An Unconventional Inequality

Transcription

1 FEATURE 10 MY FAVORITE PROBLEM A Ucovetioal Iequality 101 Itroductio Ameya Veligker Harvard Uiversity 10 Cambridge, MA aveligk@fasharvardedu Problem solvig has bee a importat aspect of mathematics i my life It is the challege of tacklig a math problem ad experiecig the momets of both isight ad perplexity associated with the problem solvig process that have draw me to the field of mathematics Whe asked to write about my favorite problem, I foud it rather difficult to sigle out a particular problem from my repository of iterestig mathematics problems Though I do ot have a sigle favorite problem, I tried to choose a problem which meets several criteria that make a mathematics problem iterestig First, the problem should have a simple/elegat solutio; at the same time, the solutio should have some key step which is clever ad difficult The ideal problem is the oe that looks itractable upo first sight but, after oe has read the solutio, should evoke the respose, Ah, that was simple! Why did t I thik of that? Moreover, the problem should have iterestig geeralizatios or coectios to other problems or ideas i mathematics Keepig i mid these characteristics of a good problem, I have selected oe which I believe meets the criteria But first, I will preset some backgroud o the problem The problem was created by Reid Barto ad submitted as a problem for the 2003 Iteratioal Mathematical Olympiad (IMO) Though it was ot selected as a questio o the IMO, it was icluded o the IMO Shortlist, a aual list of twety to thirty problems which are i cotetio for a place o the IMO exam The problem was first preseted to me durig the 2004 USA Mathematical Olympiad Summer Program (MOSP), where it was give as a problem o a practice test As I recall, o oe was able to solve it, ad I was quite fasciated after learig its clever solutio 102 Problem The problem is stated below: Problem Let be a positive iteger ad let (x 1,, x ), (y 1,, y ) be two sequeces of positive real umbers Suppose (z 2,, z 2) is a sequece of positive real umbers such that Let M = max{z 2,, z 2} Prove that z 2 i+j x iy j for all 1 i, j (101) «2 M + z2 + + z 2 x1 + + x y1 + + y (102) 2 The first thig that strikes the reader about this problem is how ucovetioal it is The coditio give i (101) is certaily bizarre, as is the appearace of M i the iequality Provig Ameya Veligker, Harvard 10, is a mathematics ad physics cocetrator livig i Currier House His academic iterests spa a wide rage of subjects, icludig umber theory, aalysis, mathematical physics, ad algorithms Outside of academics, he ejoys teis, pool, ad Idia classical music 90

2 AMEYA VELINGKER AN UNCONVENTIONAL INEQUALITY 91 iequalities is a commo type of questio i the olympiad exams, ad ay experieced olympiad problem solver has i his arseal a umber of tools to attack such questios, such as AM-GM iequality, Cauchy-Schwarz iequality, ad Muirhead s iequality (see [HLP]), to ame a few The trouble with this iequality is that oe of the stadard tricks seem to work, as we will highlight The first questios that emerge regardig the iequality are how strict it is ad what equality cases, if ay, there are Upo quick ispectio, oe otices that choosig x 1 = x 2 = = x = y 1 = y 2 = = y = z 2 = z 3 = = z 2 satisfies (101) ad yields equality i our iequality This equality coditio, combied with the ature of the left-had side of (102), is remiiscet of the Iequality of Arithmetic ad Geometric Meas (or AM-GM, for short): Theorem 1 (AM-GM) Give a list of k oegative real umbers a 1, a 2,, a k, the followig iequality holds: a 1 + a a k k a 1a 2 a k, k with equality if ad oly if a 1 = a 2 = = a k A simple proof of this iequality ca be foud i ([HLP]) Sice the equality case of (102) appears to require M = z 2 = z 3 = = z, we are tempted to try applyig the AM-GM iequality o the left-had side of (102): M + z 2 + z z 2 2 Thus, provig (102) reduces to showig that Mz2z 3 z 2 x1 + + x 2 Mz 2z 3 z 2 y1 + + y However, the problem is that the above iequality is actually false, i geeral Oe ca come up with umerous couterexamples; for istace, take = 3 with x 1 = y 1 = 1, x 2 = y 2 = 2, x 3 = y 3 = 3, z 2 = 1, z 3 = 2, z 4 = 2, z 5 = 6, ad z 6 = 3, ad the above iequality is ot satisfied Thus, this approach does ot work, as AM-GM is too weak for the left-had side Aother approach is to use the Cauchy-Schwarz iequality, which i sequece form, states the followig: Theorem 2 (Cauchy-Schwarz Iequality) Give a 1, a 2,, a, b 1, b 2,, b R, we have X x 2 i! X y 2 i!! 2 X x iy i The trouble is that the oly apparet cadidate for applicatio of Cauchy-Schwarz would be the right side of (102), i the form of x1 + + x y1 + + y x1y «xy 2 However, this iequality goes i the wrog directio, so we are forced to abado this idea Aother idea uses a differet applicatio of AM-GM, this time o the right side of (102) The right side is a product of two quatities ad leds itself to AM-GM: x1 + + x y1 + + y x 1 + +x + y 1+ +y 2! 2 x1 + + x + y y 2 = 2

3 92 THE HARVARD COLLEGE MATHEMATICS REVIEW 22 I light of the above iequality, it suffices to establish M + z 2 + z z 2 x x + y y, which seems to be a simpler iequality The reader ca try some examples ad covice himself that the above iequality appears to be true Thus, this lie of attack seems promisig However, i attemptig to prove (102), oe of the stadard tricks appear to work The tricky part lies i usig the coditio (101) effectively Oe could reasoably expect to use z k x iy j for some i + j ad prove some iequality of the form M + X xiy j x x + y y, where the sum rages over certai pairs (i, j) However, the directio of the iequality makes it early itractable to tackle via iequalities such as AM-GM or Muirhead Justifiably so, it is at this jucture that a brilliat maeuver is required First, we take advatage of the homogeeity of (101) ad (102) Let x = max{x 1,, x } ad y = max{y 1,, y } The without loss of geerality, oe may replace each x i with x i = x i/x, each y i with y i = y i/y, ad each z i with z i = z i/ xy without affectig the statemet of the problem Thus, it suffices to prove (102) uder the added assumptio that max{x 1, x 2,, x } = max{y 1, y 2,, y } = 1 Now, the critical igrediet i the proof is the followig lemma: Lemma 3 Let a 1, a 2,, a k, b 1, b 2,, b k be positive reals Suppose that, for ay r > 0, the followig property is satisfied: (i) The umber of i for which a i > r is at least the umber of i for which b i > r The, a a b b Proof Without loss of geerality, oe ca assume that a 1 a 2 a k ad b 1 b 2 b k Note that if there exists k for which b k > a k, the, the umber of i for which b i > (a k + b k )/2 is at least k + 1 (sice i = k + 1,, satisfy the relatio), while the umber of i for which a i > (a k + b k )/2 is at most k, cotradictig the iitial assumptio Hece, we must have a i b i for all i, ad so, a a b b The lemma seems rather obvious, but it is powerful eough to establish (102) ad thus provide a complete solutio 103 Solutio We ow preset a complete solutio (provided by [DJMP]) to the problem alog these lies: Proof Let x = max{x 1,, x } ad y = max{y 1,, y } The, without loss of geerality, we may assume that x = y = 1, for we ca always replace x i by x i/x, y i by y i/y, ad z i by z i/ xy without affectig the statemet of the problem It suffices to show that M + z z 2 (x x ) + (y y ) because applyig the AM-GM iequality to the sum of two terms o the right side of the above iequality would give us the desired result Now, by the previous lemma, we eed oly show that, for ay r > 0, the umber of terms o the left side of (102) that are greater tha r is at least the umber of those o the right side Observe that if r 1, this property clearly holds, as o terms o the right side are greater tha r Next, suppose r < 1 Let X = {i : x i > r}, Y = {i : y i > r}, ad Z = {i : z i > r} Note that if x i, y j > r, the z i+j x iy j > r Thus, {i + j : i X, j Y } Z However, ote that X ad Y are oempty (sice r < 1 ad x = y = 1) Thus, if X = {a 1, a 2,, a l } ad Y = {b 1, b 2,, b m} with a 1 < a 2 < < a l ad b 1 < b 2 < < b m,

4 AMEYA VELINGKER AN UNCONVENTIONAL INEQUALITY 93 the {a 1 + b 1, a 1 + b 2,, a 1 + b m, a 2 + b m, a 3 + b m,, a m + b m} Z, which shows that Z X + Y 1 But the, we also have that M > r Hece, there are at least X + Y elemets o the left side of (102) that are greater tha r This cocludes the proof The proof is everythig we wat it to be: short, elegat, ad clever 104 Further Coectios What I fid remarkable about the problem, apart from the simple (albeit difficult) ature of its solutio, is the multitude of coectios it has with the field of discrete geometry As it turs out, the iequality we have discussed is related to a importat iequality kow as the Prékopa- Leidler Iequality: Theorem 4 (Prékopa-Leidler Iequality) Let 0 < λ < 1, ad let f, g, h : R [0, ) be measurable fuctios such that h(λx + (1 λ)y) f(x) λ g(y) 1 λ for all x, y R The, Z Z «λ Z «1 λ h(x) dx f(x) dx g(x) dx R R R Of course, the most strikig differece betwee the give problem ad the statemet of the Prékopa-Leidler Iequality is the fact that the former deals with sequeces while the latter deals with fuctios I fact, our iequality ca be viewed as a discrete specializatio of the Prékopa- Leidler iequality for = 1 ad λ = 1/2 The Prékopa-Leidler iequality has may importat applicatios, such as probability theory, optimal mass trasportatio [Vi], ad the theory of diffusio [BL] Perhaps its most importat cosequece is the Bru-Mikowski iequality (see [Ba]), which ca be stated as follows (ote that there exist alterative formulatios of the iequality): Theorem 5 If A ad B are compact subsets of R, the λa + (1 λ)b 1/ λ A 1/ + (1 λ) B 1/, where X deotes the Lebesgue measure of X, ad λa + (1 λ)b deotes the Mikowski sum {λa + (1 λ)b : a A, b B} The Bru-Mikowski iequality ca be used to provide a simple proof (from [Ba]) of the famous isoperimetric iequality: Theorem 6 (Isoperimetric Iequality) Amog simple closed bodies of a give volume i R, Euclidea balls have the least surface area Proof Let C R be a compact set with volume equal to that of B, the Euclidea ball of radius 1 The, the surface area of C is give by C + ɛb C C = lim ɛ 0 ɛ By the Bru-Mikowski iequality, we have C + ɛb C 1/ + ɛ B 1/ C + ɛ C ( 1)/ B 1/ It the follows that C C ( 1)/ B 1/ = B Usig the well-kow fact B = B (see page 4 of [Ba]), we obtai C B, as desired

5 94 THE HARVARD COLLEGE MATHEMATICS REVIEW 22 Refereces [Ba] K M Ball: A elemetary itroductio to moder covex geometry (1997) [BL] H J Brascamp ad E H Lieb: O extesios of the Bru-Mikowski ad Prékopa- Leidler theorems, icludig iequalities for log cocave fuctios, ad with a applicatio to the diffusio equatio, J Fuctioal Aalysis 22 (1976) [DJMP] D Djukić, V Jaković, I Matić, ad N Petrović: The IMO Compedium New York: Spriger Sciece+Busiess Media, Ic (2006) [HLP] G H Hardy, J E Littlewood, ad G Polya: Iequalities, Cambridge: Cambridge Uiv Press (1952) [Vi] C Villai: Topics i Optimal Trasportatio, America Mathematical Society (2003) (Grad Stud i Math 58)