Hojoo lee不等式相关竞赛题.pdf

TOPICS IN INEQUALITIESHojoo LeeVersion 0.5 2005/10/30IntroductionInequalities are useful in all fields of Mathematics.The purpose in this book is to present st an dard t ec hn iq ues in the theory of inequalities.The readers will meet classical theorems including Schurs inequality,Muirheads theorem,the Cauchy-Schwartz inequality,AM-GM inequality,and Holders theorem,etc.There are many problems from Mathematical olympiads and competitions.The book is available athttp:/ wish to express my appreciation to Stanley Rabinowitz who kindly sent me his paper On The Comput er Sol ut ion of Symmet ric Homogen eous Trian gl e In eq ual it ies.This is an unfinished manuscript.I would greatly appreciate hearing about any errors in the book,even minor ones.You can send all comments to the author at .To StudentsThe given techniques in this book are j ust the tip of the inequalities iceberg.What young students read this book should be aware of is that they should find t heir own c reat ive met hods to attack problems.Ifs impossible to present al l techniques in a small book.I dont even claim that the methods in this book are mathematically beautiful.For instance,although Muirheads theorem and Schurs theorem which can be found at chapter 3 are extremely powerful to attack homogeneous symmetric polynomial inequalities,its not a good idea for beginners to learn how to apply them to problems.(Why?)However,after mastering homogenization method using Muirheads theorem and Schurs theorem,you can have a more broad mind in the theory of inequalities.Thafs why I include the methods in this book.Have fun!Recommended Reading List1.K.S.Kedlaya,A 0)aI 2 3 b2 1-1-a+1 b+1-3I 2.(Columbia 2001)3 g e R)3(x+?/+l)2+1 3x yI 3.(0 x,?/1I 4.(APMC 1993)(a,b 0)y/a+Vb a+a2b+ab2+b a+Vab+b(2-4-3-(2I 5.(Czech and Slovakia 2000)(a,b0)I 6.(Die y/WURZEL,Heinz-Jurgen Seiffert)x y 0,x,e R)2岔y力+,_ x-V y+%一I 7.(Crux Mathematicorum,Problem 2645,Hojoo Lee)(a,6,c 0)2(+。3)9(。+b+o)2abc(a2+庐+c2)一I 8.(x,n,z0)I 2|+|2-n|N+g+23 3I 9.(q)a g y：z 0)(a+/)(b+g)(c+z)/abc-x yz3I 10.(x,g,z 0)x y z-H-X+/(力+)(+:y+z)(u+6)Z+/(z+)(+/I 11.(力+g+z=1,瑞 g,z0)I 12.(Iran 1998)(1 十 七+=2,为必 z 1)y/x+y-z Vx-1+，g _ 1+z I 13.(KMO Winter Program Test 2001)(a,6,c 0)/(a2b+b2c+c2a)(ab2+be2+ca2)abc+(a3+abc)(63+abc)(c3+abc)I 14.(KMO Summer Program Test 2001)(a,6,c 0)+64+C4+Ja2b2+12。2+。202 7a3b+於。+J就3+儿3+I 15.(Gazeta Matematica,Hojoo Lee)(a,6,c 0),04+2b2+.4+稔4+。2。2+/+，+。22+4 ay/2a2+be+b，2b2+c a+c/2c2+abI 16.(a,b,ceR)+b)2+(1 0)2+(1 十产 I 17.(a,b,c0)y/a2 ab b2+y/b2 be+c2 /a2 ac c?I 18.(Belarus 2002)(a,6,c,d 0)VQ+c)2+(b+d)2 H 2叫 她 ,2+匕2+近2+12 (q+0)2+产，(q+c)2+(b+d)2I 19.(Hong Kong 1998)(a,b,c 1)7a +y/b 1+/c 1 0)y(a+b)(b+c)(c+a)Jab+be+cqI 21.(Korea 1998)(/+y+z=x yz,x,y,z 0)1 1 1 3,H-,-,-0)a b c-+1/。2+8bc y/b2 Bea a/c2+Bab I 23.(IMO Short List 2004)abbec a=1,a,b,c 0)4I 24.(a,6,c 0)!ab(a+b)+y/bc(b+c)+ca(c+a)，4abo+(a+b)(b+c)(c+a)I 25.I 26.(Macedonia 1995)(a,b,c 0)a/b Ic6+c-V c-aa-b(Nesbitfs inequality)(a,6,c 0)a b c 3-1-1-b c c-a a-b 2I 27.(IMO 2000)(abc=1,a,b,c 0)a 1-bUliI 28.(ONI,Vasile Cirtoaje)(a,b,c 0)(a+1)(b T-1)+(b T-)(c T-)+(c T-)(a+)3I 29.(IMO Short List 1998)x yz=1,x,y,z 0)X3 O3?3 3 1-1 -(1+1/)(1+2)-(1+N)(l+/)-(1+N)(l+g)4I 30.(IMO Short List 1996)(abc=1,a,6,c 0)ab be ca 0)1 1 1 3-1-1-a3(b+c)63(c+a)c3(a+b)2I 32.(IMO Short List 1993)(a,b,c,d 0)a b c d 2-+-+-+-b+2c+3d c+2d+3a d+2q+3b a+2b+3c-3I 33.(IMO Short List 1990)(ab+be+cd+da=1,a,b,c,d0)a3 63 c3 t/3 1-1-1-1-b+c+d c+d+a d+a+b a b c 3I 34.(IMO 1968)01,62 0,7/1,7/2,1,2 e 凡为切 zj,力 2g2 炎2)1 I 1 8Xiyi-32 22g2-22(6 1+力2)(沙1+02)一(力+的产I 35.(Romania 1997)(a,b,c 0)a2 b2”be c a aba2+2bc+按+2c a c2+2ab a2+2bc+接+2c a c2+2abI 36.(Canada 2002)(a,6,c 0)q3/C3 7互+益+靛2a+b+。5I 37.(USA 1997)(a,6,0 0)1 1 1 1-1-1-0)(b+c a)2(c+a 6)2(a+6 c)2 3(b+c)2+涓+(+0)2+接+(q+b)2+02 5I 39.(USA 2003)(a,65c 0)(2a+6+c)2(26+c+a)2(2c+a+6)22q2+(b+c)2+2b2+(c+a)2+2c2+(a+6)2 I 40.(Crux Mathematicorum,Problem 2580,Hojoo Lee)(a,6,c 0)1 1 1 b+c c+a a+b-|-k-1-1-a b c a2+be b2+c a c2+abI 41.(Crux Mathematicorum,Problem 2581,Hojoo Lee)(a,b,c 0)a2+be b2+c a(?+ab b-c c-a a-b a+6+cI 42.(Crux Mathematicorum,Problem 2532,Hojoo Lee)(a2+62+c2=1,a,6,c 0)工工工 2(03+。3)。2 y c2-abcI 43.(Belarus 1999)(a2+62+c2=3,a,6,c 0)1113-+-+-1 ab 1 be 1+ca-2I 44.(Crux Mathematicorum,Problem 3032,Vasile Cirtoaje)(a2+62+c2=1,a,b,c 0)1119-1-1-0)1 1 1-1-1-0)7 2 I 1+2 L 1 T2 C 1-7+bE+c/c)+l c2+1 展+i 4 7I 47.(Iran 1996)(a,6,c 0)/7 7 X(1 1 1 9(Qb+bc+。)-4I 48.(Albania 2002)(a,b,c0)+庐+；+a+6+c+/a2+62+c2I 49.(Belarus 1997)(a,6,c 0)a b c a-b bc c-a-H-b-1-1-b c a c+a a+b b c6I 50.(Belarus 1998,I.Gorodnin)(a,6,c 0)a b c b+c+aa-bb+c b+c a+bI 51.(Poland 1996)(a+6+c=1,a,6,c 1)a b cq2+1+/+i+。2+iI 52.(Bulgaria 1997)(abc=1,a,6,c 0)1 1 1 111 +0)/9+*熄+/9/+7/6+3g3 _|_ y6+炉 z6 Z6 Z3X3+X6 I 54.(Vietnam 1991)(x y z 0)999 x y y z z xz x y、2 2 2 x+y+zI 55.(Iran 1997)(61/27374=1,g,62,3,4 0)x1 Xo+x l x i max x2+X4,-力1/2 13 力4I 56.(Hong Kong 2000)(abc=1,a,b,c0)1+ab2 1+b(?1+c o2 18c3+a3+*-q3+a+c3I 57.(Hong Kong 1997)0)3+/3 x yz x+g+n+y/x2+y2+z2)9 一(2+02+22)(g+gz+21)I 58.(Czech-Slovak Match 1999)(a,6,c 0)a b c、r-1-1-1b+2c c+2a a+2bI 59.(Moldova 1999)(a,6,c 0)ab+be+c a。+匕 十。c(c+a)a(a+b)b(b+c)-c+a b+a c-bI 60.(Baltic Way 1995)(a,b,c,d 0)a+c b+d c-a a b b+c c-dI 61.(ONI,Vasile Cirtoaje)(a,6,c,J 0)a b b c c d-1-：-I 7-v b c c+d a a a+bI 62.(Poland 1993)(x,yyu,v 0)x y x v uy uvx y uv x+y u v x y u-vd-b+一*4d a7I 63.(Belarus 1997)a,x,y,z 0)a y a z a x a+z x H y H z x y z -x+a-x-a-x-a+y-a+za+x。+V-y h z a+y-a-zI 64.(Lithuania 1987)n 0)x2+x y+y2+y2+yz z2z3+,2+,x+x2 x+y+z 3I 65.(Klamkins inequality)(1 x yz 2(i-)(1-g)(i 2)(i+z)(i+g)(i+z)一I 66.(x y+yz+z x=1,x,y,z 0)/y z 26(1-2g(1-y?)2z(l-01+72 1 y2 1+22 (1+X2)2(1+y2)2(1+22)2I 67.(Russia 2002)(/+g+z=3,x yz 0)Vx y/y+/z x y+yz+z xI 68.(APMO 1998)(a,6,c 0)220+*abc69.(Elemente der Mathematik,Problem 1207,Sefket Arslanagic)(x,g,z 0)三+2+*力+沙+一 y z x-y/x yz70.(Die y/WURZEL Walther Janous)(/+g+z=1,x yz 0)(1+)(1+)(1+Z)之(1 7)2+(1。)2+(J 一/)271.(United Kingdom 1999)(p q+r=1,p,q,T0)7(pq+qr+rp)0)I 73.(IMO 1984)(7+y+n=1,x,y,z 0)70 x y yz z x 2x yz 0)abc+bed+eda+dab (a2+b2 c2)(62+c2 a2)(c2+a2,31163+23+6/yz 8I 76.(Canada 1999)(1+g+z=1,x.yz 0)2 2 2/4x y+y z+z x 0)力(1 -)(1-z2)+7/(1-z2)(l-x2)+z(l-)(1-y2)0)2a-b c-d-(ab+ac+ad+be+bd+c d)oI 79.(Poland 1998)(a+b+c+d+e+/=1,ac e+bdf 壶 a,b,c,d,e,f 0)abc+bed+c de+def+efa+fab 36I 80.(Italy 1993)(0 a,6,c 1)a2+62+c2 lI 82.(Ireland 1997)(a+6+c abc,a,6,c 0)a2+62+c2 abcI 83.(BMO 2001)(a+6+c abc,a,b,c 0)a2+62+c2 VabcI 84.(Bearus 1996)x y z=y/x yz,x)y,z0)x y+yz+z x 9(i+g+z)I 85.(Poland 1991)(x2,+y2 z2=2 x,y,z R)x-y-z 2-x yzI 86.(Mongolia 1991)(a2+62+c2=2,a,6,c 6 R)|a3+63+c3-abc 2血I 87.(Vietnam 2002,Dung Tran Nam)(a2+62+c2=9,a,6,c 6 R)2(q+b+c)abc 0)(a+b)4+(b+c)4+(c+a)4 (a4+於+c)I 89.(x,0)x yz (g+z /)(?+c +y z)I 90.(Latvia 2002)Q,4+Ja+11C4+144=L Q,A c,d。)abed 39I 91.(Proposed for 1999 USAMO,AB,pp.25)1)Xx2-2yzyy2-2zxZz2+2xy.yxy+yz+zxI 92.(APMO 2004)(a,6,c 0)(a2+2)(ft2+2)(c2+2)9(ab+be+c o)I 93.(USA 2004)(a,6,c 0)(?-a2+3)(户+3)(c5-c2+3)(a+6+c)3I 94.(USA 2001)(a2+62+c2+abc=4,a,6,c 0)0 +6c+ca abc 6 a 0)(a+3b)(b+4c)(c+2a)60abcI 96.(Macedonia 1999)(a2+62+c2=1,a,b,c 0)a+b+c+-之 4：y/3 abc I 97.(Poland 1999)(a+b+c=1,a,6,c 0)a2+62+c2+2a/3abe 0)x2-h y2-h z2 V2(x y+yz)I 99.(APMC 1995)(m,n e N,x.y 0)(n-l)(m-1)(/+馆+7/n+m)+(n+m-1)(短1gm+/y71)n m(*n*Ty+刈口+吁1)I 100.(ONI,Gabriel Dospinescu,Mircea Lascu,Marian Tetiva)(a,6,c 0)a?+b?+c?+2abe+3 之(1+a)(l+b)(l+c)10Chapter 2Substitutions2.1 Eulers Theorem and the Ravi SubstitutionMany inequalities are simplified by some suitable substitutions.We begin with a classical inequality in triangle geometry.What is the first1 n on t rivial geometric inequality?In 1765,Euler showed thatTheorem 1.Let R and r den ot e t he radii of t he c irc umc irc l e and in c irc l e of t he t rian gl e ABC.Then,we have R 2r and t he eq ual it y hol ds if and on l y if ABC is eq uil at eral.Proof.Let BC=a,CA=6,AB=c,s=and S=ABC.2 Recall the well-known identities:S=嗡)S=rs,S2=s(s q)(s b)(s c).Hence,R is equivalent to 2、or abc 8?or abc 8(s a)(s b)(s c).We need to prove the following.Theorem 2.(AP,A.Padoa)Let a,b,c be t he l en gt hs of a t rian gl e.Then,we haveabc 8(s a)(s b)(s c)or abc (b+c a)(c+a +b c)and t he eq ual it y hol ds if and on l y if a=b=c.First Proof.We use the Ravi Substitution:Since a,b,c are the lengths of a triangle,there are positive reals 瑞 y,z such that a=y-z,b=z+%c=x-y.(Why?)Then,the inequality is +z)(z+6)(x+g)Sx yz for sc,y,z 0.However,we get y+?)(?+x)(x+?/)8x yz=x y z)2+y(z x)2+z x y)2 0.Sec on d Proof.(RI)We may assume that a b c.Its equivalent toa3+63+c3+3abc a2(b+c)+b2(c+a)+c2(a+6).Since c(a+b c)a b)c(a b c)3,applying the Rearrangement inequality,we obtaina-a(b+c a)+b +a b)+c c(a+6 c)a-a(b+c a)+c +a b)+a c(a+6 c),a a(b+c a)+b +a b)+c c(a+6 c)(1+When does t he eq ual it y hol d?Ifs natural to ask that the inequality in the theorem 2 holds for arbitrary positive reals a,b,c?Yes!Ifspossible to prove the inequality without the additional condition that a,6,c are the lengths of a triangle:xThe first geometric inequality is the Triangle Inequality:AB+BC AC2In this book,P stands for the area of the polygon P.3For example,we have c(a+b c)b(c a b)=(6 c)(b+c a)0.11Theorem 3.Let x,y,z 0.Then,we have x yz (g+z /)(2+力一+g z).The eq ual it y hol ds if and on l y if x=y=z.Proof.Since the inequality is symmetric in the variables,without loss of generality,we may assume that x y z.Then,we have x y z and z-V x y.If。then x,y z are the lengths of the sides of a triangle.And by the theorem 2,we get the result.Now,we may assume that y z 0 N(y+z 力)(z+力一+g z).The inequality in the theorem 2 holds when some of g,z are zeros:Theorem 4.Let x,y,z 0.Then,we have x yz (y+z +z 4)(力+y z).Proof.Since x yz 0,we can find posit iv e sequences xn,yn n for whichlim xn=x,lim yn=y,lim zn=z.nToe n 一 oo n 一 oo(For example,take xn=x n=1,2,),etc.)Applying the theorem 2 yields/nyn 2n 2(n+n)?/n)+Vn Now,taking the limits to both sides,we get the result.Clearly,the equality holds when x=y=z.However,x yz=(y-z x)(z-x y)(x-y z)and x,y,z 0 does not guarantee that x=y=z.In fact,for 弱 y,z 0,the equality x yz=+z 6)(z+/g)(/+0-z)is equivalent tox=y=z or N=g,z=0 or y=z x=0 or z=x y=Ifs straightforward to verify the equalityx yz (y+z +x-y)(x y-z)=x(x-y)(x z)+y(y-z)(y-x)+z z-x)(z-y).Hence,the theorem 4 is a particular case of Schurs inequality.4Problem 1.(IMO 2000/2)Let ab,c be posit ive n umbers suc h t hat abc=1.Prove t hat。-1+力(1+：)(1+力First Sol ut ion.Since abc=1,we make the substitution a=b=c=ior x,y,z 0.5 We rewrite the given inequality in the terms of x,y,z:x z /y x(z y/、/一、1+-1+-)1+-(y-z-+x-y)(x-y-z).y y)z/x x JThe Ravi Substitution is useful for inequalities for the lengths a,6,c of a triangle.After the Ravi Substitution,we can remove the condition that they are the lengths of the sides of a triangle.Problem 2.(IMO 1983/6)Let a,b,c be t he l en gt hs of t he sides of a t rian gl e.Prove t hata2b(a b)+b2c(b c)+(?a(c a)0.Sol ut ion.After setting a=y z,b=z+瑞 c=力+g for 瑞 g,z 0,it becomeso o o 2 q Q 6 J2 2?x z-y x z y x yz+x y z+x yz or-1-1-之力+0+汽y z xwhich follows from the Cauchy-Schwartz inequality(/+g+z)2.4See the theorem 10 in the chapter 3.Take r=1.5For example,take x=1,y=,z=Q+z+n)(乙+力+上)y 2 x j12Problem 3.(IMO 1961/2,Weitzenbock inequality)Let a,b,c be t he l en gt hs of a t rian gl e wit h area S.Show t hata2+62+c2 435.Sol ut ion.Write a=y+b=z+跖 c=x y ior z 0.Ifs equivalent to+z)2+(z+6)2+(+v)2)2 48(6+g+2)x yZ,which can be obtained as following:(+2)2+(2+x)2+(x+.)2)2 16Q/2+27+x y?16-3(x y yz+yz-z x+x y yz).6Exercise 2.(Hadwiger-Finsler inequality)Show t hat,for an y t rian gl e wit h sides a,b,c and area S,2ab+2bc+2ca (a2+62+c2)4VS.Exercise 3.(Pedoes inequality)Let cii,6i,ci den ot e t he sides of t he t rian gl e ABiCi wit h area Let den ot e t he sides of t he t rian gl e 4262c2 3it h area F?.Show t hatQ2(Q2?+电2 C22)+b2 供2+C22。2?)+。/?+022匕2?)16 田R2.6Here,we used the well-known inequalities j?2+q2 2pq and(p+q+r)2 3(pq q r rp).132.2 Trigonometric SubstitutionsIf you are faced with an integral that contains square root expressions such as/4力也/l+?/2 dy,/7力 一 dzthen trigonometric substitutions such as x=sin力，y=tan力，z sec t are very useful.When dealing with square root expressions,making a suitable t rigon omet ric substitution simplifies the given inequality.Problem 4.(Latvia 2002)Let a,b,c,d be t he posit ive real n umbers suc h t hat1 1 1 11+。4+1+a+1+04+1+d4=Prove t hat abed 3.Sol ut ion.We can write a2=tan A,b2=tanB,c2=tan C,d2=tan。,where A,B,C,D e(0,受).Then,the algebraic identity becomes the following trigonometric identity:cos2 A+cos2 B+cos2 C+cos2 D=1.Applying the AM-GM inequality,we obtainsin2 A=1 cos2 A=cos2 B+cos2 C+cos2 D 3(cos B cos C cos Dy.Similarly,we obtainsin2 B 3(cos C cos D cos A),sin2 C 3(cos D cos A cos 8户,and sin2 D 3(cos A cos B cos。户.Multiplying these inequalities,we get the result!Exercise 4.(ONI,Titu Andreescu,Gabriel Dosinescu)Let a,b,c,d be t he real n umbers suc h t hat(1+a2)(l+62)(1+c2)(l+/)=16.Prove t hat 3 S ab+ac+ad+be+bd+cd abed 5.Problem 5.(Korea 1998)Let x,y,z be t he posit ive real s wit h x+y+z=x yz.Show t hat1 1 1 3,H-,H,-1+力2+.2-1+*-2Since the function/is not concave down on R+,we cannot apply Jensens inequality to the function However,the function/(tan。)is concave down on(0,!Sol ut ion.We can write x=tan A,y=tanB,z tan C,where A,B,C G(0,J).Using the fact that 1+tan2 0=where cos0*0,we rewrite it in the terms of A,B,C:3 cos A+cos B+cos C .It follows from tan(7r C)=z=tan(A+B)and from 7r C,A+B e(0,7r)that tt C=A+Bor A+B+C=7r.Hence,it suffices to show the following.Theorem 5.In an y ac ut e t rian gl e ABC,we have cos A+cos B+cos C 1.Proof.Since cosx is concave down on(0,its a direct consequence of Jensens inequality.We note that the function cos力 is not concave down on(0,7r).In fact,ifs concave up on(表).One may think that the inequality cos A+cos B+cos C|doesnt hold for any triangles.However,its known that it also holds for any triangles.14Theorem 6.In an y t rian gl e ABC,we have cos A+c osB+cosC 0.Sec on d Proof.Let BC=a,CA=b,AB=c.Use the Cosine Law to rewrite the given inequality in the terms of a,b,c:62+c2 a2 c2+a2 b2 a2 b2 c2 3-+-+-a俨+，_ _2)+乂02+.2 庐)+庐一c2),which is equivalent to abc (b+c a)(c+a b)(a+b c)in the theorem 2.In case even when there is no condition such as 力+g+z=x yz or x y yz+z x=the trigonometricsubstitutions are useful.Problem 6.(APMO 2004/5)Prove t hat,for al l posit ive real n umbers a,6,c,(a2+2)(62+2)(c2+2)9(ab+be+c o).Proof.Choose A,B,C G(0,引 with a=/2 tan A,b=/2 tanB,and c=Using the well-knowntrigonometric identity 1+tan2 0=c o2g,one may rewrite it as4-cos A cos B cos C(cos A sin B sin C+sin A cos B sin C+sin A sin B cos C).One may easily check the following trigonometric identitycos(A+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 above trigonometric inequality takes the form49 cos A cos B cos C(cos A cos B cos C cos(A+8+C).Let 0=.Applying the AM-GM inequality and Jesens inequality,we havecos A cos B cos C cos A+cos B+cos C33 J cos3 6(cos3 0 cos 30).Using