Topics in Inequalities - Theorems and Techniques. Hojoo Lee

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1 Topis in Inequlities - Theorems nd Tehniques Hojoo Lee Introdution Inequlities re useful in ll fields of Mthemtis The im of this problem-oriented book is to present elementry tehniques in the theory of inequlities The reders will meet lssil theorems inluding Shur s inequlity, Muirhed s theorem, the Cuhy-Shwrz inequlity, the Power Men inequlity, the AM- GM inequlity, nd Hölder s theorem I would gretly ppreite hering bout omments nd orretions from my reders You n send emil to me t ultrmetri@gmilom To Students My trget reders re hllenging high shools students nd undergrdute students The given tehniques in this book re just the tip of the inequlities ieberg Young students should find their own methods to ttk vrious problems A gret Hungrin Mthemtiin Pul Erdös ws fond of sying tht God hs trnsfinite book with ll the theorems nd their best proofs I strongly enourge reders to send me their own retive solutions of the problems in this book Hve fun! Aknowledgement I m indebted to Orlndo Döhring nd Drij Grinberg for providing me with TeX files inluding olletions of interesting inequlities I d like to thnk Mrin Muresn for his exellent olletion of problems I m lso plesed tht Co Minh Qung sent me vrious vietnm problems nd nie proofs of Nesbitt s inequlity I owe gret debts to Stnley Rbinowitz who kindly sent me his pper On The Computer Solution of Symmetri Homogeneous Tringle Inequlities Resoures on the Web MthLinks, Art of Problem Solving, MthPro Press, 4 K S Kedly, A < B, 5 T J Mildorf, Olympid Inequlities, I

2 Contents Geometri Inequlities Rvi Substitution Trigonometri Methods 7 Applitions of Complex Numbers Four Bsi Tehniques Trigonometri Substitutions Algebri Substitutions 5 Inresing Funtion Theorem 0 4 Estblishing New Bounds Homogeniztions nd Normliztions 6 Homogeniztions 6 Shur s Inequlity nd Muirhed s Theorem 9 Normliztions 4 4 Cuhy-Shwrz Inequlity nd Hölder s Inequlity 8 4 Convexity 4 4 Jensen s Inequlity 4 4 Power Mens 45 4 Mjoriztion Inequlity Supporting Line Inequlity 48 5 Problems, Problems, Problems 50 5 Multivrible Inequlities 50 5 Problems for Putnm Seminr 58 II

3 Chpter Geometri Inequlities It gives me the sme plesure when someone else proves good theorem s when I do it myself E Lndu Rvi Substitution Mny inequlities re simplified by some suitble substitutions We begin with lssil inequlity in tringle geometry Wht is the first nontrivil geometri inequlity? In 746, Chpple showed tht Theorem Chpple 746, Euler 765 Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC Then, we hve R r nd the equlity holds if nd only if ABC is equilterl Proof Let BC =, CA = b, AB =, s = +b+ nd S = [ABC] Rell the well-known identities : S = b 4R, S = rs, S = ss s bs Hene, R r is equivlent to b 4S S S s or b 8 s or b 8s s bs We need to prove the following Theorem [AP], A Pdo Let, b, be the lengths of tringle Then, we hve b 8s s bs or b b + + b + b nd the equlity holds if nd only if = b = Proof We use the Rvi Substitution : Sine, b, re the lengths of tringle, there re positive rels x, y, z suh tht = y + z, b = z + x, = x + y Why? Then, the inequlity is y + zz + xx + y 8xyz for x, y, z > 0 However, we get y + zz + xx + y 8xyz = xy z + yz x + zx y 0 Exerise Let ABC be right tringle Show tht R + r When does the equlity hold? It s nturl to sk tht the inequlity in the theorem holds for rbitrry positive rels, b,? Yes! It s possible to prove the inequlity without the dditionl ondition tht, b, re the lengths of tringle : Theorem Let x, y, z > 0 Then, we hve xyz y + z xz + x yx + y z The equlity holds if nd only if x = y = z Proof Sine the inequlity is symmetri in the vribles, without loss of generlity, we my ssume tht x y z Then, we hve x + y > z nd z + x > y If y + z > x, then x, y, z re the lengths of the sides of tringle In this se, by the theorem, we get the result Now, we my ssume tht y + z x Then, xyz > 0 y + z xz + x yx + y z The inequlity in the theorem holds when some of x, y, z re zeros : Theorem 4 Let x, y, z 0 Then, we hve xyz y + z xz + x yx + y z The first geometri inequlity is the Tringle Inequlity : AB + BC AC In this book, [P ] stnds for the re of the polygon P

4 Proof Sine x, y, z 0, we n find positive sequenes {x n }, {y n }, {z n } for whih Applying the theorem yields lim x n = x, n lim y n = y, lim z n = z n n x n y n z n y n + z n x n z n + x n y n x n + y n z n Now, tking the limits to both sides, we get the result Clerly, the equlity holds when x = y = z However, xyz = y+z xz +x yx+y z nd x, y, z 0 does not gurntee tht x = y = z In ft, for x, y, z 0, the equlity xyz = y +z xz +x yx+y z is equivlent to x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0 It s strightforwrd to verify the equlity xyz y + z xz + x yx + y z = xx yx z + yy zy x + zz xz y Hene, the theorem 4 is prtiulr se of Shur s inequlity Problem IMO 000/, Proposed by Titu Andreesu Let, b, be positive numbers suh tht b = Prove tht + b + + b First Solution Sine b =, we mke the substitution = x y, b = y z, = z x for x, y, z > 0 We rewrite the given inequlity in the terms of x, y, z : x y + z y y z + x z z x + y xyz y + z xz + x yx + y z x The Rvi Substitution is useful for inequlities for the lengths, b, of tringle After the Rvi Substitution, we n remove the ondition tht they re the lengths of the sides of tringle Problem IMO 98/6 Let, b, be the lengths of the sides of tringle Prove tht b b + b b + 0 First Solution After setting = y + z, b = z + x, = x + y for x, y, z > 0, it beomes x z + y x + z y x yz + xy z + xyz or x y + y z + z x x + y + z, whih follows from the Cuhy-Shwrz inequlity x y + z + x y + y z + z x + y + z x Exerise Let, b, be the lengths of tringle Show tht For exmple, tke x =, y =, z = b b + + b b <

5 Exerise Drij Grinberg Let, b, be the lengths of tringle Show the inequlities + b + + b b b 0, nd b + b + b b b 0 We now disuss Weitzenbök s inequlity nd relted inequlities Problem IMO 96/, Weitzenbök s inequlity Let, b, be the lengths of tringle with re S Show tht + b + 4 S Solution Write = y + z, b = z + x, = x + y for x, y, z > 0 It s equivlent to whih n be obtined s following : y + z + z + x + x + y 48x + y + zxyz, y + z + z + x + x + y 6yz + zx + xy 6 xy yz + yz zx + xy yz Here, we used the well-known inequlities p + q pq nd p + q + r pq + qr + rp Theorem 5 Hdwiger-Finsler inequlity For ny tringle ABC with sides, b, nd re F, the following inequlity holds b + b + + b + 4 F First Proof After the substitution = y + z, b = z + x, = x + y, where x, y, z > 0, it beomes whih follows from the identity xy + yz + zx xyzx + y + z, xy + yz + zx xyzx + y + z = xy yz + yz zx + zx xy Seond Proof We give onvexity proof There re mny wys to dedue the following identity: b + b + + b + 4F Sine tn x is onvex on 0, π, Jensen s inequlity shows tht b + b + + b + 4F = tn A + tn B + tn C tn A + B + C = Tsintsifs proved simultneous generliztion of Weitzenbök s inequlity nd Nesbitt s inequlity Theorem 6 Tsintsifs Let p, q, r be positive rel numbers nd let, b, denote the sides of tringle with re F Then, we hve p q + r + q r + p b + r p + q F

6 Proof V Pmbuin By Hdwiger-Finsler inequlity, it suffies to show tht p q + r + q r + p b + r p + q + b + + b + or p + q + r p + q + r p + q + r + b + + b + q + r r + p p + q or q + r + r + p + p + q q + r + r + p b + p + q + b + However, this is strightforwrd onsequene of the Cuhy-Shwrz inequlity Theorem 7 Neuberg-Pedoe inequlity Let, b, denote the sides of the tringle A B C with re F Let, b, denote the sides of the tringle A B C with re F Then, we hve b + + b + b + + b 6F F Notie tht it s generliztion of Weitzenbök s inequlitywhy? In [GC], G Chng proved Neuberg- Pedoe inequlity by using omplex numbers For very interesting geometri observtions nd proofs of Neuberg-Pedoe inequlity, see [DP] or [GI, pp9-9] Here, we offer three lgebri proofs Lemm + b + b b b > 0 Proof Observe tht it s equivlent to + b + + b + > + b b + From Heron s formul, we find tht, for i =,, 6F i = i + b i + i i 4 + b i 4 + i 4 > 0 or i + b i + i > i4 + b i 4 + i4 The Cuhy-Shwrz inequlity implies tht + b + + b + > 4 + b b b b + First Proof [LC], Crlitz By the lemm, we obtin Hene, we need to show tht L = b + + b + b + + b > 0, One my esily hek the following identity where L 6F 6F 0 L 6F 6F = 4UV + V W + W U, U = b b, V = nd W = b b Using the identity U + b V + W = 0 or W = U b V, one my lso dedue tht UV + V W + W U = U b V 4 b b V 4 It follows tht UV + V W + W U = U b V 6F 4 V 0 4

7 Crlitz lso observed tht the Neuberg-Pedoe inequlity n be dedued from Azél s inequlity Theorem 8 Azél s inequlity Let,, n, b,, b n be positive rel numbers stisfying Then, the following inequlity holds b b + + n b n + + n nd b b + + b n + + n b b + + b n Proof [AI] The Cuhy-Shwrz inequlity shows tht b + + n b + + b n b + + n b n Then, the bove inequlity is equivlent to b b + + n b n + + n b b + + b n In se + + n = 0, it s trivil Hene, we now ssume tht + + n > 0 The min trik is to think of the following qudrti polynomil n n n n P x = x b i x b i = i x + b i b i x + b b i i= i= Sine P b = n i= b i b i 0 nd sine the oeffiient of x in the qudrti polynomil P is positive, P should hve t lest one rel root Therefore, P hs nonnegtive disriminnt It follows tht b n i b i 4 i= n i= i b i= n i= b i 0 i= Seond Proof of Neuberg-Pedoe inequlity [LC], Crlitz We rewrite it in terms of, b,,, b, : + b + + b + + b b + + b b b b We employ the following substitutions As in the proof of the lemm 5, we hve x = + b +, x =, x = b, x 4 =, y = + b +, y =, y = b, y 4 = We now pply Azél s inequlity to get the inequlity x > x + y + x 4 nd y > y + y + y 4 x y x y x y x 4 y 4 x x + y + x 4 y y + y + y 4 We lose this setion with very simple proof by former student in KMO 4 summer progrm 4 Koren Mthemtil Olympids 5

8 Third Proof Toss two tringles A B C nd A B C on R : A 0, p, B p, 0, C p, 0, A 0, q, B q, 0, nd C q, 0 It therefore follows from the inequlity x + y xy tht b + + b + b + + b = p p q + q q + p + p q q q + p + p q q q = p p q + q q p + p q p q p p q + q q p 4 p p q q q p = 6F F 6

9 Trigonometri Methods In this setion, we employ trigonometri methods to ttk geometri inequlities Theorem Erdös-Mordell Theorem If from point P inside given tringle ABC perpendiulrs P H, P H, P H re drwn to its sides, then P A + P B + P C P H + P H + P H This ws onjetured by Pul Erdös in 95, nd first proved by Mordell in the sme yer Severl proofs of this inequlity hve been given, using Ptolemy s theorem by André Avez, ngulr omputtions with similr tringles by Leon Bnkoff, re inequlity by V Komornik, or using trigonometry by Mordell nd Brrow Proof [MB], Mordell We trnsform it to trigonometri inequlity Let h = P H, h = P H nd h = P H Apply the Sine Lw nd the Cosine Lw to obtin P A sin A = H H = h + h h h osπ A, P B sin B = H H = h + h h h osπ B, P C sin C = H H = h + h h h osπ C So, we need to prove tht h + h h h osπ A h + h + h sin A The min trouble is tht the left hnd side hs too hevy terms with squre root expressions Our strtegy is to find lower bound without squre roots To this end, we express the terms inside the squre root s the sum of two squres H H = h + h h h osπ A = h + h h h osb + C = h + h h h os B os C sin B sin C Using os B + sin B = nd os C + sin C =, one finds tht H H = h sin C + h sin B + h os C h os B Sine h os C h os B is lerly nonnegtive, we get H H h sin C + h sin B It follows tht h + h h h osπ A h sin C + h sin B sin A sin A = sin B sin C + sin C h sin B sin B sin C sin C sin B h = h + h + h We use the sme tehniques to ttk the following geometri inequlity Problem 4 IMO Short-list 005 In n ute tringle ABC, let D, E, F, P, Q, R be the feet of perpendiulrs from A, B, C, A, B, C to BC, CA, AB, EF, F D, DE, respetively Prove tht where pt denotes the perimeter of tringle T pabcpp QR pdef, 7

10 Solution Let s euler 5 this problem Let ρ be the irumrdius of the tringle ABC It s esy to show tht BC = ρ sin A nd EF = ρ sin A os A Sine DQ = ρ sin C os B os A, DR = ρ sin B os C os A, nd F DE = π A, the Cosine Lw gives us or where QR = DQ + DR DQ DR osπ A [ ] = 4ρ os A sin C os B + sin B os C + sin C os B sin B os C osa QR = ρ os A fa, B, C, fa, B, C = sin C os B + sin B os C + sin C os B sin B os C osa So, wht we need to ttk is the following inequlity: ρ sin A ρ os A fa, B, C ρ sin A os A or sin A os A fa, B, C sin A os A Our job is now to find resonble lower bound of fa, B, C One gin, we express fa, B, C s the sum of two squres We observe tht fa, B, C = sin C os B + sin B os C + sin C os B sin B os C osa = sin C os B + sin B os C + sin C os B sin B os C [ + osa] = sin C + B sin C os B sin B os C sin A = sin A [ 4 sin B sin C os B os C] So, we shll express 4 sin B sin C os B os C s the sum of two squres The trik is to reple with sin B + os B sin C + os C Indeed, we get 4 sin B sin C os B os C = sin B + os B sin C + os C 4 sin B sin C os B os C = sin B os C sin C os B + os B os C sin B sin C = sin B C + os B + C = sin B C + os A It therefore follows tht fa, B, C = sin A [ sin B C + os A ] sin A os A so tht os A fa, B, C sin A os A So, we n omplete the proof if we estblish tht sin A sin A os A sin A os A Indeed, one sees tht it s diret onsequene of the Cuhy-Shwrz inequlity p + q + rx + y + z px + qy + rz, where p, q, r, x, y nd z re positive rel numbers 5 euler v in Mthemtis trnsform the problems in tringle geometry to trigonometri ones 8

11 Alterntively, one my obtin nother lower bound of fa, B, C: fa, B, C = sin C os B + sin B os C + sin C os B sin B os C osa = sin C os B sin B os C + sin C os B sin B os C [ + osa] = sin B C + sinb sinc os A os A sinb sinc Then, we n use this to offer lower bound of the perimeter of tringle P QR: pp QR = ρ os A fa, B, C ρ os A sin B sin C So, one my onsider the following inequlity: pabc ρ os A sin B sin C pdef or or ρ sin A ρ os A sin B sin C ρ sin A os A sin A os A sin B sin C sin A os A However, it turned out tht this doesn t hold Try to disprove this! Problem 5 IMO 00/ Let ABC be n ute-ngled tringle with O s its irumenter Let P on line BC be the foot of the ltitude from A Assume tht BCA ABC + 0 Prove tht CAB + COP < 90 Proof The ngle inequlity CAB + COP < 90 n be written s COP < P CO This n be shown if we estblish the length inequlity OP > P C Sine the power of P with respet to the irumirle of ABC is OP = R BP P C, where R is the irumrdius of the tringle ABC, it beomes R BP P C > P C or R > BC P C We euler this It s n esy job to get BC = R sin A nd P C = R sin B os C Hene, we show the inequlity R > R sin A R sin B os C or sin A sin B os C < 4 Sine sin A <, it suffies to show tht sin A sin B os C < 4 Finlly, we use the ngle ondition C B + 0 to obtin the trigonometri inequlity sin B os C = sinb + C sinc B sinc B sin 0 = 4 We lose this setion with Brrows inequlity stronger thn Erdös-Mordell Theorem following trigonometri inequlity: We need the Proposition Let x, y, z, θ, θ, θ be rel numbers with θ + θ + θ = π Then, x + y + z yz os θ + zx os θ + xy os θ Proof Using θ = π θ + θ, it s n esy job to hek the following identity x + y + z yz os θ + zx os θ + xy os θ = z x os θ + y os θ + x sin θ y sin θ 9

12 Corollry Let p, q, nd r be positive rel numbers Let θ, θ, nd θ be rel numbers stisfying θ + θ + θ = π Then, the following inequlity holds p os θ + q os θ + r os θ qr p + rp q + pq r Proof Tke x, y, z = qr p, rp q, pq r nd pply the bove proposition Theorem Brrow s Inequlity Let P be n interior point of tringle ABC nd let U, V, W be the points where the bisetors of ngles BP C, CP A, AP B ut the sides BC,CA,AB respetively Prove tht P A + P B + P C P U + P V + P W Proof [MB] nd [AK] Let d = P A, d = P B, d = P C, l = P U, l = P V, l = P W, θ = BP C, θ = CP A, nd θ = AP B We need to show tht d + d + d l + l + l It s esy to dedue the following identities l = d d d + d os θ, l = d d d + d os θ, nd l = d d d + d os θ, By the AM-GM inequlity nd the bove orollry, this mens tht l + l + l d d os θ + d d os θ + d d os θ d + d + d As nother pplition of the bove trigonometri proposition, we estblish the following inequlity Corollry [AK], Abi-Khuzm Let x,, x 4 be positive rel numbers Let θ,, θ 4 be rel numbers suh tht θ + + θ 4 = π Then, x x + x x 4 x x + x x 4 x x 4 + x x x os θ + x os θ + x os θ + x 4 os θ 4 x x x x 4 Proof Let p = x +x x x + x +x 4 x x 4 q = xx+xx4 nd λ = p q In the view of θ + θ + θ + θ 4 = π nd θ + θ 4 + θ + θ = π, the proposition implies tht x os θ + x os θ + λ osθ + θ 4 pλ = pq, nd x os θ + x 4 os θ 4 + λ osθ + θ q λ = pq Sine osθ + θ 4 + osθ + θ = 0, dding these two bove inequlities yields x os θ + x os θ + x os θ + x 4 os θ 4 x x + x x 4 x x + x x 4 x x 4 + x x pq = x x x x 4 0

13 Applitions of Complex Numbers In this setion, we disuss some pplitions of omplex numbers to geometri inequlity Every omplex number orresponds to unique point in the omplex plne The stndrd symbol for the set of ll omplex numbers is C, nd we lso refer to the omplex plne s C The min tool is pplitions of the following fundmentl inequlity Theorem If z,, z n C, then z + + z n z + + z n Proof Use indution on n with the tringle inequlity Theorem Ptolemy s Inequlity For ny points A, B, C, D in the plne, we hve AB CD + BC DA AC BD Proof Let, b, nd 0 be omplex numbers tht orrespond to A, B, C, D in the omplex plne It beomes b + b b Applying the Tringle Inequlity to the identity b + b = b, we get the result Problem 6 [TD] Let P be n rbitrry point in the plne of tringle ABC with the entroid G Show the following inequlities BC P B P C + AB P A P B + CA P C P A BC CA AB nd P A BC + P B CA + P C AB P G BC CA AB Solution We only hek the first inequlity Regrd A, B, C, P s omplex numbers nd ssume tht P orresponds to 0 We re required to prove tht B CBC + A BAB + C ACA B CC AA B It remins to pply the Tringle Inequlity to the identity B CBC + A BAB + C ACA = B CC AA B Problem 7 IMO Short-list 00 Let ABC be tringle for whih there exists n interior point F suh tht AF B = BF C = CF A Let the lines BF nd CF meet the sides AC nd AB t D nd E, respetively Prove tht AB + AC 4DE Solution Let AF = x, BF = y, CF = z nd let ω = os π + i sin π We n toss the pitures on C so tht the points F, A, B, C, D, nd E re represented by the omplex numbers 0, x, yω, zω, d, nd e It s n esy exerise to estblish tht DF = xz x+z nd EF = xy xz xy x+y This mens tht d = x+z ω nd e = x+y ω We re now required to prove tht x yω + zω x 4 zx z + x ω + xy x + y ω Sine ω = nd ω =, we hve zω x = ωzω x = z xω Therefore, we need to prove x yω + z xω 4zx z + x 4xy x + y ω More strongly, we estblish tht x yω + z xω 4zx z+x 4xy x+y ω or p qω r sω, where p = z + x, q = y + x, r = 4zx 4xy z+x nd s = x+y It s ler tht p r > 0 nd q s > 0 It follows tht p qω r sω = p qωp qω r sωr sω = p r + pq rs + q s 0 It s esy to hek tht the equlity holds if nd only if ABC is equilterl

14 Chpter Four Bsi Tehniques Differentite! Shiing-shen Chern Trigonometri Substitutions If you re fed with n integrl tht ontins squre root expressions suh s x dx, + y dy, z dz then trigonometri substitutions suh s x = sin t, y = tn t, z = se t re very useful We will lern tht mking suitble trigonometri substitution simplifies the given inequlity Problem 8 APMO 004/5 Prove tht, for ll positive rel numbers, b,, + b + + 9b + b + First Solution Choose A, B, C 0, π with = tn A, b = tn B, nd = tn C Using the well-known trigonometri identity + tn θ = os θ, one my rewrite it s 4 os A os B os C os A sin B sin C + sin A os B sin C + sin A sin B os C 9 One my esily hek the following trigonometri identity osa + B + C = os A os B os C os A sin B sin C sin A os B sin C sin A sin B os C Then, the bove trigonometri inequlity tkes the form 4 os A os B os C os A os B os C osa + B + C 9 Let θ = A+B+C Applying the AM-GM inequlity nd Jesen s inequlity, we hve os A + os B + os C os A os B os C os θ We now need to show tht Using the trigonometri identity 4 9 os θos θ os θ os θ = 4 os θ os θ or os dgnsθ os θ = os θ os θ, it beomes 4 7 os4 θ os θ,

15 whih follows from the AM-GM inequlity os θ os θ os θ os θ + os θ + os θ = One find tht the equlity holds if nd only if tn A = tn B = tn C = if nd only if = b = = Problem 9 Ltvi 00 Let, b,, d be the positive rel numbers suh tht Prove tht bd b d 4 = Solution We n write = tn A, b = tn B, = tn C, d = tn D, where A, B, C, D 0, π Then, the lgebri identity beomes the following trigonometri identity : Applying the AM-GM inequlity, we obtin Similrly, we obtin os A + os B + os C + os D = sin A = os A = os B + os C + os D os B os C os D sin B os C os D os A, sin C os D os A os B, nd sin D os A os B os C Multiplying these four inequlities, we get the result! Problem 0 Kore 998 Let x, y, z be the positive rels with x + y + z = xyz Show tht x + y + z Sine the funtion f is not onve on R +, we nnot pply Jensen s inequlity to the funtion ft = +t However, the funtion ftn θ is onve on 0, π! First Solution We n write x = tn A, y = tn B, z = tn C, where A, B, C 0, π Using the ft tht + tn θ = os θ, we rewrite it in the terms of A, B, C : os A + os B + os C It follows from tnπ C = z = x+y xy = tna + B nd from π C, A + B 0, π tht π C = A + B or A + B + C = π Hene, it suffies to show the following Theorem In ny ute tringle ABC, we hve os A + os B + os C Proof Sine os x is onve on 0, π, it s diret onsequene of Jensen s inequlity We note tht the funtion os x is not onve on 0, π In ft, it s onvex on π, π One my think tht the inequlity os A + os B + os C doesn t hold for ny tringles However, it s known tht it holds for ll tringles Theorem In ny tringle ABC, we hve os A + os B + os C First Proof It follows from π C = A + B tht os C = osa + B = os A os B + sin A sin B or os A + os B + os C = sin A sin B + os A + os B 0

16 Seond Proof Let BC =, CA = b, AB = Use the Cosine Lw to rewrite the given inequlity in the terms of, b, : b + b Clering denomintors, this beomes + + b + + b b b b + + b + b + + b, whih is equivlent to b b + + b + b in the theorem In the first hpter, we found tht the geometri inequlity R r is equivlent to the lgebri inequlity b b + + b + b We now find tht, in the proof of the bove theorem, b b + + b + b is equivlent to the trigonometri inequlity os A + os B + os C One my sk tht In ny tringles ABC, is there nturl reltion between os A + os B + os C nd R r, where R nd r re the rdii of the irumirle nd inirle of ABC? Theorem Let R nd r denote the rdii of the irumirle nd inirle of the tringle ABC Then, we hve os A + os B + os C = + r R Proof Use the identity b + +b + b + +b = b+b+ + b+b We leve the detils for the reders Exerise 4 Let p, q, r be the positive rel numbers suh tht p + q + r + pqr = Show tht there exists n ute tringle ABC suh tht p = os A, q = os B, r = os C b Let p, q, r 0 with p + q + r + pqr = Show tht there re A, B, C [ 0, π ] with p = os A, q = os B, r = os C, nd A + B + C = π Problem USA 00 Let, b, nd be nonnegtive rel numbers suh tht + b + + b = 4 Prove tht 0 b + b + b Solution Notie tht, b, > implies tht +b + +b > 4 If, then we hve b+b+ b b 0 We now prove tht b + b + b Letting = p, b = q, = r, we get p + q + r + pqr = By the bove exerise, we n write [ = os A, b = os B, = os C for some A, B, C 0, π ] with A + B + C = π We re required to prove os A os B + os B os C + os C os A os A os B os C One my ssume tht A π or os A 0 Note tht os A os B + os B os C + os C os A os A os B os C = os Aos B + os C + os B os C os A We pply Jensen s inequlity to dedue os B + os C os A Note tht os B os C = osb C + osb + C os A These imply tht os A os Aos B + os C + os B os C os A os A os A + os A However, it s esy to verify tht os A os A + os A os A = 4

17 Algebri Substitutions We know tht some inequlities in tringle geometry n be treted by the Rvi substitution nd trigonometri substitutions We n lso trnsform the given inequlities into esier ones through some lever lgebri substitutions Problem IMO 00/ Let, b, be positive rel numbers Prove tht + 8b + b b b First Solution To remove the squre roots, we mke the following substitution : x = + 8b, y = b b + 8, z = + 8b Clerly, x, y, z 0, Our im is to show tht x + y + z We notie tht 8b = x x, b 8 = Hene, we need to show tht y y, 8b = z z = x y z 5 = x y z x + y + z, where 0 < x, y, z < nd x y z = 5xyz However, > x + y + z implies tht, by the AM-GM inequlity, x y z > x + y + z x x + y + z y x + y + z z = x + x + y + zy + z x + y + y + zz + xx + y + z + zx + y 4x yz 4 yz 4y zx 4 zx 4z xy 4 xy = 5xyz This is ontrdition! Problem IMO 995/ Let, b, be positive numbers suh tht b = Prove tht b + + b b First Solution After the substitution = x, b = y, = z, we get xyz = The inequlity tkes the form x y + z + y z + x + It follows from the Cuhy-Shwrz inequlity tht x [y + z + z + x + x + y] y + z + so tht, by the AM-GM inequlity, x y + z + y z + x + z x + y y z + x + z x + y x + y + z xyz z x + y + z x + y = Kore 998 Let x, y, z be the positive rels with x + y + z = xyz Show tht x + y + z 5

18 Seond Solution The strting point is letting = x, b = y, = z We find tht + b + = b is equivlent to = xy + yz + zx The inequlity beomes x x + + y y + + z z + or or x x + xy + yz + zx + x x + yx + z + y y + xy + yz + zx + y y + zy + x + z z + xy + yz + zx z z + xz + y By the AM-GM inequlity, we hve x = x x + yx + z x[x + y + x + z] = x x + yx + z x + yx + z x + yx + z x + z + x x + z In like mnner, we obtin y y + zy + x y y + z + Adding these three yields the required result y y + x nd z z z + xz + y z + x + z z + y We now prove lssil theorem in vrious wys Theorem Nesbitt, 90 For ll positive rel numbers, b,, we hve b + + b b Proof After the substitution x = b +, y = +, z = + b, it beomes y + z x x whih follows from the AM-GM inequlity s following: or y + z x 6, y + z x = y x + z x + z y + x y + x z + y y z 6 x z x z y x y x z y 6 = 6 z Proof We mke the substitution It follows tht x = fx = b +, y = b +, z = + b =, where ft = t + b + Sine f is onve on 0,, Jensen s inequlity shows tht f = = x + y + z fx f or f f + t x + y + z Sine f is monotone inresing, this implies tht x + y + z or b + = x + y + z 6

19 Proof As in the previous proof, it suffies to show tht T, where T = x + y + z nd x + x = One n esily hek tht the ondition x + x = beomes = xyz + xy + yz + zx By the AM-GM inequlity, we hve = xyz + xy + yz + zx T + T T + T 0 T T + 0 T IMO 000/ Let, b, be positive numbers suh tht b = Prove tht + b + + b Seond Solution [IV], Iln Vrdi Sine b =, we my ssume tht b It follows tht + b + + = + b + b b + Third Solution As in the first solution, fter the substitution = x y, b = y z, = z x for x, y, z > 0, we n rewrite it s xyz y + z xz + x yx + y z Without loss of generlity, we n ssume tht z y x Set y x = p nd z x = q with p, q 0 It s strightforwrd to verify tht xyz y + z xz + x yx + y z = p pq + q x + p + q p q pq Sine p pq + q p q 0 nd p + q p q pq = p q p + q 0, we get the result Fourth Solution From the IMO 000 Short List Using the ondition b =, it s strightforwrd to verify the equlities = + + b +, b = b = b + + +, + + b + In prtiulr, they show tht t most one of the numbers u = + b, v = b +, w = + is negtive If there is suh number, we hve + b + + = uvw < 0 < b And if u, v, w 0, the AM-GM inequlity yields = u + v uv, = b v + w b vw, = w + w b wu Thus, uv, vw b, wu b, so uvw b b = Sine u, v, w 0, this ompletes the proof Why? Note tht the inequlity is not symmetri in the three vribles Chek it! For verifition of the identity, see [IV] 7

20 Problem 4 Let, b, be positive rel numbers stisfying + b + = Show tht + b + b b b b + 4 Solution We wnt to estblish tht Set x = b, y = b, z = + b + + b + b + b b We need to prove tht x + + y + z + z + where x, y, z > 0 nd xy + yz + zx = It s not hrd to show tht there exists A, B, C 0, π with The inequlity beomes 4, x = tn A, y = tn B, z = tn C, nd A + B + C = π or or + tn A + + tn C tn B + + tn C + os A + os B + sin C + 4 os A + os B + sin C + 4 Note tht os A + os B = os A+B os A B Sine A B < π, this mens tht A + B π C os A + os B os = os It will be enough to show tht π C os + sin C, where C 0, π This is one-vrible inequlity It s left s n exerise for the reder Problem 5 Irn 998 Prove tht, for ll x, y, z > suh tht x + y + z =, x + y + z x + y + z First Solution We begin with the lgebri substitution = x, b = y, = z Then, the ondition beomes b + + = b + b + + b = nd the inequlity is equivlent to + b b + b + b + Let p = b, q =, r = b Our job is to prove tht p + q + r where p + q + r + pqr = By the exerise 7, we n mke the trigonometri substitution p = os A, q = os B, r = os C for some A, B, C 0, π with A + B + C = π Wht we need to show is now tht os A + os B + os C It follows from Jensen s inequlity Differentite! Shiing-shen Chern 8

21 Problem 6 IMO Short-list 00 Let x,, x n be rbitrry rel numbers Prove the inequlity x + x + x + x + x + + x n + x + + x n < n First Solution We only onsider the se when x,, x n re ll nonnegtive rel numberswhy? 4 Let x 0 = After the substitution y i = x x i for ll i = 0,, n, we obtin x i = y i y i We need to prove the following inequlity n yi y i < n y i Sine y i y i for ll i =,, n, we hve n upper bound of the left hnd side: n yi y i n yi y i n = y i yi y i y i y i i=0 i=0 We now pply the Cuhy-Shwrz inequlity to give n upper bound of the lst term: n n n = n y i y i y i y i y 0 y n i=0 Sine y 0 = nd y n > 0, this yields the desired upper bound n i=0 i=0 i=0 Seond Solution We my ssume tht x,, x n re ll nonnegtive rel numbers Let x 0 = 0 We mke the following lgebri substitution t i = x i x0 + + x i, i = + ti nd s i = t i + ti for ll i = 0,, n It s n esy exerise to show tht desired inequlity beomes x i x 0 + +x i n n < n Sine 0 < i for ll i =,, n, we hve n 0 i i i= n 0 i i = i= = 0 i s i Sine s i = i, the n 0 i 0 i i Sine 0 =, by the Cuhy-Shwrz inequlity, we obtin n 0 i 0 i i n n [ 0 i 0 i i ] = n [ 0 n ] i= i= i= 4 x +x + x +x +x + + x n +x + +x x n +x + x +x +x + + x n +x + +x n 9

22 Inresing Funtion Theorem Theorem Inresing Funtion Theorem Let f :, b R be differentible funtion If f x 0 for ll x, b, then f is monotone inresing on, b If f x > 0 for ll x, b, then f is stritly inresing on, b Proof We first onsider the se when f x > 0 for ll x, b Let < x < x < b We wnt to show tht fx < fx Applying the Men Vlue Theorem, we find some x, x suh tht fx fx = f x x Sine f > 0, this eqution mens tht fx fx > 0 In se when f x 0 for ll x, b, we n lso pply the Men Vlue Theorem to get the result Problem 7 Irelnd 000 Let x, y 0 with x + y = Prove tht x y x + y First Solution After homogenizing it, we need to prove 6 x + y x y x + y or x + y 6 x y x + y Now, forget the onstrint x+y =! In se xy = 0, it lerly holds We now ssume tht xy 0 Beuse of the homogeneity of the inequlity, this mens tht we my normlize to xy = Then, it beomes x + x 6 x + x or p p where p = x + x 4 Our job is now to minimize F p = p p on [4, Sine F p = p 0, where p, F is monotone inresing on [4, So, F p F 4 = 0 for ll p 4 Seond Solution As in the first solution, we prove tht x + y 6 x + y xy for ll x, y 0 In se x = y = 0, it s ler Now, if x + y > 0, then we my normlize to x + y = Setting p = xy, we hve 0 p x +y = nd x + y = x + y + xy = + p It now beomes + p 64p or p 5p + p + 0 We wnt to minimize F p = p 5p + p + on [0, ] We ompute F p = p p We find tht F is monotone inresing on [0, ] nd monotone deresing on [, ] Sine F 0 = nd F = 0, we onlude tht F p F = 0 for ll p [0, ] Third Solution We show tht x + y 6 x + y xy where x y 0 We mke the substitution u = x + y nd v = x y Then, we hve u v 0 It beomes u u 6 + v u v or u 6 u + v u v 4 Note tht u 4 u 4 v 4 0 nd tht u u v 0 So, u 6 u 4 v 4 u v = u + v u v Problem 8 IMO 984/ Let x, y, z be nonnegtive rel numbers suh tht x + y + z = Prove tht 0 xy + yz + zx xyz 7 7 First Solution Let fx, y, z = xy + yz + zx xyz We my ssume tht 0 x y z Sine x + y + z =, this implies tht x It follows tht fx, y, z = xyz + xyz + zx + xy 0 Applying the AM-GM inequlity, we obtin yz y+z = x Sine x 0, this implies tht x fx, y, z = xy + z + yz x x x + x = x + x + 4 Our job is now to mximize one-vrible funtion F x = 4 x + x +, where x [ 0, ] Sine F x = x x 0 on [ ] 0,, we onlude tht F x F = 7 7 for ll x [ 0, ] 0

23 IMO 000/ Let, b, be positive numbers suh tht b = Prove tht + b + + b Fifth Solution bsed on work by n IMO 000 ontestnt from Jpn Sine b =, t lest one of, b, is greter thn or equl to Sy b Putting = b, it beomes + b + b b b + or Setting x = b, it beomes f b x 0, where b b b b + b b + b b b + 0 f b t = t + b b t bt + bt t b t b + Fix positive number b We need to show tht F t := f b t 0 for ll t 0 It follows from b tht the ubi polynomil F t = t b + t b b + hs two rel roots b + 4b 7b + 4 nd λ = b + + 4b 7b + 4 Sine F hs lol minimum t t = λ, we find tht F t Min {F 0, F λ} for ll t 0 We hve to prove tht F 0 0 nd F λ 0 We hve F 0 = b b b + = b b + 0 It remins to show tht F λ 0 Notie tht λ is root of F / t After long division, we get F t = F t t b + + 8b + 4b 8t + 8b 7b 7b Putting t = λ, we hve F λ = 9 8b + 4b 8λ + 8b 7b 7b + 8 Thus, our job is now to estblish tht, for ll b 0, 8b b + + 4b + 4b 8 7b b 7b 7b + 8 0, whih is equivlent to 6b 5b 5b + 6 8b 4b + 8 4b 7b + 4 Sine both 6b 5b 5b + 6 nd 8b 4b + 8 re positive, 5 it s equivlent to or 6b 5b 5b + 6 8b 4b + 8 4b 7b b 5 75b b 75b + 864b 0 or 864b 4 75b + 50b 75b Let Gx = 864x 4 75x + 50x 75x We prove tht Gx 0 for ll x R We find tht G x = 456x 05x x 75 = x 456x 6669x + 75 Sine 456x 6669x + 75 > 0 for ll x R, we find tht Gx nd x hve the sme sign It follows tht G is monotone deresing on, ] nd monotone inresing on [, We onlude tht G hs the globl minimum t x = Hene, Gx G = 0 for ll x R 5 It s esy to hek tht 6b 5b 5b + 6 = 6b b b + + b + b > 6b b 0 nd 8b 4b + 8 = 8b + b > 0

24 4 Estblishing New Bounds We first give two lterntive wys to prove Nesbitt s inequlity Nesbitt For ll positive rel numbers, b,, we hve b + + Proof 4 From b+ 0, we dedue tht It follows tht Proof 5 We lim tht b + 4 b b+ b+ + = b + + b 8 b 4 + b + 8 b 4 + b + = b + or + b + b + + b + The AM-GM inequlity gives +b +b b nd + + Adding these two inequlities yields + b + b +, s desired Therefore, we hve b + + b + = Some inequlities n be proved by finding new bounds Suppose tht we wnt to estblish tht F x, y, z C If funtion G stisfies F x, y, z Gx, y, z for ll x, y, z > 0, nd Gx, y, z = C for ll x, y, z > 0, then, we dedue tht F x, y, z Gx, y, z = C For exmple, if funtion F stisfies F x, y, z x x + y + z for ll x, y, z > 0, then, tking the sum yields F x, y, z As we sw in the bove two proofs of Nesbitt s inequlity, there re vrious lower bounds Problem 9 Let, b, be the lengths of tringle Show tht b + + b b <

25 Proof We don t employ the Rvi substitution It follows from the tringle inequlity tht b + < = + b + One dy, I tried finding new lower bound of x + y + z where x, y, z > 0 There re well-known lower bounds suh s xy + yz + zx nd 9xyz But I wnted to find quite different one I tried breking the symmetry of the three vribles x, y, z Note tht x + y + z = x + y + z + xy + xy + yz + yz + zx + zx I pplied the AM-GM inequlity to the right hnd side exept the term x : It follows tht y + z + xy + xy + yz + yz + zx + zx 8x y 4 z 4 x + y + z x + 8x y 4 z 4 = x x + 8y 4 z 4 IMO 00/ Let, b, be positive rel numbers Prove tht + 8b + b b b Seond Solution We find tht the bove inequlity lso gives nother lower bound of x + y + z, tht is, x + y + z x x + 8y 4 z 4 It follows tht x 4 x + 8y 4 z 4 x x + y + z = After the substitution x = 4, y = b 4, nd z = 4, it now beomes + 8b Problem 0 IMO 005/ Let x, y, nd z be positive numbers suh tht xyz Prove tht x 5 x x 5 + y + z + y5 y y 5 + z + x + z5 z z 5 + x + y 0 First Solution It s equivlent to the following inequlity x x 5 y x 5 + y + z + y 5 + y 5 + z + x + z z 5 + z 5 + x + y + or x + y + z x 5 + y + z + x + y + z y 5 + z + x + x + y + z z 5 + x + y With the Cuhy-Shwrz inequlity nd the ft tht xyz, we hve x 5 + y + z yz + y + z x + y + z or x + y + z x 5 + y + z yz + y + z x + y + z Tking the sum nd x + y + z xy + yz + zx give us x + y + z x 5 + y + z + x + y + z y 5 + z + x + x + y + z xy + yz + zx z 5 + x + + y x + y + z

26 Seond Solution The min ide is to think of s follows : x 5 x 5 + y + z + y 5 y 5 + z + x + z 5 z 5 + x + y x x 5 + y + z + y y 5 + z + x + z z 5 + x + y We first show the left-hnd It follows from y 4 + z 4 y z + yz = yzy + z tht xy 4 + z 4 xyzy + z y + z or x 5 x 5 + y + z x 5 x 5 + xy 4 + xz 4 = x 4 x 4 + y 4 + z 4 Tking the sum, we hve the required inequlity It remins to show the right-hnd [First Wy] As in the first solution, the Cuhy-Shwrz inequlity nd xyz imply tht x 5 + y + z yz + y + z x + y + z or x yz + y + z x + y + z x x 5 + y + z Tking the sum, we hve x yz + y + z x + y + z Our job is now to estblish the following homogeneous inequlity x yz + y + z x + y + z x + y + z However, by the AM-GM inequlity, we obtin [Seond Wy] We lim tht x 4 = x 4 + y 4 x y = x x 5 + y + z x y + x yz x 4 x yz y x + z x yz x 4 + y 4 + z 4 + 4x y + 4x z x 4x + y + z x 5 + y + z We do this by proving x 4 + y 4 + z 4 + 4x y + 4x z x yz 4x + y + z x 4 + y z + yz beuse xyz implies tht Hene, we need to show the homogeneous inequlity x yz x 4 + y z + yz = x x 5 xyz + y + z x x 5 + y + z x 4 + y 4 + z 4 + 4x y + 4x z x 4 + y z + yz 4x yzx + y + z However, this is strightforwrd onsequene of the AM-GM inequlity x 4 + y 4 + z 4 + 4x y + 4x z x 4 + y z + yz 4x yzx + y + z = x 8 + x 4 y 4 + x 6 y + x 6 y + y 7 z + y z 5 + x 8 + x 4 z 4 + x 6 z + x 6 z + yz 7 + y 5 z +x 6 y + x 6 z 6x 4 y z 6x 4 yz x 6 yz 6 6 x 8 x 4 y 4 x 6 y x 6 y y 7 z y z x 8 x 4 z 4 x 6 z x 6 z yz 7 y 5 z = 0 + x 6 y x 6 z 6x 4 y z 6x 4 yz x 6 yz Tking the sum, we obtin = x 4 + y 4 + z 4 + 4x y + 4x z 4x + y + z x x 5 + y + z 4

27 Third Solution by n IMO 005 ontestnt Iurie Boreio 6 from Moldov We estblish tht It follows immeditely from the identity x 5 x x 5 + y + z x 5 x x x + y + z x 5 x x 5 + y + z x 5 x x x + y + z = x x y + z x x + y + z x 5 + y + z Tking the sum nd using xyz, we hve x 5 x x 5 + y + z x 5 + y + z x x x 5 + y + z x yz 0 Exerise 5 USAMO Summer Progrm 00 Let, b, be positive rel numbers Prove tht Hint [TJM] Estblish the inequlity b + + b b b+ Exerise 6 APMO 005 b = 8,, b, > 0 +b+ + + b + b + b Hint Use the inequlity +x +x to give lower bound of the left hnd side 6 He reeived the speil prize for this solution 5

28 Chpter Homogeniztions nd Normliztions Every Mthemtiin Hs Only Few Triks A long time go n older nd well-known number theorist mde some disprging remrks bout Pul Erdös s work You dmire Erdos s ontributions to mthemtis s muh s I do, nd I felt nnoyed when the older mthemtiin fltly nd definitively stted tht ll of Erdos s work ould be redued to few triks whih Erdös repetedly relied on in his proofs Wht the number theorist did not relize is tht other mthemtiins, even the very best, lso rely on few triks whih they use over nd over Tke Hilbert The seond volume of Hilbert s olleted ppers ontins Hilbert s ppers in invrint theory I hve mde point of reding some of these ppers with re It is sd to note tht some of Hilbert s beutiful results hve been ompletely forgotten But on reding the proofs of Hilbert s striking nd deep theorems in invrint theory, it ws surprising to verify tht Hilbert s proofs relied on the sme few triks Even Hilbert hd only few triks! Gin-Crlo Rot, Ten Lessons I Wish I Hd Been Tught, Noties of the AMS, Jnury 997 Homogeniztions Mny inequlity problems ome with onstrints suh s b =, xyz =, x+y +z = A non-homogeneous symmetri inequlity n be trnsformed into homogeneous one Then we pply two powerful theorems : Shur s inequlity nd Muirhed s theorem We begin with simple exmple Problem Hungry 996 Let nd b be positive rel numbers with + b = Prove tht + + b b + Solution Using the ondition + b =, we n redue the given inequlity to homogeneous one, i e, + b + + b + b + bb + + b or b + b + b, whih follows from +b b+b = b +b 0 The equlity holds if nd only if = b = The bove inequlity b + b + b n be generlized s following : Theorem Let,, b, b be positive rel numbers suh tht + = b + b nd mx, mxb, b Let x nd y be nonnegtive rel numbers Then, we hve x y + x y x b y b + x b y b Proof Without loss of generlity, we n ssume tht, b b, b If x or y is zero, then it lerly holds So, we ssume tht both x nd y re nonzero It follows from + = b + b tht = b + b It s esy to hek x y + x y x b y b x b y b = x y x + y x b y b x b y b = x y x b y b x b y b = x y x b y b x b y b 0 6

29 Remrk When does the equlity hold in the theorem 8? We now introdue two summtion nottions nd sym Let P x, y, z be three vribles funtion of x, y, z Let us define : P x, y, z = P x, y, z + P y, z, x + P z, x, y, P x, y, z = P x, y, z + P x, z, y + P y, x, z + P y, z, x + P z, x, y + P z, y, x sym For exmple, we know tht x y = x y + y z + z x, x = x + y + z sym x y = x y + x z + y z + y x + z x + z y, sym xyz = 6xyz Problem IMO 984/ Let x, y, z be nonnegtive rel numbers suh tht x + y + z = Prove tht 0 xy + yz + zx xyz 7 7 Seond Solution Using the ondition x + y + z =, we redue the given inequlity to homogeneous one, i e, 0 xy + yz + zxx + y + z xyz 7 7 x + y + z The left hnd side inequlity is trivil beuse it s equivlent to 0 xyz + sym x y sym The right hnd side inequlity simplifies to 7 In the view of 7 x + 5xyz 6 sym x + 5xyz 6 sym x y = x sym x y 0 x y + 5 xyz + x sym x y, it s enough to show tht x x y nd xyz + x x y sym sym We note tht x sym x y = x + y x y + xy = x + y x y xy 0 The seond inequlity n be rewritten s xx yx z 0, whih is prtiulr se of Shur s theorem in the next setion After homogenizing, sometimes we n find the right pproh to see the inequlities: 7

30 Irn 998 Prove tht, for ll x, y, z > suh tht x + y + z =, x + y + z x + y + z Seond Solution After the lgebri substitution = x, b = y, = z, we re required to prove tht + b + b + +, b where, b, 0, nd +b+ = Using the onstrint +b+ =, we obtin homogeneous inequlity or + b + + b + + b + +b+ + +b+ b + b +b+ + b + b + + b + b + +, b whih immeditely follows from the Cuhy-Shwrz inequlity [b b + + b ] + b + b + + b + b + + b 8

31 Shur s Inequlity nd Muirhed s Theorem Theorem Shur Let x, y, z be nonnegtive rel numbers For ny r > 0, we hve x r x yx z 0 Proof Sine the inequlity is symmetri in the three vribles, we my ssume without loss of generlity tht x y z Then the given inequlity my be rewritten s x y[x r x z y r y z] + z r x zy z 0, nd every term on the left-hnd side is lerly nonnegtive Remrk When does the equlity hold in Shur s Inequlity? Exerise 7 Disprove the following proposition: For ll, b,, d 0 nd r > 0, we hve r b d + b r b b db + r d + d r d d bd 0 The following speil se of Shur s inequlity is useful : xx yx z 0 xyz + x sym x y sym xyz + sym x sym Corollry Let x, y, z be nonnegtive rel numbers Then, we hve xyz + x + y + z xy + yz + zx Proof By Shur s inequlity nd the AM-GM inequlity, we hve xyz + x x y + xy xy x y We now use Shur s inequlity to give n lterntive solution of APMO 004/5 Prove tht, for ll positive rel numbers, b,, Seond Solution After expnding, it beomes 8 + b + + b + + 9b + b + b b From the inequlity b + b + 0, we obtin 6 + b 4 b Hene, it will be enough to show tht + b b Sine + b + b + b +, it will be enough to show tht + b + b, whih is prtiulr se of the following result for t = 9

32 Corollry Let t 0, ] For ll, b, 0, we hve t + tb t + b In prtiulr, we obtin non-homogeneous inequlities 5 + b4 + + b + b + b +, + b + + b + b + b +, + b + + b + b + b + Proof After setting x =, y = b, z =, it beomes t + txyz t + By the orollry, it will be enough to show tht x xy t + txyz t xyz, whih is strightforwrd onsequene of the weighted AM-GM inequlity : t + t xyz t t One my hek tht the equlity holds if nd only if = b = = xyz t t = xyz IMO 000/ Let, b, be positive numbers suh tht b = Prove tht + b + + b Seond Solution It is equivlent to the following homogeneous inequlity : b / + b/ b b b / + b/ b / + b/ b After the substitution = x, b = y, = z with x, y, z > 0, it beomes x xyz + y xyz xyz + z xyz xyz + xyz x y z, whih simplifies to y z x y y z + z x y z z x + x y z x x y + y z x y z x or x y z + x 6 y x 4 y 4 z + x 5 y z or x yy zz x + x y x y y z sym whih is speil se of Shur s inequlity Here is nother inequlity problem with the onstrint b = For n lterntive homogeniztion, see the problem in the hpter 0

33 Problem Tournment of Towns 997 Let, b, be positive numbers suh tht b = Prove tht + b + + b Solution We n rewrite the given inequlity s following : + b + b + / b + + b + / + + b / b / We mke the substitution = x, b = y, = z with x, y, z > 0 Then, it beomes x + y + xyz + y + z + xyz + z + x + xyz xyz whih is equivlent to xyz x + y + xyzy + z + xyz x + y + xyzy + z + xyzz + x + xyz or We pply the theorem 9 to obtin sym x 6 y sym x 5 y z! x 6 y = x 6 y + y 6 x sym x 5 y 4 + y 5 x 4 = x 5 y 4 + z 4 x 5 y z + y z = sym x 5 y z Exerise 8 [TZ], pp4 Prove tht for ny ute tringle ABC, ot A + ot B + ot C + 6 ot A ot B ot C ot A + ot B + ot C Exerise 9 Kore 998 Let I be the inenter of tringle ABC Prove tht IA + IB + IC BC + CA + AB Exerise 0 [IN], pp0 Let, b, be the lengths of tringle Prove tht b + + b + b + + b > + b + + b Exerise Surányi s inequlity Show tht, for ll x,, x n 0, n x n + x n n + nx x n x + x n x n n + x n Theorem Muirhed Let,,, b, b, b be rel numbers suh tht 0, b b b 0, b, + b + b, + + = b + b + b Let x, y, z be positive rel numbers Then, we hve sym x y z sym xb y b z b

34 Proof Cse b : It follows from + b nd from b tht mx + b, b so tht mx, = mx + b, b From + b b + b = nd + b b b, we hve mx + b, mxb, b Apply the theorem 8 twie to obtin x y z = z x y + x y sym z x + b y b + x b y + b = x b y + b z + y z + b x b y b z b + y b z b = sym x b y b z b Cse b : It follows from b b + b + b = + + b + + tht b + b nd tht b + b Therefore, we hve mx, mxb, + b nd mx, + b mxb, b Apply the theorem 8 twie to obtin x y z = x y z + y z sym x y b z + b + y + b z b = y b x z + b + x + b z y b x b z b + x b z b = sym x b y b z b Remrk The equlity holds if nd only if x = y = z However, if we llow x = 0 or y = 0 or z = 0, then one my esily hek tht the equlity holds when,, > 0 nd b, b, b > 0 if nd only if x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0 We n use Muirhed s theorem to prove Nesbitt s inequlity Nesbitt For ll positive rel numbers, b,, we hve b + + b b Proof 6 Clering the denomintors of the inequlity, it beomes + b + + bb + + or sym sym IMO 995 Let, b, be positive numbers suh tht b = Prove tht b + + b b b

35 Seond Solution It s equivlent to b + + b b b 4/ Set = x, b = y, = z with x, y, z > 0 Then, it beomes denomintors, this beomes or sym x y sym sym x y + sym x y 9 z + sym x 9 y 9 z 6 sym x y 8 z 5 + sym x y 9 z sym x y 8 z 5 + x 9 y +z x 4 y 4 z 4 x y 8 z 5 + 6x 8 y 8 z 8 nd every term on the left hnd side is nonnegtive by Muirhed s theorem sym x 9 y 9 z 6 sym Problem 4 Irn 996 Let x, y, z be positive rel numbers Prove tht xy + yz + zx x + y + y + z + z + x 9 4 Proof It s equivlent to We rewrite this s following 4 x 5 y + x 4 yz + 6x y z x 4 y 6 x y x y z 0 sym sym sym sym x 5 y sym x 4 y + sym x 5 y sym x y + xyz xyz + x sym By Muirhed s theorem nd Shur s inequlity, it s sum of three nonnegtive terms Problem 5 Let x, y, z be nonnegtive rel numbers with xy + yz + zx = Prove tht x + y + y + z + z + x 5 Proof Using xy + yz + zx =, we homogenize the given inequlity s following : x 8 y 8 z 8 0, x y 0 Clering or xy + yz + zx x + y + y + z + z + x 4 x 5 y + x 4 yz + 4 sym sym sym x y z + 8x y z sym 5 x 4 y + sym x y or sym x 5 y sym x 4 y + sym x 5 y sym x y + xyz sym x + 4 sym x y + 8xyz By Muirhed s theorem, we get the result In the bove inequlity, without the ondition xy + yz + zx =, the equlity holds if nd only if x = y, z = 0 or y = z, x = 0 or z = x, y = 0 Sine xy + yz + zx =, the equlity ours when x, y, z =,, 0,, 0,, 0,, 0

36 Normliztions In the previous setions, we trnsformed non-homogeneous inequlities into homogeneous ones On the other hnd, homogeneous inequlities lso n be normlized in vrious wys We offer two lterntive solutions of the problem 8 by normliztions : IMO 00/ Let, b, be positive rel numbers Prove tht + 8b + b b b Third Solution We mke the substitution x = +b+, y = b +b+, z = xfx + 8yz + yfy + 8zx + zfz + 8xy, +b+ The problem is where ft = t Sine f is onvex on R + nd x + y + z =, we pply the weighted Jensen s inequlity to obtin xfx + 8yz + yfy + 8zx + zfz + 8xy fxx + 8yz + yy + 8zx + zz + 8xy Note tht f = Sine the funtion f is stritly deresing, it suffies to show tht xx + 8yz + yy + 8zx + zz + 8xy Using x + y + z =, we homogenize it s x + y + z xx + 8yz + yy + 8zx + zz + 8xy However, this is esily seen from x + y + z xx + 8yz yy + 8zx zz + 8xy = [xy z + yz x + zx y ] 0 In the bove solution, we normlized to x + y + z = We now prove it by normlizing to xyz = Fourth Solution We mke the substitution x = b, y = b, z = b Then, we get xyz = nd the inequlity beomes x + 8y + 8z whih is equivlent to + 8x + 8y + 8x + 8y + 8z After squring both sides, it s equivlent to 8x + y + z + + 8x + 8y + 8z + 8x 50 Rell tht xyz = The AM-GM inequlity gives us x + y + z, + 8x + 8y + 8z 9x 8 9 9y 8 9 9z 8 9 = 79 nd Using these three inequlities, we get the result + 8x IMO 98/6 Let, b, be the lengths of the sides of tringle Prove tht b b + b b + 0 Dividing by + b + gives the equivlent inequlity P r +b+ +b+ + 8b +b+ 9x 8 9 9xyz 4 7 = 9 4

37 Seond Solution After setting = y + z, b = z + x, = x + y for x, y, z > 0, it beomes x z + y x + z y x yz + xy z + xyz or x y + y z + z x x + y + z Sine it s homogeneous, we n restrit our ttention to the se x + y + z = Then, it beomes x y z yf + zf + xf, y z x where ft = t Sine f is onvex on R, we pply the weighted Jensen s inequlity to obtin x y z yf + zf + xf f y y z x x y + z y z + x z = f = x Problem 6 KMO Winter Progrm Test 00 Prove tht, for ll, b, > 0, b + b + b + b + b + + b b + b + b First Solution Dividing by b, it beomes + b + b + b + b b b + b + + b + After the substitution x = b, y = b, z =, we obtin the onstrint xyz = It tkes the form x x + y + z xy + yz + zx + y z z + x + y + From the onstrint xyz =, we find two identities x y z + x + z x + z y + x z + y y + = = z + xx + yy + z, z x y x + y + z xy + yz + zx = x + yy + zz + x + xyz = x + yy + zz + x + Letting p = x + yy + zz + x, the inequlity now beomes p + + p Applying the AM-GM inequlity, we hve p xy yz zx = It follows tht p + +p = pp+p 0 Problem 7 IMO 999/ Let n be n integer with n Determine the lest onstnt C suh tht the inequlity i<j n x i x j x i + x j C holds for ll rel numbers x,, x n 0 b For this onstnt C, determine when equlity holds i n First Solution Mrin E Kuzm For x = = x n = 0, it holds for ny C 0 Hene, we onsider the se when x + +x n > 0 Sine the inequlity is homogeneous, we my normlize to x + +x n = We denote F x,, x n = x i x j x i + x j I slightly modified his solution in [Au99] i<j n x i 4 5