Inequalities Marathon

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<p>MathLinks.roInequalities Marathonversion 1.00http://www.mathlinks.ro/viewtopic.php?t=299899Author:MathLink ersEditor:Hassan Al-SibyaniOctober 10, 2009IntroductionHello everyone, this le contain 100 problem in inequalities, generally at Pre-Olympiad level. Thoseproblems has been collected from MathLinks.ro at the topic called Inequalities Marathon. I decide tomake this work to be a good reference for young students who are interested in inequalities. This workprovides some of the nicest problems among international Olympiads, as well as some amazing problems bythe participants. Some of the problems contains more than one solution from the marathon and even outsidethe marathon from other sources.The file description:I- The next page contain all the members that participate with URLs to their MathLinks.ro user-nameinformation and also that page contain some of the real names of the participants that like to have their realnames on this le.II- Then there are some pages under the title of Important Denitions, this page contain some importantdenition that you may see them in the problems so if you dont know one of those denitions you can lookat it there.III- Then the problems section comes each problem with its solution(s).A- The problem format: Problem No. Author (Proposer):B- The solution format: Solution No. (Propser): (Note that the proposer of the solution is not nec-essary the one who make this solution)I would like to all the participants of this marathon for making this work, also I would like to give aspecial thank to Endrit Fejzullahu and Popa Alexandru for their contributing making this le.If you have another solution to any of those problem or a suggestion or even if you nd any mistake ofany type on this le please PM me at my MathLinks.ro user or e-mail me at hasan4444@gawab.com.Editor1Participant sorted alphabetically by thier Mathlink username:Agr 94 Mathalex2008apratimdefermatbeautifulliarbokagadhaBrut3Forc3can hang2007dgreenb801Dimitris Xenndb0xEvariste-GaloisFantasyLovergeniusblissgreat mathhasan4444keyree10Mateescu ConstantinmathinequsMaths MechanicMircea LascuPain rinneganpeinePotlaR.MathsRoersaifsocratesVirgil NiculaZheroList of names:Name Mathlink UserApratim De apratimdefermatAravind Srinivas Agr 94 MathDimitris Charisis Dimitris XEndrit Fejzullahu enndb0xHassan Al-Sibyani hasan4444Hoang Quoc Viet great mathMateescu Constantin Mateescu ConstantinMharchi Abdelmalek Evariste-GaloisMircea Lascu Mircea LascuMohamed El-Alami peineRaghav Grover Maths MechanicToang Huc Khein Pain RinneganPopa Alexandru alex2008Redwane Khyaoui R.MathsSayan Mukherjee PotlaVirgil Nicula Virgil NiculaVo Quoc Ba Can can hang20072Important DenitionsAbel Summation FormulaLet a1, a2, , an and b1, b2, , bn be two nite sequances of number. Thena1b1+a2b2+...+anbn = (a1a2)b1+(a2a3)(b1+b2)+ +(an1an)(b1+b2+ +bn1)+an(b1+b2+ +bn)AM-GM (Arithmetic Mean-Geometric Mean) InequalityIf a1, a2, , an are nonnegative real numbers, thena1, a2 + +ann na1a2 anwith equality if and only if a1 = a2 = = an.Cauchy-Schwarz InequalityFor any real numbers a1, a2, , an and b1, b2, , bn(a21 +a22 + +a2n)(b21 +b22 + +b2n) (a1b1 +a2b2 + +anbn)2with equality if and only if ai and bi are proportional, i = 1, 2, , n.Chebyshevs InequalityIf (a1, a2, , an) and (b1, b2, , bn) are two sequences of real numbers arranged in the same order then :n(a1b1 +a2b2 +... +anbn) (a1 +a2 +... +an)(b1 +b2 +... +bn)Cyclic SumLet n be a positive integer. Given a function f of n variables, dene the cyclic sum of variables (a1, a2, , an)as</p> <p>cycf(a1, a2, a3, . . . an) = f(a1, a2, a3, . . . an)+f(a2, a3, a4, . . . an, a1)+f(a3, a4, . . . an, a1, a2)+. . .+f(an, a1, a2, . . . an1)QM-AM-GM-HM (General Mean) InequalityIf a1, a2, , an are nonnegative real numbers, then_a21 +a22 + +a2nn a1, a2 + +ann na1a2 an n1a1 + 1a2 + + 1anwith equality if and only if a1 = a2 = = an.3GM-HM (Geometric Mean-Harmonic Mean) InequalityIf a1, a2, , an are nonnegative real numbers, thenna1a2 an n1a1 + 1a2 + + 1anwith equality if and only if a1 = a2 = = an.Hlawkas InequalityLet z1, z2, z3 be three complex numbers. Then:|z1 +z2| +|z2 +z3| +|z3 +z1| |z1| +|z2| +|z3| +|z1 +z2 +z3|Holders InequalityGiven mn (m, n N) positive real numbers (xij)(i1, m , j1, n) , then :m</p> <p>i1__m</p> <p>j1xij__1mn</p> <p>j1_m</p> <p>i1x1mij_Jensens InequalityLet F be a convex function of one real variable. Let x1, . . . , xn R and let a1, . . . , an 0 satisfya1 + +an = 1. Then:F(a1x1 + +anxn) a1F(x1) + +anF(xn)Karamatas InequalityGiven that F(x) is a convex function in x, and that the sequence {x1, . . . , xn} majorizes the sequence{y1, . . . , yn}, the theorem states thatF(x1) + +F(xn) F(y1) + +F(yn)The inequality is reversed if F(x) is concave, since in this case the function F(x) is convex.Lagranges IdentityThe identity:_ n</p> <p>k=1akbk_2=_ n</p> <p>k=1a2k__ n</p> <p>k=1b2k_</p> <p>1kjn(akbj ajbk)2applies to any two sets {a1, a2, , an} and {b1, b2, , bn} of real or complex numbers.4Minkowskis InequalityFor any real number r 1 and any positive real numbers a1, a2, , an and b1, b2, , bn. Then((a1 +b1)r+ (a2 +b2)r+ + (an +bn)r)1r (ar1 +ar2 + +arn)1r+ (br1 +br2 + +brn)1rMuirheads InequalityIf a sequence A majorizes a sequence B, then given a set of positive reals x1, x2, , xn</p> <p>symx1a1x2a2 xnan</p> <p>symx1b1x2b2 xnbnQM-AM (Quadratic Mean-Arithmetic Mean) InequalityAlso called root-mean-square and state that if a1, a2, , an are nonnegative real numbers, then_a21 +a22 + +a2nn a1, a2 + +annwith equality if and only if a1 = a2 = = an.The Rearrangement InequalityLet a1 a2 an and b1 b2 bn be real numbers. For any permutation (a</p> <p>1, a</p> <p>2, , a</p> <p>n) of(a1, a2, , an), we havea1b1 +a2b2 + +anbn a</p> <p>1b1 +a</p> <p>2b2 + +a</p> <p>nbn anb1 +an1b2 + +a1bnwith equality if and only if (a</p> <p>1, a</p> <p>2, , a</p> <p>n) is equal to (a1, a2, , an) or (an, an1, , a1) respectively.Schurs InequalityFor all non-negative a, b, c R and r &gt; 0:ar(a b)(a c) +br(b a)(b c) +cr(c a)(c b) 0The four equality cases occur when a = b = c or when two of a, b, c are equal and the third is 0.Common CasesThe r = 1 case yields the well-known inequality:a3+b3+c3+ 3abc a2b +a2c +b2a +b2c +c2a +c2bWhen r = 2, an equivalent form is: a4+b4+c4+abc(a +b +c) a3b +a3c +b3a +b3c +c3a +c3bWeighted AM-GM InequalityFor any nonnegative real numbers a1, a2, , an if w1, w2, , wn are nonnegative real numbers (weights)with sum 1, thenw1a1 +w2a2 + +wnan aw11 aw22 awnnwith equality if and only if a1 = a2 = = an.5Problem 1 India 2002 (Hassan Al-Sibyani): For any positive real numbers a, b, c show that the followinginequality holdsab + bc + ca c +ac +b + a +ba +c + b +cb +aFirst Solution (Popa Alexandru): Ok. After not so many computations i got that:ab + bc + ca a +bc +a b +ca +b c +ab +c= abc(a +b)(b +c)(c +a)_a2b2 + b2c2 + c2a2 ab bc ca_+ abc(a +b)(b +c)(c +a)_abc2 + bca2 + cab2 3_So in order to prove the above inequality we need to prove a2b2 + b2c2 + c2a2 ab + bc + ca and abc2 + bca2 + cab2 3The second inequality is obvious by AM-GM , and the for the rst we have:_a2b2 + b2c2 + c2a2_2 3_a2b2 + b2c2 + c2a2__ab + bc + ca_2where i used AM-GM and the inequality 3(x2+y2+z2) (x +y +z)2for x = ab , y = bc , z = caSo the inequality is proved.Second Solution (Raghav Grover): Substitute ab = x, bc = y, ca = z So xyz = 1. The inequality aftersubstitution becomesx2z +y2x +z2x +x2+y2+z2 x +y +z + 3x2z +y2x +z2x 3 So now it is left to prove that x2+y2+z2 x +y +z which is easy.Third Solution (Popa Alexandru): Bashing out it givesa4c2+b4a2+c4b2+a3b3+b3c3+a3c3 abc(ab2+bc2+ca2+ 3abc)which is true because AM-GM gives :a3b3+b3c3+a3c3 3a2b2c2and by Muirhead :a4c2+b4a2+c4b2 abc(ab2+ +bc2+ca2)Fourth Solution (Popa Alexandru): Observe that the inequality is equivalent with:</p> <p>cyca2+bca(a +b) 36Now use AM-GM:</p> <p>cyca2+bca(a +b) 3 3 (a2+bc)abc</p> <p>(a +b)So it remains to prove:</p> <p>(a2+bc) abc</p> <p>(a +b)Now we prove(a2+bc)(b2+ca) ab(c +a)(b +c) a3+b3 ab2+a2b (a +b)(a b)2 0Multiplying the similars we are done.Problem 2 Maxim Bogdan (Popa Alexandru): Let a, b, c, d &gt; 0 such that a b c d and abcd = 1 .Then show that:(a + 1)(d + 1) 3 + 34d3Solution (Mateescu Constantin):From the condition a b c d we get that a 1d3.= (a + 1)(d + 1) _ 1d3 + 1_(d + 1)Now lets prove that_1 + 1d3_(d + 1) 3 + 34d3This is equivalent with: (d3+ 1)(d + 1) 3d3+ 34 [d(d 1)]2[d(d 1)] + 1 34 _d(d 1) 122 0.Equality holds for a = 1d3 and d(d 1) 12 = 0 d = 1 +32Problem 3 Darij Grinberg (Hassan Al-Sibyani): If a, b, c are three positive real numbers, thena(b +c)2 + b(c +a)2 + c(a +b)2 94 (a +b +c)First Solution (Dimitris Charisis):</p> <p> a2ab2+ac2+ 2abc (a +b +c)2</p> <p>syma2b + 6abcSo we only have to prove that:4(a + b + c)3 9</p> <p>syma2b + 54abc 4(a3+ b3+ c3) + 12</p> <p>syma2b + 24abc 9</p> <p>syma2b + 54abc 4(a3+b3+c3) + 3</p> <p>syma2b 30abcBut</p> <p>syma2b 6abc and a3+b3+c3 3abcSo 4(a3+b3+c3) + 3</p> <p>syma2b 30abc7Second Solution (Popa Alexandru): Use Cauchy-Schwartz and Nesbitt:(a +b +c)_ a(b +c)2 + b(c +a)2 + c(a +b)2__ ab +c + bc +a + ca +b_2 94Problem 4 United Kingdom 1999 (Dimitris Charisis): For a, b, c 0 and a + b + c = 1 prove that7(ab +bc +ca) 2 + 9abcFirst Solution (Popa Alexandru):Homogenize to2(a +b +c)3+ 9abc 7(ab +bc +ca)(a +b +c)Expanding it becomes :</p> <p>syma3+ 6</p> <p>syma2b + 21abc 7</p> <p>syma2b + 21abcSo we just need to show:</p> <p>syma3</p> <p>syma2bwhich is obvious bya3+a3+b3 3a2b and similars.Second Solution (Popa Alexandru): Schur gives 1 + 9abc 4(ab + bc + ca) and use also 3(ab + bc + ca) (a +b +c)2= 1 Suming is done .Problem 5 Gheorghe Szollosy, Gazeta Matematica (Popa Alexandru): Let x, y, z R+. Prove that:_x(y + 1) +_y(z + 1) +_z(x + 1) 32_(x + 1)(y + 1)(z + 1)Solution (Endrit Fejzullahu): Dividing with the square root on the RHS we have :_ x(x + 1)(z + 1) +_ y(x + 1)(y + 1) +_ z(y + 1)(z + 1) 32By AM-GM_ x(x + 1)(z + 1) 12_ xx + 1 + 1y + 1__ y(x + 1)(y + 1) 12_ yy + 1 + 1x + 1__ z(y + 1)(z + 1) 12_ zz + 1 + 1y + 1_Summing we obtainLHS 12__ xx + 1 + 1x + 1_+_ yy + 1 + 1y + 1_+_ zz + 1 + 1z + 1__= 328Problem 6 Romanian Regional Mathematic Olympiad 2006 (Endrit Fejzullahu): Let a, b, c be positivenumbers , then prove that1a + 1b + 1c 4a2a2+b2+c2 + 4ba2+ 2b2+c2 + 4ca2+b2+ 2c2First Solution (Mateescu Constantin): By AM GM we have 2a2+b2+c2 4abc= 4a2a2+b2+c2 4a4abc= 1bcAddind the similar inequalities = RHS 1ab+ 1bc+ 1ca (1)Using Cauchy-Schwarz we have_ 1ab+ 1bc+ 1ca_2_1a + 1b + 1c_2So 1ab+ 1bc+ 1ca 1a + 1b + 1c (2)From (1), (2) we obtain the desired result .Second Solution (Popa Alexandru): By Cauchy-Schwatz :4a2a2+b2+c2 aa2+b2 + aa2+c2Then we haveRHS </p> <p>cyca +ba2+b2 </p> <p>cyc2a +b </p> <p>cyc_ 12a + 12b_= LHSThird Solution (Popa Alexandru): Since 2a2+ b2+ c2 a2+ ab + bc + ca = (a + b)(c + a). There wehave:LHS </p> <p>cyc4a(a +b)(c +a) = 8(ab +bc +ca)(a +b)(b +c)(c +a) 1a + 1b + 1cThe last one is equivalent with (a +b)(b +c)(c +a) 8abc .Problem 7 Pham Kim Hung (Mateescu Constantin): Let a, b, c, d, e be non-negative real numbers suchthat a +b +c +d +e = 5 . Prove that:abc +bcd +cde +dea +eab 5Solution (Popa Alexandru): Assume e min{a, b, c, d}. Then AM-GM gives :e(c +a)(b +d) +bc(a +d e) e(5 e)24 + (5 2e)227 5the last one being equivalent with:(e 1)2(e + 8) 09Problem 8 Popa Alexandru (Popa Alexandru): Let a, b, c be real numbers such that 0 a b c .Prove that:(a +b)(c +a)2 6abcFirst Solution (Popa Alexandru): Letb = xa , c = yb = xya x, y 1Then:(a +b)(a +c)23 2abc (x + 1)(xy + 1)2 a3 6x2ya3 (x + 1)(xy + 1)2 6x2y (x + 1)(4xy + (xy 1)2) 6x2y 4xy + (xy 1)2 x + (xy 1)22x2y 0We have that:4xy + (xy 1)2 x + (xy 1)22x2y 4xy + 2(xy 1)22x2y( because x 1)= 2x2y2+ 2 2x2y = 2xy(y 1) + 2 &gt; 0done.Second Solution (Endrit Fejzullahu): Let b = a +x, c = b +y = a +x +y,sure x, y 0Inequality becomes(2a +x)(x +y + 2a)26a(a +x)(a +x +y) 0But(2a +x)(x +y + 2a)26a(a +x)(a +x +y) = 2a3+ 2a2y + 2axy + 2ay2+x3+ 2x2y +xy2which is clearly positive.Problem 9 Nesbitt (Raghav Grover): Prove for positive realsab +c + bc +d + cd +a + da +b 2Solution (Dimitris Charisis):From andreescu LHS (a +b +c +d)2</p> <p>symab + (ac +bd)So we only need to prove that:(a +b +c +d)2 2</p> <p>symab + 2(ac +bd) (a c)2+ (b d)2 0....10Problem 10 Vasile Cartoaje (Dimitris Charisis): Let a, b, c, d be REAL numbers such that a2+b2+c2+d2=4 Prove that:a3+b3+c3+d3 8Solution (Popa Alexandru): Just observe thata3+b3+c3+d3 2(a2+b2+c2+d2) = 8because a, b, c, d 2Problem 11 Mircea Lascu (Endrit Fejzullahu): Let a, b, c be positive real numbers such that abc = 1.Prove that</p> <p>cyc1a2+ 2b2+ 3 12Solution (Popa Alexandru): Using AM-GM we have :LHS =</p> <p>cyc1(a2+b2) + (b2+ 1) + 2 </p> <p>cyc12ab + 2b + 2= 12</p> <p>cyc1ab +b + 1 = 12because1bc +c + 1 = 1bc +c +abc = 1c 1ab +b + 1 = abab +b + 1and1ca +a + 1 = 11b +a + 1 = bab +b + 1so</p> <p>cyc1ab +b + 1 = 1ab +b + 1 + abab +b + 1 + bab +b + 1 = 1Problem 12 Popa Alexandru (Popa Alexandru): Let a, b, c &gt; 0 such that a +b +c = 1. Prove that:1 +a +b2 +c + 1 +b +c2 +a + 1 +c +a2 +b 157First Solution (Dimitris Charisis):</p> <p>1 +a +b2 +c + 1 157 + 3 </p> <p>3 + (a +b +c)2 +c 367 </p> <p> 42 +c 36711But</p> <p> 222 +c (2 + 2 + 2)22 + 2 + 2 +a +b +c = 367Second Solution (Endrit Fejzullahu): Let a b c then by Chebyshevs inequality we haveLHS 13 (1 + 1 + 1 + 2(a +b +c))</p> <p>cyc12 +a = 53</p> <p>cyc12 +aBy Titus Lemma</p> <p>cyc12 +a 97, then LHS 157Problem 13 Titu Andreescu, IMO 2000 (Dimitris Charisis): Let a, b, c be positive so that abc = 1_a 1 + 1b__b 1 + 1c__c 1 + 1a_ 1Solution (Endrit Fejzullahu):_a 1 + 1b__b 1 + 1c__c 1 + 1a_ 1Substitute a = xy, b = yzInequality is equivalent with_xy 1 + zy__yz 1 + xz__zx 1 + yx_ 1 (x +z y)(y z +x)(z x +y) xyzWLOG ,Let x &gt; y &gt; z, then x +z &gt; y, x +y &gt; z.If y +z &lt; x, then we are done because(x +z y)(y z +x)(z x +y) 0 and xyz 0Otherwise if y +z &gt; x , then x, y, z are side lengths of a triangle ,and then we can make the substitutionx = m+n, y = n +t and z = t +mInequality is equivalent with8mnt (m+n)(n +t)(t +m), this is true by AM-GMm+n 2mn, n +t 2nt and t +m 2tm, multiply and were done.Problem 14 Korea 1998 (Endrit Fejzullahu): Let a, b, c &gt; 0 and a +b +c = abc....</p>