Topics in Inequalities - Theorems and Techniques

Transcription

1 Topics in Inequlities - Theorems nd Techniques Hojoo Lee The st edition 8th August, 007 Introduction Inequlities re useful in ll fields of Mthemtics The im of this problem-oriented book is to present elementry techniques in the theory of inequlities The reders will meet clssicl theorems including Schur s inequlity, Muirhed s theorem, the Cuchy-Schwrz inequlity, the Power Men inequlity, the AM-GM inequlity, nd Hölder s theorem I would gretly pprecite hering bout comments nd corrections from my reders You cn send emil to me t ultrmetric@gmilcom To Students My trget reders re chllenging high schools students nd undergrdute students The given techniques in this book re just the tip of the inequlities iceberg Young students should find their own methods to ttck vrious problems A gret Hungrin Mthemticin Pul Erdös ws fond of sying tht God hs trnsfinite book with ll the theorems nd their best proofs I strongly encourge reders to send me their own cretive solutions of the problems in this book Hve fun! Acknowledgement I m indebted to Orlndo Döhring nd Drij Grinberg for providing me with TeX files including collections of interesting inequlities I d like to thnk Mrin Muresn for his excellent collection of problems I m lso plesed tht Co Minh Qung sent me vrious vietnm problems nd nice proofs of Nesbitt s inequlity I owe gret debts to Stnley Rbinowitz who kindly sent me his pper On The Computer Solution of Symmetric Homogeneous Tringle Inequlities Resources 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 Geometric Inequlities Rvi Substitution Trigonometric Methods 7 Applictions of Complex Numbers Four Bsic Techniques 5 Trigonometric Substitutions 5 Algebric Substitutions 9 Incresing Function Theorem 6 4 Estblishing New Bounds 9 Homogeniztions nd Normliztions 4 Homogeniztions 4 Schur s Inequlity nd Muirhed s Theorem 7 Normliztions 4 4 Cuchy-Schwrz Inequlity nd Hölder s Inequlity 48 4 Convexity 54 4 Jensen s Inequlity 54 4 Power Mens 58 4 Mjoriztion Inequlity 6 44 Supporting Line Inequlity 6 5 Problems, Problems, Problems 66 5 Multivrible Inequlities 66 5 Problems for Putnm Seminr 75 II

3 Chpter Geometric 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 clssicl inequlity in tringle geometry Wht is the first nontrivil geometric inequlity? In 746, Chpple showed tht Theorem Chpple 746, Euler 765 Let R nd r denote the rdii of the circumcircle nd incircle 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 = c, s = +b+c nd S = [ABC] Recll the well-known identities : S = bc 4R, S = rs, S = ss s bs c Hence, R r is equivlent to bc 4S S S s or bc 8 s or bc 8s s bs c We need to prove the following Theorem [AP], A Pdo Let, b, c be the lengths of tringle Then, we hve bc 8s s bs c or bc b + c c + b + b c nd the equlity holds if nd only if = b = c Proof We use the Rvi Substitution : Since, b, c re the lengths of tringle, there re positive rels x, y, z such tht = y + z, b = z + x, c = 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 Exercise Let ABC be right tringle Show tht When does the equlity hold? R + r The first geometric inequlity is the Tringle Inequlity : AB + BC AC In this book, [P ] stnds for the re of the polygon P

4 It s nturl to sk tht the inequlity in the theorem holds for rbitrry positive rels, b, c? Yes! It s possible to prove the inequlity without the dditionl condition tht, b, c 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 Since the inequlity is symmetric 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 cse, 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 Proof Since x, y, z 0, we cn find positive sequences {x n }, {y n }, {z n } for which 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 fct, 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 Hence, the theorem 4 is prticulr cse of Schur s inequlity Problem IMO 000/, Proposed by Titu Andreescu Let, b, c be positive numbers such tht bc = Prove tht + b + c + b c First Solution Since bc =, we mke the substitution = x y, b = y z, c = 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 For exmple, tke x =, y =, z = b

5 The Rvi Substitution is useful for inequlities for the lengths, b, c of tringle After the Rvi Substitution, we cn remove the condition tht they re the lengths of the sides of tringle Problem IMO 98/6 Let, b, c be the lengths of the sides of tringle Prove tht b b + b cb c + c c 0 First Solution After setting = y + z, b = z + x, c = x + y for x, y, z > 0, it becomes x z + y x + z y x yz + xy z + xyz or x y + y z + z x x + y + z, which follows from the Cuchy-Schwrz inequlity x y + z + x y + y z + z x + y + z x Exercise Let, b, c be the lengths of tringle Show tht b + c + b c + + c + b < Exercise Drij Grinberg Let, b, c be the lengths of tringle Show the inequlities nd + b + c + bc b c b c 0, b + b c + c bc b c b c 0 We now discuss Weitzenböck s inequlity nd relted inequlities Problem IMO 96/, Weitzenböck s inequlity Let, b, c be the lengths of tringle with re S Show tht + b + c 4 S Solution Write = y + z, b = z + x, c = x + y for x, y, z > 0 It s equivlent to which cn 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, c nd re F, the following inequlity holds b + bc + c + b + c 4 F First Proof After the substitution = y + z, b = z + x, c = x + y, where x, y, z > 0, it becomes which follows from the identity xy + yz + zx xyzx + y + z, xy + yz + zx xyzx + y + z = xy yz + yz zx + zx xy

6 Second Proof We give convexity proof There re mny wys to deduce the following identity: b + bc + c + b + c 4F Since tn x is convex on 0, π, Jensen s inequlity shows tht b + bc + c + b + c 4F = tn A + tn B + tn C tn A + B + C = Tsintsifs proved simultneous generliztion of Weitzenböck s inequlity nd Nesbitt s inequlity Theorem 6 Tsintsifs Let p, q, r be positive rel numbers nd let, b, c denote the sides of tringle with re F Then, we hve p q + r + q r + p b + r p + q c F Proof V Pmbuccin By Hdwiger-Finsler inequlity, it suffices to show tht or or p q + r + p + q + r q + r q r + p b + + p + q + r r p + q c + b + c + b + c r + p b + p + q + r p + q c + b + c q + r + r + p + p + q q + r + r + p b + p + q c + b + c However, this is strightforwrd consequence of the Cuchy-Schwrz inequlity Theorem 7 Neuberg-Pedoe inequlity Let, b, c denote the sides of the tringle A B C with re F Let, b, c denote the sides of the tringle A B C with re F Then, we hve b + c + b c + b + c + b c 6F F Notice tht it s generliztion of Weitzenböck s inequlitywhy? In [GC], G Chng proved Neuberg-Pedoe inequlity by using complex numbers For very interesting geometric observtions nd proofs of Neuberg-Pedoe inequlity, see [DP] or [GI, pp9-9] Here, we offer three lgebric proofs Lemm + b c + b b + c + c c + b > 0 Proof Observe tht it s equivlent to + b + c + b + c > + b b + c c From Heron s formul, we find tht, for i =,, 6F i = i + b i + c i i 4 + b i 4 + c i 4 > 0 or i + b i + c i > 4 i 4 + b i 4 + c i 4

7 The Cuchy-Schwrz inequlity implies tht +b +c +b +c > 4 + b 4 + c b 4 + c 4 +b b +c c First Proof [LC], Crlitz By the lemm, we obtin L = b + c + b c + b + c + b c > 0, Hence, we need to show tht One my esily check the following identity where Using the identity one my lso deduce tht L 6F 6F 0 L 6F 6F = 4UV + V W + W U, U = b c b c, V = c c nd W = b b U + b V + c W = 0 or W = c U b c V, UV + V W + W U = U c b V 4 b c b c 4 V c It follows tht UV + V W + W U = U c b V 6F c 4 c V 0 Crlitz lso observed tht the Neuberg-Pedoe inequlity cn be deduced from Aczél s inequlity Theorem 8 Aczél 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 Cuchy-Schwrz 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 5

8 In cse + + n = 0, it s trivil Hence, we now ssume tht + + n > 0 The min trick is to think of the following qudrtic 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= Since P b = n i= b i b i 0 nd since the coefficient of x in the qudrtic polynomil P is positive, P should hve t lest one rel root Therefore, P hs nonnegtive discriminnt It follows tht n n n b i b i 4 i b b i 0 i= i= i= i= i= Second Proof of Neuberg-Pedoe inequlity [LC], Crlitz We rewrite it in terms of, b, c,, b, c : + b + c + b + c + b b + c c + b + c 4 + b 4 + c 4 + b + c 4 + b 4 + c 4 We employ the following substitutions x = + b + c, x =, x = b, x 4 = c, y = + b + c, y =, y = b, y 4 = c As in the proof of the lemm 5, we hve x > x + y + x 4 nd y > y + y + y 4 We now pply Aczél s inequlity to get the inequlity x y x y x y x 4 y 4 x x + y + x 4 y y + y + y 4 We close this section with very simple proof by former student in KMO 4 summer progrm 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 + c + b c + b + c + b c = 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 4 Koren Mthemticl Olympids 6

9 Trigonometric Methods In this section, we employ trigonometric methods to ttck geometric inequlities Theorem Erdös-Mordell Theorem If from point P inside given tringle ABC perpendiculrs 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 conjectured 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 computtions 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 trigonometric inequlity Let h = P H, h = P H nd h = P H Apply the Since Lw nd the Cosine Lw to obtin P A sin A = H H = h + h h h cosπ A, P B sin B = H H = h + h h h cosπ B, P C sin C = H H = h + h h h cosπ C So, we need to prove tht h + h h h cosπ 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 cosπ A = h + h h h cosb + C = h + h h h cos B cos C sin B sin C Using cos B + sin B = nd cos C + sin C =, one finds tht H H = h sin C + h sin B + h cos C h cos B Since h cos C h cos B is clerly nonnegtive, we get H H h sin C + h sin B It follows tht h + h h h cosπ 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 7

10 We use the sme techniques to ttck the following geometric inequlity Problem 4 IMO Short-list 005 In n cute tringle ABC, let D, E, F, P, Q, R be the feet of perpendiculrs from A, B, C, A, B, C to BC, CA, AB, EF, F D, DE, respectively Prove tht where pt denotes the perimeter of tringle T pabcpp QR pdef, Solution Let s euler 5 this problem Let ρ be the circumrdius of the tringle ABC It s esy to show tht BC = ρ sin A nd EF = ρ sin A cos A Since DQ = ρ sin C cos B cos A, DR = ρ sin B cos C cos A, nd F DE = π A, the Cosine Lw gives us QR = DQ + DR DQ DR cosπ A [ ] = 4ρ cos A sin C cos B + sin B cos C + sin C cos B sin B cos C cosa or where QR = ρ cos A fa, B, C, fa, B, C = sin C cos B + sin B cos C + sin C cos B sin B cos C cosa So, wht we need to ttck is the following inequlity: ρ sin A ρ cos A fa, B, C ρ sin A cos A or sin A cos A fa, B, C sin A cos A Our job is now to find resonble lower bound of fa, B, C Once gin, we express fa, B, C s the sum of two squres We observe tht fa, B, C = sin C cos B + sin B cos C + sin C cos B sin B cos C cosa = sin C cos B + sin B cos C + sin C cos B sin B cos C [ + cosa] = sin C + B sin C cos B sin B cos C sin A = sin A [ 4 sin B sin C cos B cos C] So, we shll express 4 sin B sin C cos B cos C s the sum of two squres The trick is to replce with sin B + cos B sin C + cos C Indeed, we get 4 sin B sin C cos B cos C = sin B + cos B sin C + cos C 4 sin B sin C cos B cos C = sin B cos C sin C cos B + cos B cos C sin B sin C = sin B C + cos B + C = sin B C + cos A 5 euler v in Mthemtics trnsform the problems in tringle geometry to trigonometric ones 8

11 It therefore follows tht so tht fa, B, C = sin A [ sin B C + cos A ] sin A cos A cos A fa, B, C sin A cos A So, we cn complete the proof if we estblish tht sin A sin A cos A sin A cos A Indeed, one sees tht it s direct consequence of the Cuchy-Schwrz inequlity p + q + rx + y + z px + qy + rz, where p, q, r, x, y nd z re positive rel numbers Alterntively, one my obtin nother lower bound of fa, B, C: fa, B, C = sin C cos B + sin B cos C + sin C cos B sin B cos C cosa = sin C cos B sin B cos C + sin C cos B sin B cos C [ + cosa] = sin B C + sinb sinc cos A cos A sinb sinc Then, we cn use this to offer lower bound of the perimeter of tringle P QR: pp QR = ρ cos A fa, B, C ρ cos A sin B sin C So, one my consider the following inequlity: pabc ρ cos A sin B sin C pdef or or ρ sin A ρ cos A sin B sin C ρ sin A cos A sin A cos A sin B sin C sin A cos A However, it turned out tht this doesn t hold Try to disprove this! Problem 5 Let I be the incenter of the tringle ABC with BC =, CA = b nd AB = c Prove tht, for ll points X, XA + bxb + cxc bc 9

12 Proof This geometric inequlity follows from the following geometric identity: XA + bxb + cxc = + b + cxi + bc 6 There re mny wys to estblish this identity To euler this, we toss the picture on the crtesin plne so tht Ac cos B, c sin B, B0, 0 nd C, 0 Letting r be the inrdius of ABC nd s = +b+c, we get Is b, r It s well-known tht r = Set Xp, q On the one hnd, we obtin s s bs c s XA + bxb + cxc = [p c cos B + q c sin B ] + b p + q + c [ p + q ] = + b + cp cp + cos B + + b + cq cq sin B + c + c = sp cp + + c b + sq cq [ ABC] c c + c + c = sp p + c + b + c b + sq 4q[ ABC] + c + c = sp ps s b + sq 4qsr + c + c = sp 4s s b p + sq 4rsq + c + c On the other hnd, we obtin It follows tht + b + cxi + bc = s [ p s b + q r ] = s [ p s bp + s b + q qr + r ] = sp 4s s b p + ss b + sq 4rsq + sr + bc XA + bxb + cxc + b + cxi bc = c + c ss b sr bc = c + c ss b s s bs c bc = c + c b ss b s s bs c = cs b ss b s s bs c = s b [c ss b s s c] However, we compute c ss b s s c = s + + b + cs = 0 Problem 6 IMO 00/ Let ABC be n cute-ngled tringle with O s its circumcenter 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 cn be written s COP < P CO This cn be shown if we estblish the length inequlity OP > P C Since the power of P with respect to the circumcircle of ABC is OP = R BP P C, where R is the circumrdius of the tringle ABC, 6 IMO Short-list 988 0

13 it becomes 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 cos C Hence, we show the inequlity R > R sin A R sin B cos C or sin A sin B cos C < 4 Since sin A <, it suffices to show tht sin A sin B cos C < 4 Finlly, we use the ngle condition C B + 0 to obtin the trigonometric inequlity sin B cos C = sinb + C sinc B sinc B sin 0 = 4 We close this section with Brrows inequlity stronger thn Erdös-Mordell Theorem We need the following trigonometric inequlity: Proposition Let x, y, z, θ, θ, θ be rel numbers with θ + θ + θ = π Then, x + y + z yz cos θ + zx cos θ + xy cos θ Proof Using θ = π θ + θ, it s n esy job to check the following identity x + y + z yz cos θ + zx cos θ + xy cos θ = z x cos θ + y cos θ + x sin θ y sin θ 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 cos θ + q cos θ + r cos 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 bisectors of ngles BP C, CP A, AP B cut the sides BC,CA,AB respectively 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 deduce the following identities l = d d d + d cos θ, l = d d d + d cos θ, nd l = d d d + d cos θ, By the AM-GM inequlity nd the bove corollry, this mens tht l + l + l d d cos θ + d d cos θ + d d cos θ d + d + d As nother ppliction of the bove trigonometric 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 such tht θ + + θ 4 = π Then, x x + x x 4 x x + x x 4 x x 4 + x x x cos θ + x cos θ + x cos θ + x 4 cos θ 4 x x x x 4

14 Proof Let p = x +x x x + x +x 4 x x 4 q = x x +x x 4 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 cos θ + x cos θ + λ cosθ + θ 4 pλ = pq, nd x cos θ + x 4 cos θ 4 + λ cosθ + θ q λ = pq Since cosθ + θ 4 + cosθ + θ = 0, dding these two bove inequlities yields x cos θ + x cos θ + x cos θ + x 4 cos θ 4 x x + x x 4 x x + x x 4 x x 4 + x x pq = x x x x 4

15 Applictions of Complex Numbers In this section, we discuss some pplictions of complex numbers to geometric inequlity Every complex number corresponds to unique point in the complex plne The stndrd symbol for the set of ll complex numbers is C, nd we lso refer to the complex plne s C The min tool is pplictions of the following fundmentl inequlity Theorem If z,, z n C, then z + + z n z + + z n Proof Induction on n 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, c nd 0 be complex numbers tht correspond to A, B, C, D in the complex plne It becomes b c + b c c b Applying the Tringle Inequlity to the identity bc + b c = cb, we get the result Problem 7 [TD] Let P be n rbitrry point in the plne of tringle ABC with the centroid 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 check the first inequlity Regrd A, B, C, P s complex numbers nd ssume tht P corresponds 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 8 IMO Short-list 00 Let ABC be tringle for which there exists n interior point F such tht AF B = BF C = CF A Let the lines BF nd CF meet the sides AC nd AB t D nd E, respectively Prove tht AB + AC 4DE Solution Let AF = x, BF = y, CF = z nd let ω = cos π + i sin π We cn toss the pictures on C so tht the points F, A, B, C, D, nd E re represented by the complex numbers 0, x, yω, zω, d, nd e It s n esy exercise to estblish tht DF = xz xy x+z nd EF = x+y This mens tht d = xz xy 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 Since 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

16 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 cler 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 check tht the equlity holds if nd only if ABC is equilterl 4

17 Chpter Four Bsic Techniques Differentite! Shiing-shen Chern Trigonometric Substitutions If you re fced with n integrl tht contins squre root expressions such s z x dx, + y dy, dz then trigonometric substitutions such s x = sin t, y = tn t, z = sec t re very useful We will lern tht mking suitble trigonometric substitution simplifies the given inequlity Problem 9 APMO 004/5 Prove tht, for ll positive rel numbers, b, c, + b + c + 9b + bc + c First Solution Choose A, B, C 0, π with = tn A, b = tn B, nd c = tn C Using the well-known trigonometric identity + tn θ =, one my rewrite it s cos θ 4 cos A cos B cos C cos A sin B sin C + sin A cos B sin C + sin A sin B cos C 9 One my esily check the following trigonometric identity cosa + B + C = cos A cos B cos C cos A sin B sin C sin A cos B sin C sin A sin B cos C Then, the bove trigonometric inequlity tkes the form 4 cos A cos B cos C cos A cos B cos C cosa + B + C 9 Let θ = A+B+C Applying the AM-GM inequlity nd Jesen s inequlity, we hve cos A + cos B + cos C cos A cos B cos C cos θ We now need to show tht Using the trigonometric identity 4 9 cos θcos θ cos θ cos θ = 4 cos θ cos θ or cos θ cos θ = cos θ cos θ, 5

18 it becomes 4 7 cos4 θ cos θ, which follows from the AM-GM inequlity cos θ cos θ cos θ cos θ + cos θ + cos θ = One find tht the equlity holds if nd only if tn A = tn B = tn C = if nd only if = b = c = Problem 0 Ltvi 00 Let, b, c, d be the positive rel numbers such tht Prove tht bcd b c d 4 = First Solution We cn write = tn A, b = tn B, c = tn C, d = tn D, where A, B, C, D 0, π Then, the lgebric identity becomes the following trigonometric identity : Applying the AM-GM inequlity, we obtin Similrly, we obtin cos A + cos B + cos C + cos D = sin A = cos A = cos B + cos C + cos D cos B cos C cos D sin B cos C cos D cos A, sin C cos D cos A cos B, nd sin D cos A cos B cos C Multiplying these four inequlities, we get the result! Problem Kore 998 Let x, y, z be the positive rels with x + y + z = xyz Show tht + x + + y + + z Since the function f is not concve on R +, we cnnot pply Jensen s inequlity to the function ft = However, the function ftn θ is concve on +t 0, π! First Solution We cn write x = tn A, y = tn B, z = tn C, where A, B, C 0, π Using the fct tht + tn θ = cos θ, we rewrite it in the terms of A, B, C : cos A + cos B + cos C x+y It follows from tnπ C = z = xy = tna + B nd from π C, A + B 0, π tht π C = A + B or A + B + C = π Hence, it suffices to show the following Theorem In ny cute tringle ABC, we hve cos A + cos B + cos C Proof Since cos x is concve on 0, π, it s direct consequence of Jensen s inequlity 6

19 We note tht the function cos x is not concve on 0, π In fct, it s convex on π, π One my think tht the inequlity cos A + cos B + cos C doesn t hold for ny tringles However, it s known tht it holds for ll tringles Theorem In ny tringle ABC, we hve cos A + cos B + cos C First Proof It follows from π C = A + B tht cos C = cosa + B = cos A cos B + sin A sin B or cos A + cos B + cos C = sin A sin B + cos A + cos B 0 Second Proof Let BC =, CA = b, AB = c Use the Cosine Lw to rewrite the given inequlity in the terms of, b, c : b + c bc + c + b c + + b c b Clering denomintors, this becomes bc b + c + bc + b + c + b c, which is equivlent to bc b + c c + b + b c in the theorem In the first chpter, we found tht the geometric inequlity R r is equivlent to the lgebric inequlity bc b + c c + b + b c We now find tht, in the proof of the bove theorem, bc b + c c + b + b c is equivlent to the trigonometric inequlity cos A + cos B + cos C One my sk tht In ny tringles ABC, is there nturl reltion between cos A + cos B + cos C nd R r, where R nd r re the rdii of the circumcircle nd incircle of ABC? Theorem Let R nd r denote the rdii of the circumcircle nd incircle of the tringle ABC Then, we hve cos A + cos B + cos C = + r R Proof Use the identity b + c + bc + b + c + b c = bc + b + c c + b + b c We leve the detils for the reders Exercise 4 Let p, q, r be the positive rel numbers such tht p + q + r + pqr = Show tht there exists n cute tringle ABC such tht p = cos A, q = cos B, r = cos C b Let p, q, r 0 with p + q + r + pqr = Show tht there re A, B, C [ 0, π ] with p = cos A, q = cos B, r = cos C, nd A + B + C = π Problem USA 00 Let, b, nd c be nonnegtive rel numbers such tht +b +c +bc = 4 Prove tht 0 b + bc + c bc Solution Notice tht, b, c > implies tht + b + c + bc > 4 If, then we hve b + bc + c bc bc 0 We now prove tht b + bc + c bc Letting = p, b = q, c = r, we get p + q + r + pqr = By the bove exercise, we cn write [ = cos A, b = cos B, c = cos C for some A, B, C 0, π ] with A + B + C = π We re required to prove cos A cos B + cos B cos C + cos C cos A cos A cos B cos C 7

20 One my ssume tht A π or cos A 0 Note tht cos A cos B+cos B cos C+cos C cos A cos A cos B cos C = cos Acos B+cos C+cos B cos C cos A We pply Jensen s inequlity to deduce cos B + cos C cos A Note tht cos B cos C = cosb C + cosb + C cos A These imply tht cos A cos Acos B + cos C + cos B cos C cos A cos A cos A + cos A However, it s esy to verify tht cos A cos A + cos A cos A = 8

21 Algebric Substitutions We know tht some inequlities in tringle geometry cn be treted by the Rvi substitution nd trigonometric substitutions We cn lso trnsform the given inequlities into esier ones through some clever lgebric substitutions Problem IMO 00/ Let, b, c be positive rel numbers Prove tht + 8bc + b b + 8c + c c + 8b First Solution To remove the squre roots, we mke the following substitution : x = + 8bc, y = b b + 8c, z = c c + 8b Clerly, x, y, z 0, Our im is to show tht x + y + z We notice tht 8bc = x x, b 8c = Hence, we need to show tht y y, c 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 contrdiction! Problem 4 IMO 995/ Let, b, c be positive numbers such tht bc = Prove tht b + c + b c + + c + b First Solution After the substitution = x, b = y, c = z, we get xyz = The inequlity tkes the form x y + z + y z + x + It follows from the Cuchy-Schwrz inequlity tht x [y + z + z + x + x + y] y + z + so tht, by the AM-GM inequlity, z x + y y z + x + z x + y + z x + y x y + z + y z + x + z x + y x + y + z xyz = 9

22 Kore 998 Let x, y, z be the positive rels with x + y + z = xyz Show tht x + y + z Second Solution The strting point is letting = x, b = y, c = z We find tht + b + c = bc is equivlent to = xy + yz + zx The inequlity becomes 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 + y y + x Adding these three yields the required result nd z z z + xz + y z + x + z z + y We now prove clssicl theorem in vrious wys Theorem Nesbitt, 90 For ll positive rel numbers, b, c, we hve b + c + b c + + c + b Proof After the substitution x = b + c, y = c +, z = + b, it becomes y + z x x or which follows from the AM-GM inequlity s following: 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 x = b + c, y = b c +, z = c + b It follows tht fx = =, where ft = t + b + c + t 0

23 Since f is concve on 0,, Jensen s inequlity shows tht f = = x + y + z fx f or f Since f is monotone incresing, this implies tht x + y + z or f b + c = x + y + z x + y + z Proof As in the previous proof, it suffices to show tht T, where T = x + y + z nd x + x = One cn esily check tht the condition x + x = becomes = 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, c be positive numbers such tht bc = Prove tht + b + c + b c Second Solution [IV], Iln Vrdi Since bc =, we my ssume tht b It follows tht + b + c + = c + c b c + b b + Third Solution As in the first solution, fter the substitution = x y, b = y z, c = z x for x, y, z > 0, we cn rewrite it s xyz y + z xz + x yx + y z Without loss of generlity, we cn 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 Since 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 condition bc =, it s strightforwrd to verify the equlities = + + c b +, b c Why? Note tht the inequlity is not symmetric in the three vribles Check it! For verifiction of the identity, see [IV]

24 = b = c b + + c +, c c + + b + c In prticulr, they show tht t most one of the numbers u = + b, v = b + c, w = c + is negtive If there is such number, we hve + b + c + = uvw < 0 < b c And if u, v, w 0, the AM-GM inequlity yields = u + cv c uv, = b v + w b vw, = c w + w b c wu Thus, uv c, vw b, wu c b, so uvw c b proof c b = Since u, v, w 0, this completes the Problem 5 Let, b, c be positive rel numbers stisfying + b + c = Show tht + bc + b bc b + c + c + b + 4 Solution We wnt to estblish tht Set x = bc, y = c b, z = + bc + + c b + b c + b c b c We need to prove tht x + + y + z + z + 4, 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 becomes x = tn A, y = tn B, z = tn C, nd A + B + C = π or or + tn A + + tn C tn B + + tn C + cos A + cos B + sin C + 4 cos A + cos B + sin C + 4

25 Note tht cos A + cos B = cos A+B cos A B A + B cos A + cos B cos Since A B = cos < π π C, this mens tht It will be enough to show tht π C cos + sin C, where C 0, π This is one-vrible inequlity It s left s n exercise for the reder Here, we give nother solution of the problem 0 Ltvi 00 Let, b, c, d be the positive rel numbers such tht Prove tht bcd b c d 4 = Second Solution given by Jeong Soo Sim t the KMO Weekend Progrm 007 We need to prove the inequlity 4 b 4 c 4 d 4 8 After mking the substitution A = + 4, B = + b 4, C = + c 4, D = + d 4, we obtin 4 = A A, b4 = B B, c4 = C C, d4 = D D The constrint becomes A + B + C + D = nd the inequlity cn be written s or or B + C + D A A A B B C + D + A B C C D + A + B C D D 8 A + B + C D 8 B + C + DC + D + AD + A + BA + B + C 8ABCD However, this is n immedite consequence of the AM-GM inequlity: B + C + DC + D + AD + A + BA + B + C BCD CDA DAB ABC Problem 6 Irn 998 Prove tht, for ll x, y, z > such tht x + y + z =, x + y + z x + y + z Differentite! Shiing-shen Chern

26 First Solution We begin with the lgebric substitution = x, b = y, c = z Then, the condition becomes b + + c = b + b c + c + b c = nd the inequlity is equivlent to + b + c + + b + c b + bc + c Let p = bc, q = c, r = b Our job is to prove tht p + q + r where p + q + r + pqr = By the exercise 7, we cn mke the trigonometric substitution p = cos A, q = cos B, r = cos C for some A, B, C 0, π with A + B + C = π Wht we need to show is now tht cos A+cos B+cos C It follows from Jensen s inequlity Problem 7 Belrus 998 Prove tht, for ll, b, c > 0, Solution After writing x = b nd y = c b, we get b + b c + c + b b + c + b + c c + + c = y x, One my rewrite the inequlity s + b b + c = x + + y, b + c c + = + y y + x Apply the AM-GM inequlity to obtin x y + x x y + x + x + y + y x y + xy + xy x y, x y + x + y + y xy, x + y xy Adding these three inequlities, we get the result The equlity holds if nd only if x = y = or = b = c Problem 8 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 consider the cse when x,, x n re ll nonnegtive rel numberswhy? 4 Let x 0 = After the substitution y i = x 0 + +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 i=0 Since 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 4 x +x + i=0 x +x +x + + i=0 x n +x + +x n x +x + i=0 x +x +x + + x n +x + +x n 4

27 We now pply the Cuchy-Schwrz 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 Since y 0 = nd y n > 0, this yields the desired upper bound n i=0 Second Solution We my ssume tht x,, x n re ll nonnegtive rel numbers Let x 0 = 0 We mke the following lgebric substitution t i = x i x0 + + x i, c i = + ti nd s i = t i + ti for ll i = 0,, n It s n esy exercise to show tht the desired inequlity becomes x i x 0 + +x i = c 0 c i s i Since s i = c i, c 0 c c + c 0 c c c + + c 0 c c n cn < n Since 0 < c i for ll i =,, n, we hve n c 0 c i ci i= n c 0 c i ci = i= n c0 c i c 0 c i c i i= Since c 0 =, by the Cuchy-Schwrz inequlity, we obtin n n c0 c i c 0 c i c i n [c 0 c i c 0 c i c i ] = n [ c 0 c n ] i= i= 5

28 Incresing Function Theorem Theorem Incresing Function Theorem Let f :, b R be differentible function If f x 0 for ll x, b, then f is monotone incresing on, b If f x > 0 for ll x, b, then f is strictly incresing on, b Proof We first consider the cse 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 c x, x such tht fx fx = f cx x Since f c > 0, this eqution mens tht fx fx > 0 In cse when f x 0 for ll x, b, we cn lso pply the Men Vlue Theorem to get the result Problem 9 Irelnd 000 Let x, y 0 with x + y = Prove tht x y x + y First Solution After homogenizing it, we need to prove x + y 6 x y x + y or x + y 6 x y x + y Now, forget the constrint x + y =! In cse xy = 0, it clerly holds We now ssume tht xy 0 Becuse of the homogeneity of the inequlity, this mens tht we my normlize to xy = Then, it becomes 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, Since F p = p 0, where p, F is monotone incresing on [4, So, F p F 4 = 0 for ll p 4 Second Solution As in the first solution, we prove tht x + y 6 x + y xy for ll x, y 0 In cse x = y = 0, it s cler 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 becomes + p 64p or p 5p + p + 0 We wnt to minimize F p = p 5p + p + on [0, ] We compute F p = p p We find tht F is monotone incresing on [0, ] nd monotone decresing on [, ] Since F 0 = nd F = 0, we conclude 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 becomes 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 0 IMO 984/ Let x, y, z be nonnegtive rel numbers such tht x + y + z = Prove tht 0 xy + yz + zx xyz 7 7 6

29 First Solution Let fx, y, z = xy +yz +zx xyz We my ssume tht 0 x y z Since 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 Since 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 function F x = 4 x + x +, where x [ 0, ] Since F x = x x 0 on [ 0, ], we conclude tht F x F = 7 7 for ll x [ 0, ] IMO 000/ Let, b, c be positive numbers such tht bc = Prove tht + b + c + b c Fifth Solution bsed on work by n IMO 000 contestnt from Jpn Since bc =, t lest one of, b, c is greter thn or equl to Sy b Putting c = b, it becomes + b + b b b + or b b b b + b b + b b b + 0 Setting x = b, it becomes f b x 0, where 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 cubic polynomil F t = t b + t b b + hs two rel roots b + 4b 7b + 4 nd λ = b + + 4b 7b + 4 Since F hs locl 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 Notice 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, which is equivlent to 6b 5b 5b + 6 8b 4b + 8 4b 7b + 4 7

30 Since 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 Since 456x 6669x + 75 > 0 for ll x R, we find tht Gx nd x hve the sme sign It follows tht G is monotone decresing on, ] nd monotone incresing on [, We conclude tht G hs the globl minimum t x = Hence, Gx G = 0 for ll x R 5 It s esy to check tht 6b 5b 5b + 6 = 6b b b + + b + b > 6b b 0 nd 8b 4b + 8 = 8b + b > 0 8

31 4 Estblishing New Bounds We first give two lterntive wys to prove Nesbitt s inequlity Nesbitt For ll positive rel numbers, b, c, we hve Proof 4 From It follows tht Proof 5 We clim tht b + c + b c + + b+c 0, we deduce tht b + c b + c 4 8 b+c b+c + = b + c + b + c c + b 8 b c 4 + b + c 8 b c 4 + b + c = or + b + c b + c The AM-GM inequlity gives + b + b b nd + c + c c Adding these two inequlities yields + b + c b + c, s desired Therefore, we hve b + c + b + c = Some inequlities cn be proved by finding new bounds Suppose tht we wnt to estblish tht F x, y, z C If function 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 deduce tht F x, y, z Gx, y, z = C For exmple, if function 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 9

32 Problem Let, b, c be the lengths of tringle Show tht b + c + b c + + c + b < Proof We don t employ the Rvi substitution It follows from the tringle inequlity tht b + c < = + b + c One dy, I tried finding new lower bound of x+y+z where x, y, z > 0 There re well-known lower bounds such 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 except the term x : y + z + xy + xy + yz + yz + zx + zx 8x y 4 z 4 It follows tht x + y + z x + 8x y 4 z 4 = x x + 8y 4 z 4 IMO 00/ Let, b, c be positive rel numbers Prove tht + 8bc + b b + 8c + c c + 8b Second 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 = c 4, it now becomes + 8bc Problem IMO 005/ Let x, y, nd z be positive numbers such tht xyz Prove tht x 5 x x 5 + y + z + y5 y y 5 + z + x + z5 z z 5 + x + y 0 0

33 First Solution It s equivlent to the following inequlity x x 5 y x 5 + y + z + y 5 z + y 5 + z + x + 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 Cuchy-Schwrz inequlity nd the fct 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 Second 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 Cuchy-Schwrz 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 x x 5 + 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 x 4 = x 4 + y 4 x y = [Second Wy] We clim tht x 4 + y 4 + z 4 + 4x y + 4x z 4x + y + z x y + x yz x 4 x yz y x + z x yz x x 5 + y + z

34 We do this by proving becuse xyz implies tht x 4 + y 4 + z 4 + 4x y + 4x z 4x + y + z x yz x 4 + y z + yz = x x 5 xyz + y + z Hence, we need to show the homogeneous inequlity x yz x 4 + y z + yz 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 consequence 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 Third Solution by n IMO 005 contestnt Iurie Boreico 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 Here is brillint solution of Problem KMO Weekend Progrm 007 Prove tht, for ll, b, c, x, y, z > 0, x + x + by b + y + 6 He received the specil prize for this solution cz c + z + b + cx + y + z + b + c + x + y + z

35 Solution by Snghoon We need the following lemm: Lemm For ll p, q, ω, ω > 0, we hve or Proof of lemm It s equivlent to pq p + q ω p + ω q ω + ω p + q ω p + ω q ω + ω pq 0 ω p ω q 0 Tking p, q, ω, ω =, x, x + y + z, + b + c in the lemm, we get Similrly, we obtin nd Adding the bove three inequlities, we get x + x x + y + z + + b + c x x + y + z + + b + c by b + y x + y + z b + + b + c y x + y + z + + b + c cz c + z x + y + z c + + b + c z x + y + z + + b + c or x + x + by b + y + cz c + z x + y + z + b + c + + b + c x + y + z x + y + z + + b + c x + x + by b + y + cz c + z + b + cx + y + z + b + c + x + y + z Exercise 5 USAMO Summer Progrm 00 Let, b, c be positive rel numbers Prove tht b c + + b + c c + + b Hint [TJM] Estblish the inequlity b+c Exercise 6 APMO 005 bc = 8,, b, c > 0 +b+c + + b + b + b + c + c + c + 4 Hint Use the inequlity +x +x to give lower bound of the left hnd side

36 Chpter Homogeniztions nd Normliztions Every Mthemticin Hs Only Few Tricks A long time go n older nd well-known number theorist mde some disprging remrks bout Pul Erdös s work You dmire Erdos s contributions to mthemtics s much s I do, nd I felt nnoyed when the older mthemticin fltly nd definitively stted tht ll of Erdos s work could be reduced to few tricks which Erdös repetedly relied on in his proofs Wht the number theorist did not relize is tht other mthemticins, even the very best, lso rely on few tricks which they use over nd over Tke Hilbert The second volume of Hilbert s collected ppers contins Hilbert s ppers in invrint theory I hve mde point of reding some of these ppers with cre It is sd to note tht some of Hilbert s beutiful results hve been completely 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 tricks Even Hilbert hd only few tricks! Gin-Crlo Rot, Ten Lessons I Wish I Hd Been Tught, Notices of the AMS, Jnury 997 Homogeniztions Mny inequlity problems come with constrints such s b =, xyz =, x + y + z = A non-homogeneous symmetric inequlity cn be trnsformed into homogeneous one Then we pply two powerful theorems : Shur s inequlity nd Muirhed s theorem We begin with simple exmple Problem 4 Hungry 996 Let nd b be positive rel numbers with + b = Prove tht + + b b + Solution Using the condition + b =, we cn reduce the given inequlity to homogeneous one, i e, + b + + b + b + bb + + b or b + b + b, which follows from + b b + b = b + b 0 The equlity holds if nd only if = b = The bove inequlity b + b + b cn be generlized s following : Theorem Let,, b, b be positive rel numbers such 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 4

37 Proof Without loss of generlity, we cn ssume tht, b b, b If x or y is zero, then it clerly holds So, we ssume tht both x nd y re nonzero It follows from + = b +b tht = b + b It s esy to check 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 Remrk When does the equlity hold in the theorem 8? We now introduce two summtion nottions nd sym Let P x, y, z be three vribles function 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 5 IMO 984/ Let x, y, z be nonnegtive rel numbers such tht x + y + z = Prove tht 0 xy + yz + zx xyz 7 7 Second Solution Using the condition x + y + z =, we reduce the given inequlity to homogeneous one, i e, sym 0 xy + yz + zxx + y + z xyz 7 7 x + y + z The left hnd side inequlity is trivil becuse it s equivlent to 0 xyz + sym x y 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, 5

38 it s enough to show tht We note tht x sym x sym x y nd xyz + x sym x y x y = x + y x y + xy = x + y x y xy 0 The second inequlity cn be rewritten s xx yx z 0, which is prticulr cse of Schur s theorem in the next section After homogenizing, sometimes we cn find the right pproch to see the inequlities: Irn 998 Prove tht, for ll x, y, z > such tht x + y + z =, x + y + z x + y + z Second Solution After the lgebric substitution = x, b = y, c = z, we re required to prove tht + b + b c c + +, b c where, b, c 0, nd + b + c = Using the constrint + b + c =, we obtin homogeneous inequlity + b + c + b + +b+c +b+c +b+c b c + + c b c or + b + c + b + b + c c + b + b c + +, c b c which immeditely follows from the Cuchy-Schwrz inequlity [b + c + c + b + + b c] + b + b + c c + b + b c + + c b c 6

39 Schur s Inequlity nd Muirhed s Theorem Theorem Schur Let x, y, z be nonnegtive rel numbers For ny r > 0, we hve x r x yx z 0 Proof Since the inequlity is symmetric 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 clerly nonnegtive Remrk When does the equlity hold in Schur s Inequlity? Exercise 7 Disprove the following proposition: For ll, b, c, d 0 nd r > 0, we hve r b c d + b r b cb db + c r c c c d + d r d d bd c 0 The following specil cse of Schur 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 Schur s inequlity nd the AM-GM inequlity, we hve xyz + x x y + xy xy x y We now use Schur s inequlity to give n lterntive solution of APMO 004/5 Prove tht, for ll positive rel numbers, b, c, + b + c + 9b + bc + c Second Solution After expnding, it becomes 8 + bc + b b From the inequlity b + bc + c 0, we obtin 6 + b 4 b Hence, it will be enough to show tht + bc b Since + b + c b + bc + c, it will be enough to show tht + bc + b, which is prticulr cse of the following result for t = 7

40 Corollry Let t 0, ] For ll, b, c 0, we hve t + tbc t + b In prticulr, we obtin non-homogeneous inequlities 5 + bc4 + + b + c b + bc + c, + bc + + b + c b + bc + c, + bc + + b + c b + bc + c Proof After setting x =, y = b, z = c, it becomes t + txyz t + x xy By the corollry, it will be enough to show tht t + txyz t xyz, which is strightforwrd consequence of the weighted AM-GM inequlity : t + t xyz t t xyz t t One my check tht the equlity holds if nd only if = b = c = = xyz IMO 000/ Let, b, c be positive numbers such tht bc = Prove tht + b + c + b c Second Solution It is equivlent to the following homogeneous inequlity : bc / + bc/ b bc / + bc/ c bc / + bc/ bc b c After the substitution = x, b = y, c = z with x, y, z > 0, it becomes x xyz + y xyz xyz + z xyz xyz + xyz x y z, y which simplifies to x y y z + z x y z z x + x y z x x y + y z x y z 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 which is specil cse of Schur s inequlity For n lterntive homogeniztion, see the problem in the chpter 8

41 Here is nother inequlity problem with the constrint bc = Problem 6 Tournment of Towns 997 Let, b, c be positive numbers such tht bc = Prove tht + b + + b + c + + c + + Solution We cn rewrite the given inequlity s following : + b + bc / + b + c + bc / + c + + bc / bc / We mke the substitution = x, b = y, c = z with x, y, z > 0 Then, it becomes x + y + xyz + y + z + xyz + z + x + xyz xyz which is equivlent to xyz x + y + xyzy + z + xyz x + y + xyzy + z + xyzz + x + xyz or sym x 6 y sym x 5 y z! We pply the theorem 9 to obtin 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 Exercise 8 [TZ], pp4 Prove tht for ny cute tringle ABC, cot A + cot B + cot C + 6 cot A cot B cot C cot A + cot B + cot C Exercise 9 Kore 998 Let I be the incenter of tringle ABC Prove tht IA + IB + IC BC + CA + AB Exercise 0 [IN], pp0 Let, b, c be the lengths of tringle Prove tht b + c + b c + b + c + c b > + b + c + bc 9

42 Exercise 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 + x n n Theorem Muirhed Let,,, b, b, b be rel numbers such 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 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 twice 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 twice 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 check 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 cn use Muirhed s theorem to prove Nesbitt s inequlity 40