Basics of Olympiad Inequalities - Williams College

1 Basics of Olympiad InequalitiesSamin RiasatiiIntroductionThe aim of this note is to acquaint students, who want to participate in mathematical Olympiads, toOlympiad level Inequalities from the Basics . Inequalities are used in all fields of mathematics. They havesome very interesting properties and numerous applications. Inequalities are often hard to solve, and it isnot always possible to find a nice solution. But it is worth approaching an inequality rather than solvingit. Most Inequalities need to be transformed into a suitable form by algebraic means before applyingsome theorem. This is what makes the problem rather difficult. Throughout this little note you will finddifferent ways and approaches to solve an inequality.

2 Most of the problems are recent and thus need afruitful combination of wisely applied took me around two years to complete this; although I didn t work on it for some months duringthis period. I have tried to demonstrate how one can use the classical Inequalities through different ex-amples that show different ways of applying them. After almost each section there are some exerciseproblems for the reader to get his/her hands dirty! And at the end of each chapter some harder problemsare given for those looking for challenges. Some additional exercises are given at the end of the book forthe reader to practice his/her skills. Solutions to some selected problems are given in the last chapter topresent different strategies and techniques of solving inequality problems.

3 In conclusion, I have tried toexplain that Inequalities can be overcome through practice and more , though this note is aimed for students participating in the Bangladesh Mathematical Olympiadwho will be hoping to be in the Bangladesh IMO team I hope it will be useful for everyone. I am reallygrateful to the MathLinks forum for supplying me with the huge collection of Riasat28 November, 2008iiiivINTRODUCTIONC ontentsIntroductioniii1 The AM-GM General AM-GM Inequality .. Weighted AM-GM Inequality .. More Challenging Problems ..72 Cauchy-Schwarz and H older s Cauchy-Schwarz Inequality .. H older s Inequality .. More Challenging Problems ..173 Rearrangement and Chebyshev s Rearrangement Inequality.

4 Chebyshev s inequality .. More Chellenging Problems ..254 Other Useful Schur s Inequality .. Jensen s Inequality .. Minkowski s Inequality .. Ravi Transformation .. Normalization .. Homogenization ..295 Supplementary Problems316 Hints and Solutions to Selected Problems33 References39vviCONTENTSC hapter 1 The AM-GM General AM-GM InequalityThe most well-known and frequently used inequality is the Arithmetic mean-Geometric mean inequalityor widely known as the AM-GM inequality. The term AM-GM is the combination of the two termsArithmetic MeanandGeometric Mean. The arithmetic mean of two numbersaandbis defined bya+ abis the geometric mean ofaandb. The simplest form of the AM-GM inequality is thefollowing:Basic AM-GM positive real numbersa, ba+b2 proof is simple.

5 Squaring, this becomes(a+b)2 4ab,which is equivalent to(a b)2 is obviously true. Equality holds if and only ifa= real numbersa, b, cprove thata2+b2+c2 ab+bc+ AM-GM inequality, we havea2+b2 2ab,b2+c2 2bc,c2+a2 the three Inequalities and then dividing by 2 we get the desired result. Equality holds if and onlyifa=b= inequality is equivalent to(a b)2+ (b c)2+ (c a)2 0,12 CHAPTER 1. THE AM-GM INEQUALITY which is obviously , the general AM-GM inequality is also true for anynpositive AM-GM positive real numbersa1, a2, .. , anthe following inequality +a2+ +ann n a1a2 an,with equality if and only ifa1=a2= = we present the well known Cauchy s proof by induction. This special kind of inductionis done by performing the following steps:i.

6 Base Pn the statement that the AM-GM is true 1: We already proved the inequality forn= 2. Forn= 3 we get the following inequality:a+b+c3 3 , b=y3, c=z3we equivalently getx3+y3+z3 3xyz is true by Example and the identityx3+y3+z3 3xyz= (x+y+z)(x2+y2+z2 xy yz zx).Equality holds forx=y=z, that is,a=b= 2: Assuming thatPnis true, we havea1+a2+ +ann n a1a2 it s not difficult to notice thata1+a2+ +a2n nn a1a2 an+nn an+1an+2 a2n 2n2n a1a2 a2nimplyingP2nis 3: First we assume thatPnis true +a2+ +ann n a1a2 this is true for all positiveais, we letan=n 1 a1a2 an 1. So now we havea1+a2+ +ann n a1a2 an 1n 1 a1a2 an 1=n (a1a2 an 1)nn 1=n 1 a1a2 an 1=an, GENERAL AM-GM INEQUALITY3which in turn is equivalent toa1+a2+ +an 1n 1 an=n 1 a1a2 an proof is thus complete.

7 It also follows by the induction that equality holds fora1=a2= = to understand yourself why this induction works. It can be useful , a2, .. , anbe positive real numbers such thata1a2 an= 1. Prove that(1 +a1)(1 +a2) (1 +an) AM-GM,1 +a1 2 a1,1 +a2 2 a2,..1 +an 2 the above Inequalities and using the facta1a2 an=1 we get our desired result. Equalityholds forai= 1, i= 1,2, .. , , b, cbe nonnegative real numbers. Prove that(a+b)(b+c)(c+a) inequality is equivalent to(a+b ab)(b+c bc)(c+a ca) 2 2 2,true by AM-GM. Equality holds if and only ifa=b= , b, c >0. Prove thata3bc+b3ca+c3ab a+b+ AM-GM we deduce thata3bc+b+c 33 a3bc b c= 3a,b3ca+c+a 33 b3ca c a= 3b,c3ab+a+b 33 c3ab a b= 1. THE AM-GM INEQUALITYA dding the three Inequalities we geta3bc+b3ca+c3ab+ 2(a+b+c) 3(a+b+c),which was what we (Samin Riasat)Leta, b, cbe positive real numbers.

8 Prove thatab(a+b) +bc(b+c) +ca(c+a) cycab ab(b+c)(c+a). AM-GM,2ab(a+b) + 2ac(a+c) + 2bc(b+c)=ab(a+b) +ac(a+c) +bc(b+c) +ab(a+b) +ac(a+c) +bc(b+c)=a2(b+c) +b2(a+c) +c2(a+b) + (a2b+b2c+a2c) + (ab2+bc2+a2c) a2(b+c) +b2(a+c) +c2(a+b) + (a2b+b2c+a2c) + 3abc=a2(b+c) +b2(a+c) +c2(a+b) +ab(a+c) +bc(a+b) +ac(b+c)=(a2(b+c) +ab(a+c))+(b2(a+c) +bc(a+b))+(c2(a+b) +ac(b+c)) 2 a3b(b+c)(a+c) + 2 b3c(a+c)(a+b) + 2 c3a(a+b)(b+c)= 2ab ab(b+c)(a+c) + 2cb bc(a+c)(a+b) + 2ac ca(a+b)(b+c).Equality holds if and only ifa=b= , b >0. Prove thatab+ba all real numbersa, b, cprove the following chain inequality3(a2+b2+c2) (a+b+c)2 3(ab+bc+ca).Exercise , b, cbe positive real numbers. Prove thata3+b3+c3 a2b+b2c+ , b, cbe positive real numbers. Prove thata3+b3+c3+ab2+bc2+ca2 2(a2b+b2c+c2a).

9 Exercise , b, cbe positive real numbers such thatabc= 1. Prove thata2+b2+c2 a+b+ WEIGHTED AM-GM INEQUALITY5 Exercise (a)Leta, b, c >0. Show that(a+b+c)(1a+1b+1c) 9.(b)For positive real numbersa1, a2, .. , anprove that(a1+a2+ +an)(1a1+1a2+ +1an) , b, cbe nonnegative real numbers such thata+b+c= 3. Prove thata2+b2+c2+ab+bc+ca , b, c, d >0. Prove thata2b+b2c+c2d+d2a a+b+c+ Weighted AM-GM InequalityThe weighted version of the AM-GM inequality follows from the original AM-GM inequality. Supposethata1, a2, .. , anare positive real numbers and letm1, m2, .. , mnbe positive integers. Then we haveby AM-GM,a1+a1+ +a1 m1+a2+a2+ +a2 m2+ +an+an+ +an mnm1+m2+ +mn a1a1.. a1 m1a2a2.. a2 m2 anan.

10 An mn 1m1+m2+ + can be written asm1a1+m2a2+ +mnanm1+m2+ +mn (am11am22 amnn)1m1+m2+ + equivalently in symbols miai mi ( amii)1 mj=mkm1+m2+ +mnfork= 1,2, .. , nwe can rewrite this as follows:Weighted AM-GM positive real numbersa1, a2, .. , anandnweightsi1, i2, .. , insuch thatn k=1ik= 1, we havea1i1+a2i2+ +anin ai11ai22 1. THE AM-GM INEQUALITYA lthough we have a proof ifi1, i2, .. , inare rational, this inequality is also true if they are positive realnumbers. The proof, however, is beyond the scope of this , b, cbe positive real numbers such thata+b+c= 3. Show thatabbcca that1 =a+b+c3 ab+bc+caa+b+c (abbcca)1a+b+c,which impliesabbcca (Nguyen Manh Dung)Leta, b, c >0 such thata+b+c= 1. Prove thataabbcc+abbcca+acbacb weighted AM-GM, we havea2+b2+c2a+b+c (aabbcc)1a+b+c= a2+b2+c2 aabbcc,ab+bc+caa+b+c (abbcca)1a+b+c= ab+bc+ca aabbcc,ac+ba+cba+b+c (acbacb)1a+b+c= ab+bc+ca up the three Inequalities we get(a+b+c)2 aabbcc+abbcca+ is,aabbcc+abbcca+acbacb few Inequalities can be solved using only the weighted AM-GM inequality.