The Zarankiewicz problem in 3-partite graphs

Transcription

1 The Zarankiewicz problem in 3-parie graphs Michael Tai Craig Timmons Absrac Le F be a graph, k 2 be an ineger, and wrie ex χ k (n, F ) for he maximum number of edges in an n-verex graph ha is k-parie and has no subgraph isomorphic o F The funcion ex χ 2 (n, F ) has been sudied by many researchers Finding ex χ 2 (n, K s, ) is a special case of he Zarankiewicz problem We prove an analogue of he Kövári-Sós-Turán Theorem by showing ex χ 3 (n, K s, ) ( ) 1 1 1/s ( ) 1 1/s + o(1) n 2 1/s 3 2 for 2 s Using Sidon ses consruced by Bose and Chowla, we prove ha his upper bound in asympoically bes possible in he case ha s = 2 and 3 is odd, ie, ex χ 3 (n, K 2,2+1 ) = 3 n3/2 + o(n 3/2 ) for 1 1 Inroducion Le G and F be graphs We say ha G is F -free if G does no conain a subgraph ha is isomorphic o F The Turán number of F is he maximum number of edges in an F -free graph wih n verices This maximum is denoed ex(n, F ) and an F -free graph wih n verices and ex(n, F ) edges is called an exremal graph for F One of he mos well-sudied cases is when F = C 4, a cycle of lengh four This problem was considered by Erdős [7] in 1938, and lies in he inersecion of exremal graph heory, finie geomery via projecive planes and difference ses, and combinaorial number heory via Sidon ses Roughly 30 years laer, Brown [3], and Erdős, Rényi, and Sós [8, 9] independenly showed ha ex(n, C 4 ) = 1 2 n3/2 + o(n 3/2 ) They consruced, for each prime power q, a C 4 -free graph wih q 2 + q + 1 verices and 1q(q + 2 1)2 edges These graphs are examples of orhogonal polariy graphs which have since been sudied and applied o oher problems in combinaorics Answering a quesion of Erdős, Füredi [11, 12] showed ha for q > 13, orhogonal polariy graphs are he only exremal graphs for C 4 when he number of verices is q 2 + q + 1 Füredi [13] also used finie fields o consruc, for each 1, K 2,+1 -free graphs wih n verices and 2 n3/2 + o(n 3/2 ) edges Deparmen of Mahemaical Sciences, Carnegie Mellon Universiy, mai@cmuedu Research is suppored by NSF gran DMS Deparmen of Mahemaics and Saisics, California Sae Universiy Sacrameno, craigimmons@csusedu Research suppored in par by Simons Foundaion Gran #

2 This consrucion, ogeher wih he famous upper bound of Kövári, Sós, and Turán [17], shows ha ex(n, K 2,+1 ) = 2 n3/2 + o(n 3/2 ) for all 1 Because of is imporance in exremal graph heory, variaions of he biparie Turán problem have been considered One such insance is o find he maximum number of edges in an F -free n by m biparie graph Wrie ex(n, m, F ) for his maximum Esimaing ex(n, n, K s, ) is he balanced case of he Zarankiewicz problem The resuls of [13, 17] show ha ex(n, n, K 2,+1 ) = n 3/2 + o(n 3/2 ) for 1 The case when F is a cycle of even lengh has also received considerable aenion Naor and Versraëe [18] sudied he case when F = C 2k More precise esimaes were obained by Füredi, Naor, and Versraëe when F = C 6 [14] For more resuls along hese lines, see [4, 5, 16] and he survey of Füredi and Simonovis [15] o name a few Now we inroduce he exremal funcion ha is he focus of his work For an ineger k 2, define ex χ k (n, F ) o be he maximum number of edges in an n-verex graph G ha is F -free and has chromaic number a mos k Thus, ex χ 2 (n, F ) is he maximum number of edges in an F -free biparie graph wih n verices (he par sizes need no be he same) Trivially, for any k In he case ha k = 2, ex χ k (n, F ) ex(n, F ) ex χ 2 (n, K 2, ) = n3/2 + o(n 3/2 ) by [13, 17] Our focus will be on ex χ 3 (n, K 2, ) and our firs resul gives an upper bound on ex χ 3 (n, K s, ) Theorem 11 For n 1 and 2 s, ex χ 3 (n, K s, ) ( ) 1 1/s ( 1/s o(1)) n 2 1/s 3 2 When s = 2, Theorem 11 improves he rivial bound 1 ex χ 3 (n, K 2, ) ex(n, K 2, ) = n 3/2 + o(n 3/2 ) 2 Allen, Keevash, Sudakov, and Versraëe [1] consruced 3-parie graphs wih n verices ha are K 2,3 -free and have 1 3 n 3/2 n edges This consrucion shows ha Theorem 11 is asympoically bes possible in he case ha s = 2, = 3 Our nex resul shows ha his asympoic formula holds for K 2,2+1 for all 1 Theorem 12 For any ineger 1, ex χ 3 (n, K 2,2+1 ) = 2 3 n3/2 + o(n 3/2 )

3 We believe ha he mos ineresing remaining open case is deermining he behavior when forbidding K 2,2 = C 4 Problem 13 Deermine he asympoic behavior of ex χ 3 (n, C 4 ) In paricular i would be very ineresing o know wheher or no ex χ 2 (n, C 4 ) ex χ 3 (n, C 4 ) We discuss his in more deail in Secion 4 A random pariion ino k pars of an n-verex C 4 -free graph wih 1 2 n3/2 + o(n 3/2 ) edges gives a lower bound of ex χ k (n, C 4 ) ( 1 1 ) n 3/2 k 2 o(n3/2 ) When k 4, his is a beer lower bound han he one provided by n3/2 + o(n 3/2 ) ex χ 2 (n, C 4 ) ex χ k (n, C 4 ) which holds for all k 2 In he nex secion we prove Theorem 11 and in Secion 3 we prove Theorem 12 2 Proof of Theorem 11 In his secion we prove Theorem 11 The proof is based on he sandard double couning argumen of Kövári, Sós, and Turán [17] Proof of Theorem 11 Le G be an n-verex 3-parie graph ha is K s, -free Le A 1, A 2, and A 3 be he pars of G, and define δ i by δ i n = A i By he Kövári-Sós-Turán Theorem [17], here is a consan β s, > 0 such ha he number of edges wih one end poin in A 1 and he oher in A 2 is a mos β s, n 2 1/s If here are o(n 2 1/s ) edges beween A 1 and A 2, hen we may remove hese edges o obain a biparie graph G ha is K s, -free which gives e(g) e(g ) o(n 2 1/s ) ex χ 2 (n, K s, ) In his case, we may apply he upper bound of Füredi [10] o see ha he conclusion of Theorem 11 holds Therefore, we may assume ha here is a posiive consan c 1,2 so ha he number of edges beween A 1 and A 2 is c 1,2 n 2 1/s Similarly, le c 1,3 n 2 1/s and c 2,3 n 2 1/s be he number of edges beween A 1 and A 3, and beween A 2 and A 3, respecively 3

4 Using he assumpion ha G is K s, -free and convexiy, we have ( ) A1 ( 1) ( ) da1 (v) + ( ) da1 (v) s s s v A 2 v A 3 ( 1 A 2 e(a ) ( A 2 1, A 2 ) 1 + A 3 e(a ) A 3 1, A 3 ) s s δ ( ) s 2n e(a1, A 2 ) s + δ ( s 3n e(a1, A 3 ) s) s! A 2 s! A 3 Afer some simplificaion we ge ( 1) (δ 1n) s s! δ 2n s! ( c1,2 n 2 1/s δ 2 n ) s s + δ ( 3n c1,3 n 2 1/s s s) s! δ 3 n For j {2, 3}, we can assume ha c 1,jn 2 1/s δ j n > s oherwise e(a 1, A j ) = c 1,j n 2 1/s sδ j n sn = o(n 2 1/s ) From he inequaliy (1 + x) s 1 + sx for x 1, we now have ( ) ( 1)δ1n s s c1,2 n 2 1/s s ( ) δ 2 n δ 2 ns 2 c1,2 n 2 1/s s 1 δ 2 n δ 2 n ( ) c1,3 n 2 1/s s ( ) + δ 3 n δ 3 ns 2 c1,3 n 2 1/s s 1 δ 3 n δ 3 n Dividing hrough by n s and rearranging gives ( 1)δ s 1 cs 1,2 δ s cs 1,3 δ3 s 1 Muliplying hrough by δ s 1 2 δ s 1 3 leads o s2 c s 1 1,2 δ2 s 2 n s2 c 1 1/s δ s 2 ( 1)δ1δ s 2 s 1 δ3 s 1 c s 1,2δ3 s 1 + c s 1,3δ2 s 1 s2 δ3 s 1 δ 2 c s 1 1,2 n 1 1/s Since δ 2 and δ 3 are boh a mos 1 and c 1,j is a mos β s,, s 1 1,3 3 n 1 1/s ( 1)δ1δ s 2 s 1 δ3 s 1 c s 1,2δ3 s 1 + c s 1,3δ2 s 1 2s2 βs, s 1 n 1 1/s By symmery beween he pars A 1, A 2, and A 3, s2 δ2 s 1 δ 3 c s 1 1,3 n 1 1/s and ( 1)δ2δ s 1 s 1 δ3 s 1 c s 1,2δ3 s 1 + c s 2,3δ1 s 1 2s2 βs, s 1 n 1 1/s ( 1)δ3δ s 1 s 1 δ2 s 1 c s 1,3δ2 s 1 + c s 2,3δ1 s 1 2s2 βs, s 1 n 1 1/s 4

5 Add hese hree inequaliies ogeher and divide by 2 o obain 1 2 δs 1 1 δ2 s 1 δ3 s 1 (δ 1 + δ 2 + δ 3 ) c s 1,2δ3 s 1 + c s 1,3δ2 s 1 + c s 2,3δ1 s 1 3s2 βs, s 1 n 1 1/s Now n = A 1 + A 2 + A 3 = (δ 1 + δ 2 + δ 3 )n so we may replace δ 1 + δ 2 + δ 3 wih 1 This leads us o he opimizaion problem of maximizing subjec o he consrains and c 1,2 + c 1,3 + c 2,3 0 δ i, 0 c i,j 1, δ 1 + δ 2 + δ 3 = 1, 1 2 δs 1 1 δ2 s 1 δ s 1 3 δ3 s 1 c s 1,2 + δ2 s 1 c s 1,3 + δ s 1 1 c s 2,3 This can be done using he mehod of Lagrange Mulipliers (see he Appendix) and gives c 1,2 + c 1,3 + c 2,3 ( ) 1 1/s ( ) 1/s We conclude ha he number of edges of G is a mos ( ) 1 1/s ( ) 1/s 1 1 n 2 1/s + o(n 2 1/s ) Proof of Theorem 12 In his secion we consruc a 3-parie K 2,2+1 -free graph wih many edges The consrucion is inspired by Füredi s consrucion of dense K 2, -free graphs [13] Le 1 be an ineger Le q be a power of a prime chosen so ha divides q 1 and le θ be a generaor of he muliplicaive group F q 2 := F q 2\{0} Le A Z q 2 1 be a Bose-Chowla Sidon se [2] Tha is, A = {a Z q 2 1 : θ a θ F q } and noe ha A = q Le H be he subgroup of Z q 2 1 generaed by ( q 1 ( ( ( q 1 q 1 q 1 H = { 0, ) (q + 1), 2 ) (q + 1),, ( 1) )(q + 1) Thus, } ) (q + 1) and furhermore, H is conained in he subgroup of Z q 2 1 generaed by q + 1 Le G q, be he biparie graph whose pars are X and Y where each of X and Y is a disjoin copy of he quoien group Z q 2 1/H A verex x + H X is adjacen o x + a + H Y for all a A We will need he following lemma, which was proved in [20] 5

6 Lemma 31 [Lemma 22 of [20]] Le A Z q 2 1 be a Bose-Chowla Sidon se Then A A = Z q 2 1 \ {q + 1, 2(q + 1), 3(q + 1),, (q 2)(q + 1)} In paricular, Lemma 31 implies ha (A A) H = Lemma 32 If 1 is an ineger and q is a power of a prime for which divides q 1, hen he graph G q, is a biparie graph wih q2 1 ( ) verices in each par, is K 2,+1 -free, and has q edges q 2 1 Proof I is clear ha G q, is biparie and has q2 1 verices in each par Le x + H be a verex in X The neighbors of x + H are of he form x + a + H where a A We show ha hese verices are all disinc If x + a + H = x + b + H for some a, b H, hen a b H By Lemma 31 (A A) H = {0} where A A = {a b : a, b A} We conclude ( ) ha a = b and so he degree of x + H is A = q This also implies ha G q, has q edges and o finish he proof, we mus q 2 1 show ha G q, has no K 2,+1 We consider wo cases depending on which par conains he par of size wo of he K 2,+1 Firs suppose ha x + H and y + H are disinc verices in X and le z + H be a common neighbor in Y Then z + H = x + a + H and z + H = y + b + H for some a, b A Therefore, z = x + a + h 1 and z = y + b + h 2 for some h 1, h 2 H From his pair of equaions we ge a b = y x + h 2 h 1 Since H is a subgroup, h 2 h 1 = h 3 for some h 3 H and now we have a b = y x + h 3 (1) The righ hand side of (1) is no zero since x + H and y + H are disinc verices in A As A is a Sidon se and y x + h 3 0, here is a mos one ordered pair (a, b) A 2 for which a b = y x + h 3 There are possibiliies for h 3 and so possible ordered pairs (a, b) A 2 for which z + H = x + a + H = y + b + H is a common neighbor of x + H and y + H This shows ha x + H and y + H have a mos common neighbors Now suppose x+h and y+h are disinc verices in Y and z+h is a common neighbor in X There are elemens a, b A such ha z+a+h = x+h and z+b+h = y+h Thus, z + a + h 1 = x and z + b + h 2 = y for some h 1, h 2 H Therefore, x a h 1 = y b h 2 so a b = x y + h 2 h 1 We can hen argue as before ha here are a mos ordered pairs (a, b) A 2 such ha z + H is a common neighbor of z + a + H = x + H and z + b + H = y + H Once again, le 1 be an ineger and q be a power of a prime for which divides q 1 Le Γ q, be he 3-parie graph wih pars X, Y, and Z where each par is a copy of he quoien group Z q 2 1/H Here H is he subgroup generaed by ( q 1 )(q + 1) A verex x + H X is adjacen o x + a + H Y for all a A Similarly, a verex y + H Y is adjacen o y + a + H Z for all a A, and a verex z + H Z is adjacen o z + a + H X for all a A 6

7 Lemma 33 The graph Γ q, is K 2,2+1 -free Proof By Lemma 32, a pair of verices in one par of Γ q, have a mos common neighbors in each of he oher wo pars Thus, here canno be a K 2,2+1 in Γ q, where he par of size wo is conained in one par Now le x+h and y +H be verices in wo differen pars Wihou loss of generaliy, assume x + H X and y + H Y Suppose z + H Z is a common neighbor of x + H and y+h There are elemens a, b A such ha z+h = y+a+h and z+b+h = x+h so we have z = y + a + h 1 and z + b = x + h 2 for some h 1, h 2 H This pair of equaions implies a + b = x y + h 2 h 1 and since H is a subgroup, h 2 h 1 H Le h 2 h 1 = h 3 where h 3 H so a + b = x y + h 3 There are possibiliies for h 3 Given h 3, he equaion a + b = x y + h 3 uniquely deermines he pair {a, b} since A is a Sidon se There are wo ways o order a and b and so x + H and y + H have a mos 2 common neighbors in Z Proof of Theorem 12 By Theorem 11, ( 1/ ex χ 3 (n, K 2,2+1 ) = + o(1))) n 3/2 = n3/2 + o(n 3/2 ) As for he lower bound, if q is any power of a prime for which divides q 1, hen by Lemmas 32 and 33, he graph Γ q, is a 3-parie graph wih q2 1 ( ) verices in each par, is K 2,2+1 -free, and has 3q edges Thus, q 2 1 ( ) 3(q 2 1) ex χ 3, K 2,2+1 3q ( q 2 1 If n = 3(q2 1), hen he above can be rewrien as ( ) n ex χ 3 (n, K 2,2+1 ) n n3/2 n A sandard densiy of primes argumen finishes he proof ) 7

8 4 Concluding Remarks We may consider a similar graph o G q, and Γ q, which does no necessarily have bounded chromaic number Le Γ be a finie abelian group wih a subgroup H of order Le A Γ be a Sidon se such ha (A A) H = {0} Then we may consruc a graph G wih verex se Γ/H where x + H y + H if and only if x + y = a + h for some a A and h H Then he proof of Lemma 32 shows ha G is a K 2,+1 -free graph on Γ / H verices and every verex has degree A or A 1 When Γ = Z q 2 1, divides q 1, and A is a Bose-Chowla Sidon se, he resuling graph G is similar o he one consruced by Füredi in [13] When = 1 he main resul of [19] shows ha hese wo graphs are isomorphic However, in general hese graphs are no isomorphic When q = 19 and {3, 6} he graph consruced above has one more edge han he graph consruced by Füredi Turning o he quesion of deermining ex χ 3 (n, C 4 ), Theorem 11 shows ha ex χ 3 (n, C 4 ) n3/2 6 Furhermore, he opimizaion shows ha if his bound is igh asympoically, hen a consrucion would have o be 3-parie wih each par of size asympoic o n and 3 average degree asympoic o n beween each par The following consrucion is due 6 o Jason Williford [21] Theorem 41 Le R be a finie ring, A R an addiive Sidon se and B = ca = {ca : a A} Then if (A A) (B B) = {0} where c is inverible, here is a graph on 3 R verices which is 3-parie, C 4 -free and is A -regular beween pars Proof We consruc a graph wih parie ses S 1, S 2, S 3 where S 1 = R, S 2 = {A + i} i R and S 3 = {B + j} j R A verex in S 1 is adjacen o a verex in S 2 or S 3 by inclusion The verex A + j S 2 is adjacen o B + i S 3 if cj + i A Since c is inverible, we have ha boh A and B are Sidon ses Therefore, he biparie graphs beween S 1 and S 2, and beween S 1 and S 3 are incidence graphs of parial linear spaces, and hus do no conain C 4 If here were a C 4 wih A + i, A + j S 2 and B + k, B + l S 3, i implies ha here exis a 1, a 2, a 3, a 4 A such ha ci + k = a 1 ci + l = a 2 cj + k = a 3 cj + l = a 4 This means ha k l = a 1 a 2 = a 3 a 4 Since A is a Sidon se his means ha a 1 = a 2 or a 1 = a 3, which implies ha eiher k = l or i = j 8

9 If here were a C 4 wih i S 1, A + j, A + k S 2 and B + l S 3, i means ha here are a 1, a 2, a 3, a 4 A such ha i = a 1 + j i = a 2 + k cj + l = a 3 ck + l = a 4 This means ha c(j k) = c(a 2 a 1 ) = a 4 a 3 Since B = ca we have ha b 2 b 1 = a 4 a 3 for some b 1, b 2 B, and herefore b 2 b 1 = a 4 a 3 = 0 This implies ha j = k The case when here are wo verices in S 3 and one each in S 1 and S 2 is similar The condiion ha (A A) (B B) = {0} and A is a Sidon se implies ha 2 A ( A 1) R 1 In Z 5, if A = {0, 1} and B = 2A = {0, 2}, we have (A A) (B B) = {0} and (A A) (B B) = Z 5 This gives a 3-parie graph on 15 verices which is C 4 -free and is 4-regular In Z 41, he se A = {1, 10, 16, 18, 37} and B = 9A have he same propery ha (A A) (B B) = {0} and (A A) (B B) = Z 41 This gives a 3-parie C 4 -free graph on 123 verices which is 10 regular In general, a (v, k, λ)-difference family in a group Γ of order v is a collecion of ses {D 1,, D } each of size k such ha he mulise (D 1 D 1 ) (D D ) conains every nonzero elemen of Γ exacly λ imes If one could find an infinie family of (2k 2 2k +1, k, 1)-difference families in Z 2k 2 2k+1 where he wo blocks are muliplicaive ranslaes of each oher by a uni, hen he resuling graph would mach he upper bound in Theorem 11 The ses A = {0, 1} and 2A in Z 5, and A = {1, 10, 16, 18, 37} and 9A in Z 41 are examples of his for k = 2 and k = 5 respecively We could no figure ou how o exend his consrucion in general, and in [6] i is shown ha no (61, 6, 1)-difference family exiss in F 61 To show Theorem 11 is igh asympoically i would suffice o find somehing weaker han a (2k 2 2k + 1, k, 1)-difference family where he wo blocks are muliplicaive ranslaes of each oher We do no need every nonzero elemen of he group o be represened as a difference of wo elemens, jus a proporion of hem ending o 1 5 Acknowledgemens The auhors would like o hank Casey Tompkins for inroducing he firs auhor o he problem We would also like o hank Cory Palmer and Jason Williford for helpful discussions References [1] P Allen, P Keevash, B Sudakov, J Versraëe, Turán numbers of biparie graphs plus an odd cycle, J Combin Theory Ser B 106 (2014),

10 [2] R C Bose, S Chowla, Theorems in he addiive heory of numbers, Commen Mah Helv / [3] W G Brown, On graphs ha do no conain a Thomsen graph, Canad Mah Bull [4] D de Caen, L A Székely, The maximum size of 4- and 6-cycle free biparie graphs on m, n verices, Ses, graphs and numbers (Budapes, 1991), , Collob Mah Soc János, Bolyai, 60, Norh-Holland, Amserdam, 1992 [5] D de Caen, L A Székely, On dense biparie graphs of girh eigh and upper bounds for cerain configuraions in planar poin-line sysems, J Combin Theory Ser A 77 (1997), no 2, [6] K Chen, L Zhu, Exisence of (q, 6, 1) Difference Families wih q a Prime Power, Des Codes Cryp 15 (1998) [7] P Erdős, On sequences of inegers no one of which dives he produc of wo ohers and some relaed problems, Mi Forsch-Ins Mah Mech Univ Tomsk 2 (1938), [8] P Erdős, A Rényi, On a problem in he heory of graphs (Hungarian) Magyar Tud Akad Ma Kuaó In Közl (1963) [9] P Erdős, A Rényi, V T Sós, On a problem of graph heory, Sudia Sci Mah Hungar [10] Z Füredi, An upper bound on Zarankiewicz problem, Combin Probab Compu 5 (1996), no 1, [11] Z Füredi, Graph wihou quadrilaerals, J Combin Theory Ser B 34 (1983), no 2, [12] Z Füredi, On he number of edges of quadrilaeral-free graphs, J Combin Theory Ser B 68 (1996), no 1, 1 6 [13] Z Füredi, New asympoics for biparie Turán numbers, J of Combin Theory Ser A, 75 (1996), no 1, [14] Z Füredi, A Naor, J Versraëe, On he Turán number for he hexagon, Adv Mah 203 (2006), no 2, [15] Z Füredi, M Simonovis, The hisory of degenerae (biparie) exremal graph problems, Bolyai Soc Mah Sud, 25, János Bolyai Mah Soc, Budapes, 2013 [16] E Győri, C 6 -free biparie graphs and produc represenaion of squares Graphs and combinaorics (Marseille, 1995) Discree Mah 165/166 (1997), [17] T Kövári, V T Sós, P Turán, On a problem of Zarankiewicz, Colloquium Mah 3, (1954)

11 [18] A Naor, J Versraëe, A noe on biparie graphs wihou 2k-cycles, Combin Probab Compu 14 (2005), no 5-6, [19] M Tai, C Timmons, Orhogonal Polariy Graphs and Sidon Ses, J Graph Theory} 82 (2016), [20] M Tai, C Timmons, Sidon ses and graphs wihou 4-cycles, J of Comb 5 (2014), no 2, [21] J Williford, privae communicaion 6 Appendix Here we solve he opimizaion problem of Theorem 11 using he mehod of Lagrange Mulipliers For convenience, we wrie x for c 1,2, y for c 1,3, and z for c 2,3 Recall ha δ 1, δ 2, and δ 3 are posiive real numbers ha saisfy δ 1 + δ 2 + δ 3 = 1 Le and g(x, y, z) = 1 f(x, y, z) = x + y + z 2 δs 1 1 δ2 s 1 δ s 1 3 δ3 s 1 x s δ2 s 1 y s δ1 s 1 z s For a parameer λ, le L(x, y, z, λ) = f(x, y, z) + λg(x, y, z) Taking parial derivaives, we ge L x = 1 sλδ3 s 1 x s 1 = 0, (2) L y = 1 sλδ s 1 2 y s 1 = 0, (3) L z = 1 sλδ1 s 1 z s 1 = 0, (4) ( ) 1 λ 2 δs 1 1 δ2 s 1 δ3 s 1 δ3 s 1 x s δ2 s 1 y s δ1 s 1 z s = 0 (5) Noe ha λ 0 oherwise we conradic (2) so by (5), From (2), (3), and (4) we have 1 2 δs 1 1 δ2 s 1 δ2 s 1 = δ3 s 1 x s + δ2 s 1 y s + δ1 s 1 z s (6) ( ) 1 1 s 1 = δ3 x = δ 2 y = δ 1 z (7) 2λ Combining his wih (6) and using δ 3 = 1 δ 1 δ 2, we ge an equaion ha can be solved for x o obain ( ) ( 1)δ s x = 1δ2 s 1/s 2(δ 1 (1 δ 1 ) + δ 2 (1 δ 2 ) δ 1 δ 2 ) Using (7), we can hen solve for y and z and ge x + y + z = ( 1)1/s 2 1/s (δ 1 (1 δ 1 ) + δ 2 (1 δ 2 ) δ 1 δ 2 ) 1 1/s 11

12 The maximum value of δ 1 (1 δ 1 ) + δ 2 (1 δ 2 ) δ 1 δ 2 over all δ 1, δ 2 0 for which 0 δ 1 + δ 2 1 is 1 and i is obained only when δ 3 1 = δ 2 = 1 3 Therefore, ( ) 1 1/s ( 1)1/s 1 x + y + z 2 1/s 3 12