Basics of olympiad inequalities

Basics of Olympiad Inequalities Samin Riasat ii Introduction The aim of this note is to acquaint students, who want to participate in mathematical Olympiads, to Olympiad level inequalities from the basics Inequalities are used in all fields of mathematics They have some very interesting properties and numerous applications Inequalities are often hard to solve, and it is not always possible to find a nice solution But it is worth approaching an inequality rather than solving it Most inequalities need to be transformed into a suitable form by algebraic means before applying some theorem This is what makes the problem rather difficult Throughout this little note you will find different ways and approaches to solve an inequality Most of the problems are recent and thus need a fruitful combination of wisely applied techniques It took me around two years to complete this; although I didn’t work on it for some months during this period I have tried to demonstrate how one can use the classical inequalities through different examples that show different ways of applying them After almost each section there are some exercise problems for the reader to get his/her hands dirty! And at the end of each chapter some harder problems are given for those looking for challenges Some additional exercises are given at the end of the book for the reader to practice his/her skills Solutions to some selected problems are given in the last chapter to present different strategies and techniques of solving inequality problems In conclusion, I have tried to explain that inequalities can be overcome through practice and more practice Finally, though this note is aimed for students participating in the Bangladesh Mathematical Olympiad who will be hoping to be in the Bangladesh IMO team I hope it will be useful for everyone I am really grateful to the MathLinks forum for supplying me with the huge collection of problems Samin Riasat 28 November, 2008 iii iv INTRODUCTION Contents Introduction The 1.1 1.2 1.3 iii AM-GM Inequality General AM-GM Inequality Weighted AM-GM Inequality More Challenging Problems 1 Cauchy-Schwarz and Hă olders Inequalities 2.1 Cauchy-Schwarz Inequality 2.2 Hăolders Inequality 2.3 More Challenging Problems 9 14 17 Rearrangement and Chebyshev’s Inequalities 3.1 Rearrangement Inequality 3.2 Chebyshev’s inequality 3.3 More Chellenging Problems 19 19 23 25 Other Useful Strategies 4.1 Schur’s Inequality 4.2 Jensen’s Inequality 4.3 Minkowski’s Inequality 4.4 Ravi Transformation 4.5 Normalization 4.6 Homogenization 27 27 27 28 28 29 29 Supplementary Problems 31 Hints and Solutions to Selected Problems 33 References 39 v vi CONTENTS Chapter The AM-GM Inequality 1.1 General AM-GM Inequality The most well-known and frequently used inequality is the Arithmetic mean-Geometric mean inequality or widely known as the AM-GM inequality The term AM-GM is the combination of the two terms Arithmetic Mean and Geometric Mean The arithmetic mean of two numbers a and b is defined by a+b √ Similarly ab is the geometric mean of a and b The simplest form of the AM-GM inequality is the following: Basic AM-GM Inequality For positive real numbers a, b a+b √ ≥ ab The proof is simple Squaring, this becomes (a + b)2 ≥ 4ab, which is equivalent to (a − b)2 ≥ This is obviously true Equality holds if and only if a = b Example 1.1.1 For real numbers a, b, c prove that a2 + b2 + c2 ≥ ab + bc + ca First Solution By AM-GM inequality, we have a2 + b2 ≥ 2ab, b2 + c2 ≥ 2bc, c2 + a2 ≥ 2ca Adding the three inequalities and then dividing by we get the desired result Equality holds if and only if a = b = c Second Solution The inequality is equivalent to (a − b)2 + (b − c)2 + (c − a)2 ≥ 0, CHAPTER THE AM-GM INEQUALITY which is obviously true However, the general AM-GM inequality is also true for any n positive numbers General AM-GM Inequality For positive real numbers a1 , a2 , , an the following inequality holds √ a1 + a2 + · · · + an ≥ n a1 a2 · · · an , n with equality if and only if a1 = a2 = · · · = an Proof Here we present the well known Cauchy’s proof by induction This special kind of induction is done by performing the following steps: i Base case ii Pn =⇒ P2n iii Pn =⇒ Pn−1 Here Pn is the statement that the AM-GM is true for n variables Step 1: We already proved the inequality for n = For n = we get the following inequality: a+b+c √ ≥ abc Letting a = x3 , b = y , c = z we equivalently get x3 + y + z − 3xyz ≥ This is true by Example 1.1.1 and the identity x3 + y + z − 3xyz = (x + y + z)(x2 + y + z − xy − yz − zx) Equality holds for x = y = z, that is, a = b = c Step 2: Assuming that Pn is true, we have √ a1 + a2 + · · · + an ≥ n a1 a2 · · · an n Now it’s not difficult to notice that √ √ √ a1 + a2 + · · · + a2n ≥ n n a1 a2 · · · an + n n an+1 an+2 · · · a2n ≥ 2n 2n a1 a2 · · · a2n implying P2n is true Step 3: First we assume that Pn is true i.e √ a1 + a2 + · · · + an ≥ n a1 a2 · · · an n √ As this is true for all positive s, we let an = n−1 a1 a2 · · · an−1 So now we have a1 + a2 + · · · + an ≥ n n a1 a2 · · · an−1 n √ n−1 n a1 a2 · · · an−1 (a1 a2 · · · an−1 ) n−1 √ = n−1 a1 a2 · · · an−1 = = an , 1.1 GENERAL AM-GM INEQUALITY which in turn is equivalent to a1 + a2 + · · · + an−1 ≥ an = n−1 √ n−1 a1 a2 · · · an−1 The proof is thus complete It also follows by the induction that equality holds for a1 = a2 = · · · = an Try to understand yourself why this induction works It can be useful sometimes Example 1.1.2 Let a1 , a2 , , an be positive real numbers such that a1 a2 · · · an = Prove that (1 + a1 )(1 + a2 ) · · · (1 + an ) ≥ 2n Solution By AM-GM, √ + a1 ≥ a1 , √ + a2 ≥ a2 , √ + an ≥ an Multiplying the above inequalities and using the fact a1 a2 · · · an =1 we get our desired result Equality holds for = 1, i = 1, 2, , n Example 1.1.3 Let a, b, c be nonnegative real numbers Prove that (a + b)(b + c)(c + a) ≥ 8abc Solution The inequality is equivalent to a+b √ ab b+c √ bc c+a √ ca ≥ · · 2, true by AM-GM Equality holds if and only if a = b = c Example 1.1.4 Let a, b, c > Prove that a3 b3 c3 + + ≥ a + b + c bc ca ab Solution By AM-GM we deduce that a3 +b+c≥3 bc b3 +c+a≥3 ca c3 +a+b≥3 ab a3 · b · c = 3a, bc b3 · c · a = 3b, ca c3 · a · b = 3c ab CHAPTER THE AM-GM INEQUALITY Adding the three inequalities we get a3 b3 c3 + + + 2(a + b + c) ≥ 3(a + b + c), bc ca ab which was what we wanted Example 1.1.5 (Samin Riasat) Let a, b, c be positive real numbers Prove that ab(a + b) + bc(b + c) + ca(c + a) ≥ ab cyc a (b + c)(c + a) b Solution By 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) + (a2 b + b2 c + a2 c) + (ab2 + bc2 + a2 c) ≥ a2 (b + c) + b2 (a + c) + c2 (a + b) + (a2 b + b2 c + a2 c) + 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 = 2ab a3 b(b + c)(a + c) + b3 c(a + c)(a + b) + a (b + c)(a + c) + 2cb b c3 a(a + b)(b + c) b (a + c)(a + b) + 2ac c c (a + b)(b + c) a Equality holds if and only if a = b = c Exercise 1.1.1 Let a, b > Prove that a b + ≥ b a Exercise 1.1.2 For all real numbers a, b, c prove the following chain inequality 3(a2 + b2 + c2 ) ≥ (a + b + c)2 ≥ 3(ab + bc + ca) Exercise 1.1.3 Let a, b, c be positive real numbers Prove that a3 + b3 + c3 ≥ a2 b + b2 c + c2 a Exercise 1.1.4 Let a, b, c be positive real numbers Prove that a3 + b3 + c3 + ab2 + bc2 + ca2 ≥ 2(a2 b + b2 c + c2 a) Exercise 1.1.5 Let a, b, c be positive real numbers such that abc = Prove that a2 + b2 + c2 ≥ a + b + c 3.3 MORE CHELLENGING PROBLEMS 25 Exercise 3.2.7 Let a, b, c > such that abc = Prove that ≤ cyc · + b) c2 (a ab ≤ cyc cyc ab + b) c2 (a Exercise 3.2.8 Let a, b, c > Prove that aa bb cc ≥ (abc) 3.3 a+b+c More Chellenging Problems Exercise 3.3.1 (Samin Riasat) Let a, b, c > Prove that min{a, b, c} a+b+c max{a, b, c} + ≥ √ − min{a, b, c} max{a, b, c} abc Exercise 3.3.2 Let a1 ≥ a2 · · · ≥ an and b1 ≥ b2 ≥ · · · ≥ bn be positive (i ) If (ci )ni=1 is a permutation of (bi )ni=1 prove that a1b1 −c1 ab22 −c2 · · · abnn −cn ≥ (ii ) Let b = b1 + b2 + · · · + bn Prove that n a1b1 −b ab22 −b · · · abnn −b ≥ Exercise 3.3.3 Let x, y, z ∈ R+ Prove that √ 3 xyz x3 + y + z + ≥ 3xyz x+y+z Exercise 3.3.4 (Samin Riasat) Let a, b, c be positive real numbers Prove that a2 + b2 + c2 ≥ ab + bc + ca a b c + + b+c c+a a+b *Exercise 3.3.5 (Samin Riasat) Let a, b, c be positive real numbers and n be a positive integer Prove that n n−1 a an a ≤ b+c bn + bn−1 c c+a cyc cyc cyc 26 CHAPTER REARRANGEMENT AND CHEBYSHEV’S INEQUALITIES Chapter Other Useful Strategies 4.1 Schur’s Inequality Let a, b, c be positive real numbers, and n be positive Then the following inequality holds: an (a − b)(a − c) + bn (b − c)(b − a) + cn (c − a)(c − b) ≥ 0, with equality if and only if a = b = c or a = b, c = and permutations The above inequality is known as Schur’s inequality, after Issai Schur Proof Since the inequality is symmetric in a, b, c WLOG we may assume that a ≥ b ≥ c Then the inequality is equivalent to (a − b)(an (a − c) − bn (b − c)) + cn (a − c)(b − c) ≥ 0, which is obviously true 4.2 Jensen’s Inequality Suppose f is a convex function in [a, b] Then the inequality f a1 + a2 + · · · + an n ≤ f (a1 ) + f (a2 ) + · · · + f (an ) n is true for all ∈ [a, b] Similarly, if f is concave in the interval the sign of inequality turns over This is called Jensen’s inequality The convexity is usually determined by checking if f (x) ≥ holds for all x ∈ [a, b] Similarly for concavity one can check if f (x) ≤ for all x ∈ [a, b] Here is an example: Example 4.2.1 Let a, b, c > Prove that aa bb cc ≥ a+b+c a+b+c Solution Consider the function f (x) = x ln x Verify that f (x) = 1/x > for all x ∈ R+ Thus f is 27 28 CHAPTER OTHER USEFUL STRATEGIES convex in R+ and by Jensen’s inequality we conclude that f (a) + f (b) + f (c) ≥ 3f a+b+c a b c ⇔ ln a + ln b + ln c ≥ ln a+b+c a+b+c , which is equivalent to ln(aa bb cc ) ≥ ln a+b+c a+b+c , which was what we wanted 4.3 Minkowski’s Inequality Minkowski’s inequality states that for positive numbers xi , yi and p the following inequality holds: p n (xi + yi ) p xpi ≤ i=1 4.4 p n i=1 p n yip + i=1 Ravi Transformation Suppose that a, b, c are the side lengths of a triangle Then positive real numbers x, y, z exist such that a = x + y, b = y + z and c = z + x To verify this, let s be the semi-perimeter Then denote z = s − a, x = s − b, y = s − c and the b+c−a > and similarly for the others conclusion is obvious since s − a = Geometrically, let D, E, F denote the points of tangency of BC, CA, AB, respectively, with the incircle of triangle ABC Then BD = BF = x, CD = CE = y and AE = AF = z implies the conclusion Here are some examples of how the Ravi transformation can transform a geometric inequality into an algebraic one: Example 4.4.1 (IMO 1964) Let a, b, c be the side lengths of a triangle Prove that a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc First Solution Verify that the inequality can be written as (a + b − c)(b + c − a)(c + a − b) ≤ abc Let a = x + y, b = y + z and c = z + x Then the above inequality becomes 8xyz ≤ (x + y)(y + z)(z + x), which is Example 1.1.3 Second Solution The inequality is equivalent to a3 + b3 + c3 + 3abc ≥ a2 b + ab2 + b2 c + bc2 + c2 a + ca2 , or, a(a − b)(a − c) + b(b − c)(b − a) + c(c − a)(c − b) ≥ 0, 4.5 NORMALIZATION 29 which is Schur’s inequality Example 4.4.2 Let a, b, c be the lengths of the sides of a triangle Prove that √ √ ab + bc + √ ca ≥ √ a+b−c+ √ b+c−a+ √ c + a − b Solution Let x, y, z > such that a = x + y, b = y + z, c = z + x Then our inequality is equivalent to √ (x + y)(y + z) ≥ cyc x cyc From Cauchy-Schwarz inequality, (x + y)(y + z) ≥ 3 cyc (y + √ zx) cyc ≥2 √ y+4 cyc √ =2 zx cyc x cyc 4.5 Normalization Homogeneous inequalities can be normalized, e.g applied restrictions with homogeneous expressions in the variables For example, in order to show that a3 + b3 + c3 − 3abc ≥ 0, one may assume, WLOG, that abc = or a + b + c = etc The reason is explained below Suppose that abc = k Let a = ka , b = kb , c = kc This implies a b c = 1, and our inequality becomes a + b + c − 3a b c ≥ 0, which is the same as before Therefore the restriction abc = doesn’t change anything of the inequality Similarly one might also assume a + b + c = The reader is requested to find out how it works 4.6 Homogenization This is the opposite of Normalization It is often useful to substitute a = x/y, b = y/z, c = z/x, when the condition abc = is given Similarly when a + b + c = we can substitute a = x/x + y + z, b = y/x + y + z, c = z/x + y + z to homogenize the inequality For an example of homogenization note that we can write the inequality in exercise 1.3.1 in the following form: a b c a+b+c + + ≥ √ b c a abc On the other hand, if we substitute a = x/y, b = y/z, c = z/x the inequality becomes, zx xy yz x y z + + ≥ + + , y z x y z x which clearly looks easier to deal with (Hint: Rearrangement) Many such substitutions exist, and the reader is urged to study them and find them using his/her own ideas 30 CHAPTER OTHER USEFUL STRATEGIES Chapter Supplementary Problems Exercise 5.1.1 Let a, b, c be nonnegative reals Prove that ab + bc + ca ≤ 3 (a + b) (b + c) (c + a) Exercise 5.1.2 For a, b, c > prove that a b c + + ≥ 4 (b + c) (c + a) (a + b) 2(a + b)(b + c)(c + a) Exercise 5.1.3 Let a, b, c be real numbers Prove that + (abc)2 + a2 + b2 + c2 ≥ 2(ab + bc + ca) Exercise 5.1.4 (Michael Rozenberg) Let a, b, c be non-negative numbers such that a + b + c = Prove that √ a 2b + c2 + b 2c + a2 + c 2a + b2 ≤ 3 Exercise 5.1.5 For any acute-angled triangle ABC show that s tan A + tan B + tan C ≥ , r where s and r denote the semi-perimeter and the inraduis, respectively Exercise 5.1.6 (Iran 2008) Find the smallest real K such that for each x, y, z ∈ R+ : √ √ √ x y + y z + z x ≤ K (x + y)(y + z)(z + x) Exercise 5.1.7 (USA 1997) Prove the following inequality for a, b, c > a3 1 1 + + ≤ 3 + b + abc b + c + abc a + c + abc abc Exercise 5.1.8 Let x ≥ y ≥ z > be real numbers Prove that x2 y y z z x + + ≥ x2 + y + z z x y 31 32 CHAPTER SUPPLEMENTARY PROBLEMS Exercise 5.1.9 (Greece 2007) Let a, b, c be sides of a triangle Show that (c + a − b)4 (a + b − c)4 (b + c − a)4 + + ≥ ab + bc + ca a(a + b − c) b(b + c − a) c(c + a − b) Exercise 5.1.10 (Samin Riasat) Let a, b, c be sides of a triangle Show that (a + b − c)4 (b + c − a)4 (c + a − b)4 + + ≥ a2 + b2 + c2 a(a + b − c) b(b + c − a) c(c + a − b) Exercise 5.1.11 (Crux Mathematicorum) If a, b and c are the sidelengths of a triangle, then prove the inequality (b + c)2 (c + a)2 (a + b)2 + + ≥ a2 + bc b + ca c + ab Exercise 5.1.12 Let a, b, c be the side-lengths of a triangle Prove that √ (a + b)(b + c) a − b + c ≥ 4(a + b + c) (−a + b + c)(a − b + c)(a + b − c) cyc Exercise 5.1.13 Prove that if a, b, c > then √ √ √ √ abc( a + b + c) + (a + b + c)2 ≥ 3abc(a + b + c) Exercise 5.1.14 Let a, b, c be non-negative real numbers such that a + b + c = Prove that ab + bc + ca ≤ (1 − ab)(1 − bc) ≤ a2 + b2 + c2 cyc Exercise 5.1.15 Let a, b and c be nonnegative real numbers such that 1 + + = a2 + b2 + c2 + Prove that ab + bc + ca ≤ Exercise 5.1.16 (Korea 1998) Let I be the incenter of triangle ABC Prove that 3(IA2 + IB + IC ) ≥ AB + BC + CA2 Exercise 5.1.17 (Samin Riasat) Let a, b, c be positive real numbers such that a6 + b6 + c6 = Prove that a7 b2 + b7 c2 + c7 a2 ≤ Exercise 5.1.18 (Samin Riasat) Let x, y, z be positive real numbers Prove that x y z + + ≥ y z x x+y + 2z y+z + 2x z+x 2y Exercise 5.1.19 (Samin Riasat) Let x, y, z be positive real numbers Prove that xy + (x + y)(y + z) yz + (y + z)(z + x) zx ≤ (z + x)(x + y) Chapter Hints and Solutions to Selected Problems 1.1.2 Expand and use Example 1.1.1 1.1.3 a3 + a3 + b3 ≥ 3a2 b 3a 1.3.1 Use ab + ab + cb ≥ √ to prove abc a b c a+b+c + + ≥ √ b c a abc 1.3.4 See hint for 1.3.1 1.3.5 Prove and use the following: a2 b2 c2 (a + b + c)(a2 + b2 + c2 ) + + ≥ b c a ab + bc + ca a a = ab+ca 2.1.2 b+c 2.1.5 Use Example 2.1.5 2.3.6 Solution: Note that √ √ a2 +abc c+ab = a(c+a)(a+b) (b+c)(c+a) Therefore our inequality is equivalent to ⇐⇒ a(c + a)(a + b) a+b+c ≤ √ (b + c)(c + a) abc a(a + b) bc(c + a)(a + b) ≤ (a + b + c)(a + b)(b + c)(c + a) By AM-GM, a(a + b) · bc(c + a)(a + b) ≤ = a(a + b)(b(c + a) + c(a + b)) a(a + b)(ab + 2bc + ca) Now a(a + b)(ab + 2bc + ca) = a2 (ab + bc + ca) + a2 bc + ab(ab + bc + ca) + ab2 c = (a2 + b2 + c2 + ab + bc + ca)(ab + bc + ca) + 2abc(a + b + c) = (a + b + c)2 (ab + bc + ca) − (ab + bc + ca)2 + 2abc(a + b + c) = (a + b + c)2 (ab + bc + ca) − (a2 b2 + b2 c2 + c2 a2 ) ≤ (a + b + c)2 (ab + bc + ca) − abc(a + b + c) = (a + b + c)(a + b)(b + c)(c + a), 33 34 CHAPTER HINTS AND SOLUTIONS TO SELECTED PROBLEMS which was what we wanted 2.3.7 Solution: For the right part, from Hăolders Inequality we have x x+y x x+y x(x + y) ≥ (x + y + z)3 (x + y + z)3 x2 + y + z + xy + yz + zx x ≥ x+y So it remains to show that (x + y + z)3 x2 + y + z + xy + yz + zx ≥ 27 (yz + zx + xy) Let x + y + z = and xy + yz + zx = t Thus we need to prove that 27t 4 ⇔ t(1 − t)2 ≤ 27 ⇔ (4 − 3t)(1 − 3t)2 ≥ 0, 1−t ≥ which is obvious using t ≤ 1/3 Now for the left part,we need to prove that cyc x √ ≤ x+y 27 (x + y + z ) 4 We have √ cyc x = x+y cyc x x+y x x+y ≤ cyc ≤ √ x cyc x3 x+y cyc x3 x+y x x+y Thus we need to show that √ x cyc x x+y ≤ 27 (x + y + z ) Or, x cyc x x+y ≤ (x2 + y + z ) 35 But we have x cyc x x+y = = ≤ √ √ √ x x y+z z+x (x + y)(y + z)(z + x) √ √ √ (x y + z)( x z + x) (x + y)(y + z)(z + x) x2 (y + z) ( x(z + x)) (x + y)(y + z)(z + x) Let p = x + y + z, q = xy + yz + zx, r = xyz Then it remains to show that (pq − 3r)(p2 − q) ≤ (p2 − 2q) pq − r 2 ⇔ 2(p q − pq − 3p r + 3qr) ≤ 3(p3 q − p2 r − 2pq + 2qr) ⇔ p3 q + 3p2 r ≥ 4pq ⇔ p2 q + 3pr ≥ 4q ⇔ (x + y + z)2 (xy + yz + zx) + 3xyz(x + y + z) ≥ 4(xy + yz + zx)2 ⇔ (x2 + y + z )(xy + yz + zx) + 2(xy + yz + zx)2 + 3xyz(x + y + z) ≥ 4(xy + yz + zx)2 ⇔ (x2 + y + z )(xy + yz + zx) + 3xyz(x + y + z) ≥ 2(xy + yz + zx)2 ⇔ (x2 + y + z )(xy + yz + zx) ≥ 2(x2 y + y z + z x2 ) + xyz(x + y + z) x3 y ≥ ⇔ sym x2 y sym xy(x − y)2 ≥ 0, ⇔ sym which is obviously true 3.1.5 a ≥ b ≥ c implies a/b + c ≥ b/c + a ≥ c/a + b 3.2.6 Use Example 3.2.3 3.3.4 Solution: WLOG assume that a ≥ b ≥ c This implies a + b ≥ c + a ≥ b + c Therefore a b c ≥ ≥ b+c c+a a+b On the other hand, we have ab ≥ ca ≥ bc Therefore a(b + c) ≥ b(c + a) ≥ c(a + b) Applying Chebyshev’s a b c inequality for the similarly sorted sequences b+c , c+a , a+b and (a(b + c), b(c + a), c(a + b)) we get a · a(b + c) ≥ b+c a b+c ⇔ 3(a2 + b2 + c2 ) ≥ 2(ab + bc + ca) which was what we wanted a(b + c) a b+c 36 CHAPTER HINTS AND SOLUTIONS TO SELECTED PROBLEMS 3.3.5 Solution: Let x = an b+c , y √ n But the sequences ( an−1 x, = n bn c+a , z bn−1 y, = cn a+b Then a = b+c n an (b + c)n = n x (b + c)n−1 = n √ n cn−1 z) and an−1 x (ca + ab)n−1 , (ca+ab)n−1 n n , (ab+bc)n−1 n (bc+ca)n−1 are oppo- sitely sorted, since the sequences (a, b, c) and (x, y, z) are similarly sorted Hence by Rearrangement inequality we get n an−1 x ≤ (ca + ab)n−1 n an−1 x = (ab + bc)n−1 n xan−1 bn−1 (c + a)n−1 Finally using Hă olders inequality a b+c n xan1 bn1 (c + a)n−1 n x bn−1 a c+a n−1 , which was what we wanted 5.1.2 The following stronger inequality holds: a b c (a + b + c)2 + + ≥ (b + c)4 (c + a)4 (a + b)4 2(ab + bc + ca)(a + b)(b + c)(c + a) You may use Chebyshev’s inequality to prove it 5.1.3 First Solution: Consider the numbers a2 − 1, b2 − 1, c2 − Two of them must be of the same sign i.e either positive or negative WLOG suppose that a2 − and b2 − are of the same sign Then (a2 − 1)(b2 − 1) ≥ ⇒ a2 b2 + ≥ a2 + b2 ≥ (a+b) Now the inequality can be written as c2 (a2 b2 + 1) − 2c(a + b) + + (a − b)2 ≥ Using the above argument, we’ll be done if we can show that c2 (a + b)2 − 2c(a + b) + + (a − b)2 ≥ Or, (ca + bc − 2)2 + (a − b)2 ≥ 0, which is obviously true Second Solution: WLOG we may assume that a, b, c are positive First we have the following inequality a2 + b2 + c2 + 3(abc) ≥ 2(ab + bc + ca) 37 This follows from Schur and AM-GM 2 2 (a )3 + (b )3 + (c )3 + 3(abc) ≥ 2 So we’ll be done if we can show that 2 b3 + a3 b3 a3 ≥ 2ab 2 + (abc)2 ≥ 3(abc) Let (abc) = t We need to show that + t3 ≥ 3t, or, (t − 1)2 (t + 2) ≥ 0, which is obviously true 5.1.7 Solution: The inequality is equivalent to a3 + b3 ≥2 a3 + b3 + abc Verify that a3 + b3 a+b ≥ 3 a + b + abc a+b+c ⇔ c(a2 + b2 − ab) ≥ abc ⇔ c(a − b)2 ≥ which is obviously true Hence we conclude that a3 + b3 ≥ a3 + b3 + abc a+b = a+b+c 5.1.12 Solution: The inequality is equivalent to (a + b)(b + c) (b + c − a)(c + a − b) ≥ 4(a + b + c) From AM-GM we get (a + b)(b + c) (b + c − a)(c + a − b) ≥ 2(a + b)(b + c) (a + b)(b + c) = b+c−a+c+a−b c Therefore it remains to show that (a + b)(b + c) ≥ 4(a + b + c) c Since the sequences { a1 , 1b , 1c } and {(c + a)(a + b), (a + b)(b + c), (b + c)(c + a)} are oppositely sorted, from Rearrangement we get (a + b)(b + c) ≥ c Therefore it remains to show that (a + b)(b + c) ca =a+b+c+ b b ca ≥ a + b + c, b 38 CHAPTER HINTS AND SOLUTIONS TO SELECTED PROBLEMS which follows from Rearrangement ca ≥ b ca = a + b + c c 5.1.15 Solution: Let x = 2a21+2 , y = 2b21+2 , z = 2c21+2 Thus a = from the given condition x + y + z = and we need to prove that cyc 1−2x 2x , b = 1−2y 2y , c (1 − 2x)(1 − 2y) ≤ , 2x · 2y or cyc (y + z − x)(z + x − y) ≤ xy But we have cyc (y + z − x)(z + x − y) ≤ xy cyc y+z−x z+x−y + y x Hence we are done Remark: The inequality holds even if a, b, c are real numbers = = 1−2z 2z Then References Pham Kim Hung, Secrets in Inequalities (Volume 1), GIL Publishing House, 2007 Radmila Bulajich Manfrino, Jos´e Antonio G´omez Ortega, Rogelio Valcez Delgado, Inequalities, Institudo de Matematicas, Universidad Nacional Aut´onoma de M´exico, 2005 http://www.mathlinks.ro 39 ... Introduction The aim of this note is to acquaint students, who want to participate in mathematical Olympiads, to Olympiad level inequalities from the basics Inequalities are used in all fields of mathematics... combination of the two terms Arithmetic Mean and Geometric Mean The arithmetic mean of two numbers a and b is defined by a+b √ Similarly ab is the geometric mean of a and b The simplest form of the... b1 b2 bn b1 + b2 + · · · + bn And the case of equality easily follows too From (2.1) we deduce another proof of the Cauchy-Schwarz inequality Third Proof We want to show that b2i c2i ≥ bi ci Let