------------------------------------------------------------- = > (a/S)*(b+c) + (b/S)*(c+d) + (c/S)*(d+e) + (d/S)*(e+a) + (e/S)*(a+b) > > 1 S^2 >= (S^2) * ----------------------------------------------- = ---, > a*(b+c) + b*(c+d) + c*(d+e) + d*(e+a) + e*(a+b) R > >where R = a*(b+c) + b*(c+d) + c*(d+e) + d*(e+a) + e*(a+b). > > S^2 >Now it is not that hard to establish 2*(S^2) >= 5*R, hence L >= --- >= 5/2. > R >[For 2*(S^2) >= 5*R, notice that 2*(S^2) - 5*R can be written, >following Prof. Israel's approach always, as >(1/2)*((a-b)^2 + (a-c)^2 + (a-d)^2 + (a-e)^2 + ... + (d-e)^2)] Unfortunately, 2*(S^2) - n*R, where S and R are appropriately defined, can no longer be written as a sum of squares as above for n = 6 (or higher); nor have my other attempts to make it positive -- by assuming (WLOG) a_1 <= ... <= a_6 for example -- been successful. (But the method does work for n = 3 and n = 4, too, I should mention, and along very similar lines.) As Pierre Bornsztein revealed in his post, this problem has some depth ... Following his leads, I got into Steven Finch's page where more references are given (http://www.mathsoft.com/asolve/constant/shapiro/shapiro.html), tracing everything to a 1954 Monthly problem submitted by H. S. Shapiro of New York University; a non-solution appeared in the March 1956 issue ("An Invalid Inequality", pp. 191-192), mentioning the existence of proofs for n = 3 and n = 4 (by the proposer) and n = 5 (by C. R. Phelps), as well as *nine* false proofs of the general case, and citing the following counterexample (for n = 20) by M. J. Lighthill: for 1+5e, 6e, 1+4e, 5e, 1+3e, 4e, 1+2e, 3e, 1+e, 2e, 1+2e, e, 1+3e, 2e, 1+4e, 3e, 1+5e, 4e, 1+6e, 5e (where e ~ 0), our "cyclic sum" is equal to 10 - e^2 + O(e^3) < 10 :-) Since then, considerable progress has been made, see for example P. J. Bushell's "Shapiro's Cyclic Sum" in Bulletin of the London Mathematical Society, 26 (1994), pp. 564-574. I will only mention here that, as far as I understand (from Pierre's posting for example), all counterexamples (for even > 12 and odd > 23) barely work: S_n > .4945*n. And it is this marginality of the conjectured inequality's failure that impresses me, rather than its failure itself! (At the intuitive level, it is hard to make 2*(S^2) - n*R negative because each of the two quantities is equal to the sum of 2*(n^2) two-factor products (including perfect squares); but, of course, the problem is much deeper and Jensen's inequality does not seem to suffice even for the few case when the inequality holds.) George Baloglou -- http://www.oswego.edu/~baloglou Michel pointed out that the talks were sponsored by NATO and asked him if he knew what NATO was. No, Grothendieck replied. Michel explained it to him and recalls Grothendieck saying, "They never told me!" [In "Grothendieck: The Genie of the Bois-Marie", AMS Notices, 3/99, p. 332] ============================================================================== From: baloglou@panix.com (George Baloglou) Subject: Re: A difficult inequality Date: 4 Sep 1999 01:28:59 -0400 Newsgroups: sci.math [reply address is baloglouAToswego.edu -- baloglou@panix.com accepts no mail] [This post discusses my adventurous non-proof of Shapiro's Inequality for n = 4: if you have followed that thread then you know that Bob Israel's proof posted here earlier this week is much more efficient, so please read below only in case you are truly interested in the fine mysteries of this problem!] In article <7qd21j$97$1@panix3.panix.com> I wrote: > >>a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) >= 2 -- equality clearly holds for >>a = b = c = d. Hmm ... Multivariable Calculus and Lagrange Multipliers, >>perhaps? Anything "quicker"? I think I will leave this to somebody else, >>it's getting a bit late here :-) > >Well, I am back and I see no clever way of settling this ... But brute force >(reduction to common denominator and cross multiplication) works, reducing >the inequality, after "combined applications" of X^2+Y^2 >= 2*X*Y, to > > (a-c)*[(a^2)*d-b*(c^2)] + (b-d)*[a*(b^2)-c*(d^2)] >= 0 *** > >An intriguing inequality indeed, at least as interesting as the original! >With some sinister manipulation, it can be shown to be equivalent to > >d*(a+c)*[(a-c)^2] + (a-c)*(b-d)*(d-c)*(c+d) + a*(b+d)*[(b-d)^2] >= 0 > > OR > >b*(a+c)*[(a-c)^2] + (a-c)*(b-d)*(b-a)*(a+b) + c*(b+d)*[(b-d)^2] >= 0 > >which are trivially true (positive middle term) in case a <= b <= c <= d: >but this is an assumption that, as another solver (of the initial, easy >inequality) has already noticed, can be made (without loss of generality) >from the very beginning. Hmm ... that other solver ("lewanj") was not quite right: the ordering of a, b, c, d does matter; if for example one needs to "verify" the inequality 3/(5+2) + 5/(2+7) + 2/(7+3) + 7/(3+5) > 2, then, no matter how the numbers are "rotated" (a = 3, b = 5, c = 2, d = 7 OR a = 5, b = 2, c = 7, d = 3 OR a = 2, b = 7, c = 3, d = 5 OR a = 7, b = 3, c = 5, d = 2), the condition a <= b <= c <= d is *not* satisfied -- nor are satisfied the other 17 (out of 24 possible) conditions that make *at least* one of those two "middle terms" above positive: I must declare my proof of Shapiro's Inequality for n = 4 to be invalid :-( Well, even though Bob Israel's fine proof for n = 4 is available, "honor" made me try again the brute force approach, which this time led me to break the full inequality into 3 inequalities, the only "challenging" one being (a^3)*d + (b^3)*a + (c^3)*b + (d^3)*c >= >= a*b*(c^2) + b*c*(d^2) + c*d*(a^2) + d*a*(b^2), which happens to be false -- for the example given right below, it leads to 6552 >= 6728 ... Well, I prudently give up -- for the time being anyway :-) >Hmm ... how about the inequality *** above? Well, sad as it is, it may >fail at times (when the right inequalities between a, b, c, d fail); >look for example at a = 1, b = 13, c = 3, d = 11 ... [Nevertheless, >a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) = 1/16 + 13/14 + 3/12 + 11/14 = >= 227/112 > 2 ... no problem! :-) ] ^^^^^^^^^^^ Well, there *is* a problem, as I explained in the beginning of this post :-( .... So, how did Shapiro and Phelps settle the cases n = 4 and n = 5, respectively? It is unlikely they applied Israel's "convexity approach", for in that case they should probably have been able to settle the n = 6 case, too ... but I guess 45 years ago these folks did know something about inequalities that I (and many others) apparently do not know (or care for). Instead of a postscipt, here is the elusive (*in a way*) "full" inequality: (a^3)*c + (b^3)*d + (c^3)*a + (d^3)*b + (a^3)*d + (b^3)*a + (c^3)*b + (d^3)*c + (a^2)*(d^2) + (b^2)*(a^2) + (c^2)*(b^2) + (d^2)*(c^2) >= 2*(a^2)*(c^2) + 2*(b^2)*(d^2) + a*b*(c^2) + b*c*(d^2) + c*d*(a^2) + d*a*(b^2) + a*(b^2)*c + b*(c^2)*d + c*(d^2)*a + d*(a^2)*b George Baloglou -- http://www.oswego.edu/~baloglou Michel pointed out that the talks were sponsored by NATO and asked him if he knew what NATO was. No, Grothendieck replied. Michel explained it to him and recalls Grothendieck saying, "They never told me!" [In "Grothendieck: The Genie of the Bois-Marie", AMS Notices, 3/99, p. 332]