Vn Inequalities Marathon

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MathLinks.roInequalities Marathonhttp://www.mathlinks.ro/viewtopic.php?t=299899Author:MathLink ersEditor:Hassan Al-SibyaniOctober 4, 2009Participant sorted alphabetically by thier Mathlink username:alex2008apratimdefermatbeautifulliarbokagadhaBrut3Forc3dgreenb801Dimitris Xenndb0xFantasyLovergeniusblissgreat mathhasan4444keyree10Mateescu ConstantinMaths MechanicPain rinneganPotlaR.MathsRoersaifsocratesVirgil NiculaList of names:Name Mathlink UserApratim De apratimdefermatEndrit Fejzullahu enndb0xHassan Al-Sibyani hasan4444Hoang Quoc Viet great mathMateescu Constantin Mateescu ConstantinRaghav Grover Maths MechanicToang Huc Khein Pain RinneganPopa Alexandru alex2008Redwane Khyaoui R.MathsSayan Mukherjee PotlaVirgil Nicula Virgil Nicula1Problem 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:

cyca2+bca(a +b) 3Now use AM-GM:

cyca2+bca(a +b) 3 3 (a2+bc)abc

(a +b)2So it remains to prove:

(a2+bc) abc

(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 > 0 such that a b c d and abcd = 1 .Then show that:(a + 1)(d + 1) 3 + 34d3First Solution (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 X):

a2ab2+ac2+ 2abc (a +b +c)2

syma2b + 6abcSo we only have to prove that:4(a + b + c)3 9

syma2b + 54abc 4(a3+ b3+ c3) + 12

syma2b + 24abc 9

syma2b + 54abc 4(a3+b3+c3) + 3

syma2b 30abcBut

syma2b 6abc and a3+b3+c3 3abcSo 4(a3+b3+c3) + 3

syma2b 30abcSecond 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 943Problem 4 United Kingdom (Dimitris X): For a, b, c 0 and a+b+c = 1 prove that 7(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 :

syma3+ 6

syma2b + 21abc 7

syma2b + 21abcSo we just need to show:

syma3

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)First 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__= 32Problem 6 (Endrit Fejzullahu): Let a, b, c be positive numbers , then prove that1a + 1b + 1c 4a2a2+b2+c2 + 4ba2+ 2b2+c2 + 4ca2+b2+ 2c24First 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

cyca +ba2+b2

cyc2a +b

cyc_ 12a + 12b_= LHSProblem 7 (Mateescu Constantin): Let a, b, c, d, e be non-negative real numbers such that a + b +c +d +e = 5 . Prove that:abc +bcd +cde +dea +eab 5First Solution (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) 0Problem 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 6x2ya35(x + 1)(xy + 1)2 6x2y(x + 1)(4xy + (xy 1)2) 6x2y4xy + (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 > 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 (Raghav Grover): Prove for positive realsab +c + bc +d + cd +a + da +b 2First Solution (Dimitris X):From andreescu LHS (a +b +c +d)2

symab + (ac +bd)So we only need to prove that:(a +b +c +d)2 2

symab + 2(ac +bd) (a c)2+ (b d)2 0....Problem 10 (Dimitris X): Let a, b, c, d be REAL numbers such that a2+ b2+ c2+ d2= 4 Provethat:a3+b3+c3+d3 8First Solution (Popa Alexandru): Just observe thata3+b3+c3+d3 2(a2+b2+c2+d2) = 8because a, b, c, d 26Problem 11 (Endrit Fejzullahu): Let a, b, c be positive real numbers such that abc = 1 .Prove that

cyc1a2+ 2b2+ 3 12First Solution (Popa Alexandru): Using AM-GM we have :LHS =

cyc1(a2+b2) + (b2+ 1) + 2

cyc12ab + 2b + 2= 12

cyc1ab +b + 1 = 12because1bc +c + 1 = 1bc +c +abc = 1c 1ab +b + 1 = abab +b + 1and1ca +a + 1 = 11b +a + 1 = bab +b + 1so

cyc1ab +b + 1 = 1ab +b + 1 + abab +b + 1 + bab +b + 1 = 1Problem 12 Popa Alexandru (Popa Alexandru): Let a, b, c > 0 such that a +b +c = 1. Prove that:1 +a +b2 +c + 1 +b +c2 +a + 1 +c +a2 +b 157First Solution (Dimitris X):

1 +a +b2 +c + 1 157 + 3

3 + (a +b +c)2 +c 367

42 +c 367But

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))

cyc12 +a = 53

cyc12 +aBy Titus Lemma

cyc12 +a 97, then LHS 157Problem 13 Titu Andreescu, IMO 2000 (Dimitris X): Let a, b, c be positive so that abc = 1_a 1 + 1b__b 1 + 1c__c 1 + 1a_ 17First Solution (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 > y > z, then x +z > y, x +y > z.If y +z < x, then we are done because(x +z y)(y z +x)(z x +y) 0 and xyz 0Otherwise if y +z > 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 > 0 and a +b +c = abc. Prove that:1a2+ 1 + 1b2+ 1 + 1c2+ 1 32First Solution (Dimitris X): Setting a = 1x, b = 1y, c = 1z the condition becomes xy + yz + zx = 1, and theinequality:

xx2+ 1 32But

xx2+ 1 =

x_x2+xy +xz +zy=

_ xx +yxx +zBut_ xx +yxx +z xx +y + xx +z2So

_ xx +yxx +z xx +y + yx +y + xz +x + zz +x + yy +z + zz +y2 = 32Second Solution (Raghav Grover):Substitute a = tan x,b = tan y and c = tan z where x +y +z = And we are left to provecos x + cos y + cos z 32Which i think is very well known..Third Solution (Endrit Fejzullahu): By AM-GM we have a + b + c 3 3abc and since a + b + c = abc =(abc)2 27We rewrite the given inequality as13_ 1a2+ 1 + 1b2+ 1 + 1c2+ 1_ 128Since function f(a) = 1a2+ 1 is concave ,we apply Jensens inequality13f(a) + 13f(b) + 13f(c) f_a +b +c3_= f_abc3_= 1_(abc)232 + 1 12 (abc)2 27, QEDProblem 15 (Dimitris X):If a, b, c R and a2+b2+c2= 3. Find the minimum value of A = ab +bc +ca 3(a +b +c).First Solution (Endrit Fejzullahu):ab +bc +ca = (a +b +c)2a2b2c22 = (a +b +c)232Let a +b +c = xThen A = x22 3x 32We consider the second degree fuction f(x) = x22 3x 32We obtain minimum for f_b2a_= f(3) = 6Then Amin = 6 ,it is attained for a = b = c = 1Problem 16 (Endrit Fejzullahu): If a, b, c a