57 Title: Secrets in Inequalities volume - advanced inequalities - free chapter Author: Pham Kim Hung Publisher: GIL Publishing House ISBN: 978-606-500-00-5 Copyright c 008 GIL Publishing House. All rights reserved. ALL RIGHTS RESERVED. This book contains material protected under International and Federal Copyright Laws and Treaties. Any unauthorized reprint or use of this material is prohibited. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system without express written permission from the author / publisher. GIL Publishing House comenzi@gil.ro PO box 44, PO 3 Zalău, Sălaj ROMANIA This PDF is the free part of the book Secrets in Inequalities volume - advanced inequalities written by Pham Kim Hung. The current PDF is free of charge and available for anyone but it is under the copyright of GIL Publishing House therefore you can not use it for commercial purpose.
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Table of contents Collected Articles on Inequalities 6 Article. GENERALIZATION OF SCHUR INEQUALITY.............. 6. Generalized Schur Inequality for Three Numbers............. 6. A Generalization of Schur Inequality for n Numbers........... 69 Article. LOOKING AT FAMILIAR EXPRESSIONS................ 76.3 On AM-GM Inequality............................ 76.4 On Nesbitt s Inequality............................ 84.5 On Schur Inequality.............................. 88 Article 3. THOUGHT BRINGS KNOWLEDGE - VARIED IDEAS.......... 9.6 Exponent Smash................................ 9.7 Unexpected Equalities............................ 96.8 Undesirable Conditions........................... 304 Article 4. CYCLIC INEQUALITIES OF DEGREE 3................. 30.9 Getting Started................................ 30.0 Application for Polynomial Inequalities.................. 3. Applications for Fraction Inequalities................... 33 Article 5. INTEGRAL AND INTEGRATED INEQUALITES............. 39. Getting started................................. 39.3 Integrated Inequalities............................ 33 Article 6. TWO IMPROVEMENTS OF THE MIXING VARIABLES METHOD... 336.4 Mixing Variables by Convex Functions................... 336.5 nsmv Theorem and Applications..................... 346 Article 7. MAJORIZATION AND KARAMATA INEQUALITY........... 358.6 Theory of Majorization............................ 358.7 Karamata Inequality............................. 36 59
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Chapter Collected Articles on Inequalities After reading the previous chapters, you should have gained a lot of insight into inequalities. The world of inequalities is really uniquely wonderful and interesting to explore. In this chapter, we will examine inequalities in a more general and larger context with the help of the mathematical techniques and methods developed in the previous chapter. This chapter contains 9 sections, organized into 8 articles. Many interesting matters will be discussed here, such as some generalizations of Schur inequality, some estimations of familiar expressions, some strange kinds of inequalities, some improvements of the classical mixing variable method and some applications of Karamata inequality. We wish to receive more comments and contributions from you, the readers. 6
6 c GIL Publishing House. All rights reserved. Article Generalization of Schur Inequality. Generalized Schur Inequality for Three Numbers We will be talking about Schur inequality in these pages. Just Schur inequality? And is it really necessary to review it now? Yes, certainly! But instead of using Schur inequality in brute force solutions eg. solutions that use long, complicated expanding), we will discover a very simple generalization of Schur inequality. An eightgrade student can easily understand this matter; however, its wide and effective influence may leave you surprised. Theorem Generalized Schur Inequality). Let a, b, c, x, y, z be six non-negative real numbers such that the sequences a, b, c) and x, y, z) are monotone, then xa b)a c) + yb a)b c) + zc a)c b) 0. PROOF. WLOG, assume that a b c. Consider the following cases i). x y z. Then, we have c a)c b) 0, so zc a)c b) 0. Moreover, xa c) yb c) xb c) yb c) = x y)b c) 0 xa b)a c) + yb a)b c) 0. Summing up these relations, we have the desired result. i). x y z. We have a b)a c) 0, so xa b)a c) 0. Moreover, za c) ya b) za b) ya b) = z y)a b) 0 zc a)c b) + yb a)b c) 0. Summing up the inequalities above, we have the desired result. Comment. Denote S = xa b)a c). By the same reasoning as above, we can prove that S 0 if at least one of the following stronger conditions is fulfilled. If a b c 0, x y 0 and z 0.
c GIL Publishing House. All rights reserved. 63. If a b c 0, y z 0 and x 0. 3. If a b c 0 and ax by 0 or by cz 0. ) and ) are quite obvious. To prove 3), just notice that if a, b, c > 0 then xa b)a c) + yb a)b c) + zc a)c b)) abc = ax a ) b a ) + by c b ) a b ) + cz c c ) a c ), b and the problem turns to a normal form of the generalized Schur inequality shown above. This was such an easy proof! But you need to know that this simple theorem always provides unexpectedly simple solutions to a lot of difficult problems. That makes the difference, not its simple solution. Let s see some examples and you will understand why many inequality solvers like to use the generalized Schur inequality in their proofs. Example... Let a, b, c be three positive real numbers. Prove that a + b + c a + bc b + c + b + ca c + a + c + ab a + b. Ho Joo Lee) SOLUTION. According to the identity a + bc b + c we can change our inequality into the form in which a b)a c) a =, b + c xa b)a c) + yb a)b c) + zc a)c b) 0, x = b + c ; y = c + a ; z = a + b. WLOG, assume that a b c, then clearly x y z. The conclusion follows from the generalized Schur inequality instantly. Example... Let a, b, c be positive real numbers with sum 3. Prove that a + b + c 3 a + bc + 3 b + ac + 3 c + ab.
64 c GIL Publishing House. All rights reserved. SOLUTION. Rewrite the inequality into the following from a ) a + b + c a 0 a b)a c) + bc a 3 0. + abc Notice that if a b c then a 3 + abc b 3 + abc c 3 + abc. The conclusion follows from the generalized Schur inequality. Example..3. Let a, b, c be the side lengths of a triangle. Prove that a + b c a + b c + b + c a b + c a + c + a b c + a b 3. SOLUTION. By a simple observation, the inequality is equivalent to Pham Kim Hung) Italian Winter Camp 007) ) a + b c 0 a + b c a + b c 0 a + b c a + b c ab ca + b c) a + b c ) a + b + c + a + b c ) 0 c a)c b) S c 0, where S c = a + b c ) a + b + c + a + b c ) ab + ca + b c) ), and S a, S b are determined similarly. It s easy to check that if b c then a + b c c + a b ; a + b + c + a + b c a + b + c + c + a b ; ab + ca + b c) ca + bc + a b) ; Therefore S c S b. The proof is finished by the generalized Schur inequality.
c GIL Publishing House. All rights reserved. 65 Example..4. Let x, y, z be positive real numbers such that x + y + z =. Prove that x + yz x y + z) + y + zx y z + x) + z + xy z x + y). APMO 007) SOLUTION. We use the following simple transformation x + yz x y + z) = = x y)x z) + xy + z) x y + z) x y)x z) x + y + z) y + z. By the generalized Schur inequality, we get that x y)x z) x 0. y + z) By AM-GM inequality, the remaining work is obvious y + z ) y + z = x =. This ends the proof. Equality holds for x = y = z = 9. Example..5. Let a, b, c be positive real numbers. Prove that a + bc b + c) + b + ac a + c) + c + ab a + b) 9 4. SOLUTION. The inequality can be rewritten as or equivalently A + B 9 4, where A = a b)a c) + ab + bc + ca) b + c) 9 4, a b)a c) b + c) ; B = ab + bc + ca b + c) ; By the generalized Schur inequality, we deduce that A 0. Moreover, B 9 4 by Iran 96 inequality. Therefore, we are done and the equality holds for a = b = c.
66 c GIL Publishing House. All rights reserved. Example..6. Let a, b, c be positive real numbers. Prove that a 3 + abc b + c) 3 + b 3 + abc c + a) 3 + c 3 + abc a + b) 3 a b + c + b c + a + SOLUTION. Notice that a 3 + abc b + c) 3 a a b + c = b + c a + bc b + c ) a aa b)a c) = b + c) a b + c + bc + ). ab + c) The inequality can be rewritten as S a a b)a c) 0 with c a + b. Nguyen Van Thach) a S a = b + c) a b + c + bc + ) ; ab + c) b S b = c + a) c c + a + ab + ) ; ca + b) c S c = a + b) c a + b + ab + ). ca + b) Now suppose that a b c, then it s easy to get that b + c) a + bc a + c) b + ac ; b + c) ab + c) a + c) ba + c). a Thus b+c) + bc + ) c ab + c) c+a) + ab + ) ca + b) and therefore we have S a S b. We can conclude that S a a b)a c) S a a b)a c) + S b b a)b c) S a S b )a b)b c) 0. This is the end of the proof. The equality holds for a = b = c. Example..7. Let a, b, c be non-negative real numbers. Prove that a a + b)a + c) + b b + c)b + a) + c c + a)c + b) 3.
c GIL Publishing House. All rights reserved. 67 SOLUTION. If c = 0, the problem is obvious. Suppose that a, b, c > 0, then we have 3 a a + b)a + c) = a a + b + c a ) a + b)a + c) = aa b)a c) a + b + c)a + b)a + c) = a a ) b a ) c a + b + c)a + b)a + c). By the generalized Schur inequality, it suffices to prove that if a b then a a + b)a + c) b b + a)b + c). But the previous inequality is equivalent to a b) aba + b + c) + ca + ab + b ) ) 0, which is obvious. The equality holds for a = b = c and a = b, c = 0 up to permutation. Example..8. Let a, b, c be positive real numbers. Prove that SOLUTION. We have ab + bc + ca a + bc ) a + bc + b + ca + a + b + c c + ab. ab + bc + ca a + b + c) ab + bc + ca = ab + bc + ca a + bc = a b)a c) Pham Huu Duc) ) + c a)c b) ab + bc + ca a + bc + ab + bc + ca WLOG, assume that a b c. By the generalized Schur inequality, it suffices to prove that ) a a + bc + ab + bc + ca b b + ca + ab + bc + ca Indeed, the difference between the left-hand side and the right-hand side is a b a b)ca + cb ab) ab + bc + ca a + bc)b + ca) ). ).
68 c GIL Publishing House. All rights reserved. = a b) a b + ca 3 + b 3 ) a c b c + ca b) a + b) ) ab + bc + ca)a + bc)b + ca) We just got the desired result. The equality holds for a = b = c. Example..9. Let a, b, c be non-negative real numbers such that ab + bc + ca =. Prove that + b c b + c) + + c a c + a) + + a b a + b) 5. Mathlinks Contest) 0. SOLUTION. First we have that + b c b + c) = ab + bc + ca) + b c b + c) = a + abc b + c + b c b + c). Therefore, our inequality can be rewritten as a ) ab + ) 4bc a b + c) + 4b c ) b + c b + c) bc 0. The left hand expression of the previous inequality is = b + bc + c ab ac)b c) b + c) = b c) b b a) c c a) ) b + c) = b c) b 3 c 3 ab c ) ) b + c) b b c)b a) b + c) + b ) b + a), and the proof is completed by the generalized Schur inequality because if b c then b b + c) + b b + a) c c + a) + c c + b).
c GIL Publishing House. All rights reserved. 69. A Generalization of Schur Inequality for n Numbers If Schur inequality for three variables and its generalized form have been discussed thoroughly in the previous section, we now go ahead to the generalization of Schur inequality for n variables. As a matter of fact, we want an estimation of F n = a a a )...a a n )+a a a )a a 3 )...a a n )+...+a n a n a )...a n a n ). The first question is if the inequality F n 0 holds. Unfortunately, it is not always true it is only true for n = 3). Furthermore, the general inequality a k a a )...a a n )+a k a a )a a 3 )...a a n )+...+a k na n a )...a n a n ) 0 is also false for all n 4 and k 0. To find a counter-example, we have to check the case n = 4 only and notice that if n > 4, we can choose a k = 0 k 4. For n = 4, consider the inequality a k a b)a c)a d)+b k b a)b c)b d)+c k c a)c b)c d)+d k d a)d b)d c) 0. Just choose a = b = c, the inequality becomes d k d a) 3 0, which is clearly false when d a). Our work now is to find another version of this inequality. To do so, we first have to find something new in the simple case n = 4. The following results are quite interesting. Example... Let a, b, c, d be non-negative real numbers such that a + b + c + d =. Prove that aa b)a c)a d)+bb a)b c)b d)+cc a)c b)c d)+dd a)d b)d c) 43. Pham Kim Hung) SOLUTION. We use the entirely mixing variable and the renewed derivative to solve this problem. Notice that our inequality is exactly 43 a + b + c + d)4 + aa b)a c)a d) 0. Notice that the inequality is clearly true if d = 0, so we only need to prove that after taking the global derivative) 7 a + b + c + d)3 + a b)a c)a d) 0.
70 c GIL Publishing House. All rights reserved. Because the expression a b)a c)a d) is unchanged if we decrease a, b, c all at once, it suffices to consider the inequality in case min{a, b, c, d} = 0. WLOG, assume that d = 0, then the inequality becomes 7 a + b + c)3 + aa b)a c) + bb a)b c) + cc a)c b) abc 0. This inequality follows from AM-GM inequality and Schur inequality immediately. We are done. Equality cannot hold. Example... Let a, b, c, d be non-negative real numbers. Prove that aa b)a c)a d)+bb a)b c)b d)+cc a)c b)c d)+dd a)d b)d c)+abcd 0. Pham Kim Hung) SOLUTION. We use the global derivative as in the previous solution. Notice that this inequality is obvious due to Schur inequality if one of four numbers a, b, c, d is equal to 0. By taking the global derivative of the left-hand side expression, we only need to prove that abc 0. a b)a c)a d) + Using the mixing all variables method, if suffices to prove it in case min{a, b, c, d} = 0. WLOG, assume that a b c d = 0. The inequality becomes aa b)a c) + bb a)b c) + cc a)c b) 0, which is exactly Schur inequality for three numbers;, so we are done. The equality holds for a = b = c, d = 0 or permutations. Example..3. Let a, b, c, d be non-negative real numbers such that a + b + c + d = 4. Prove that aa b)a c)a d) + bb a)b c)b d) + cc a)c b)c d)+ +dd a)d b)d c) abcd. Pham Kim Hung)
c GIL Publishing House. All rights reserved. 7 SOLUTION. We have to prove that 6 aa b)a c)a d) + a + b + c + d ) 6abcd 0. If d = 0, the inequality is obvious due to AM-GM inequality and Schur inequality for three numbers). According to the mixing all variables method and the global derivative, it suffices to prove that 6 a b)a c)a d) + 4a + b + c + d)a + b + c + d ) 6 abc 0. or 4 a b)a c)a d) + a + b + c + d)a + b + c + d ) 4 abc 0 ) If one of four numbers a, b, c, d, say d, is equal to 0, then the previous inequality becomes aa b)a c) + a + b + c)a + b + c ) 8abc 0, a,b,c 4 which is obvious due to the following applications of Schur inequality and AM-GM inequality a,b,c 4 aa b)a c) 0 ; a + b + c)a + b + c ) 8abc 9abc 8abc 0 ; Therefore, according to the mixing all variables method, in order to prove ) by taking the global derivative, it suffices to prove that a,b,c,d 4 a + Clearly, AM-GM inequality yields that a,b,c,d a ) 8 sym ab ) a,b,c,d 4 a 8 ab ; 3 sym a,b,c,d a ) 6 3 ab ; sym Adding up the results above, we get ) and then ). The conclusion follows and the equality holds for a = b = c = d =. This should satisfy anyone who desperately wanted a Schur inequality in 4 variables. What happens for the case n = 5? Generalizations are a bit more complicated.
7 c GIL Publishing House. All rights reserved. Example..4. Let a, b, c, d, e be non-negative real numbers such that a+b+c+d+e =. Prove that aa b)a c)a d)a e)+bb a)b c)b d)b e)+cc a)c b)c d)c e)+ +dd a)d b)d c)d e) + ee a)e b)e c)e d) 430. Pham Kim Hung) SOLUTION. To prove this problem, we have to use two of the previous results. Our inequality is equivalent to 430 a + b + c + d + e)5 + aa b)a c)a d)a e) 0. Taking the global derivative, we have to prove that 5 860 a + b + c + d + e)4 + a b)a c)a d)a e) 0. Due to the mixing all variables method, we only need to check this inequality in case min{a, b, c, d, e} = 0. WLOG, assume that a b c d e = 0. The inequality becomes a,b,c,d 5 860 a + b + c + d)4 + aa b)a c)a d) 0. This inequality is true according to example.. because 5 860. This shows 43 that it suffices to consider the first inequality in case a b c d e = 0. In this case, the inequality becomes a,b,c,d 430 a + b + c + d)5 + a a b)a c)a d) 0. If one of the numbers a, b, c, d is equal to 0, the inequality is true by Schur inequality so we only need to prove that by taking the global derivative) or a,b,c,d 6 a + b + c + d)4 + aa b)a c)a d) 0 a,b,c,d 43 a + b + c + d)4 + aa b)a c)a d) 0. This inequality is exactly the inequality in example... The proof is completed and we cannot have equality.
c GIL Publishing House. All rights reserved. 73 Example..5. Let a, b, c, d, e be non-negative real numbers. Prove that aa b)a c)...a e) + bb a)b c)...b e) + cc a)c b)c d)c e)+ +dd a)...d c)d e)+ee a)e b)...e d)+a bcd+b cde+c dea+d eab+e abc 0. SOLUTION. This problem is easier than the previous problem. Taking the global derivative for a first time, we obtain an obvious inequality a b)a c)a d)a e) + abcd + a bc + cd + da) 0 which is true when one of the numbers a, b, c, d, e is equal to 0. Now we only need to prove the initial inequality in case min{a, b, c, d, e} = 0. WLOG, assume that e = 0, then the inequality becomes a,b,c,d a a b)a c)a d) + a bcd 0. If one of a, b, c, d is equal to 0, the inequality is true due to Schur inequality. Therefore we only need to prove that taking the global derivative for the second time) a,b,c,d aa b)a c)a d) + abcd + a bc + cd + da) 0. This inequality is true according to example.. and we are done immediately. The equality holds if and only if three of the five numbers a, b, c, d, e are equal to each other and the two remaining numbers are equal to 0. These problems have given us a strong expectation of something similar in the general case of n variables. Of course, everything becomes much harder in this case, and we will need to use induction. Example..6. Let a, a,..., a n be non-negative real numbers such that a +a +...+a n =. For c = 9 n 7 nn )n ), prove that a a a )...a a n )+a a a 3 )...a a n )+...+a n a n a )a n a )...a n a n ) c. Pham Kim Hung) SOLUTION. To handle this problem, we need to prove it in the general case, that means, find an estimation of F k,n = n n a k i j=,j i a i a j ),
74 c GIL Publishing House. All rights reserved. where the non-negative real numbers a, a,..., a n have sum. After a process of guessing and checking induction steps, we find out n n a k i a i a j ) 3 9 n k n + k )n + k )n + k 3). j=,j i Let s construct the following sequence for all k, n 4 c k,n = 9 n+k 9 n + k )n + k )n + k 3). We will prove the following general result by induction n ) k+n n n a i + a i a j ) 0 ) c k,n a k i j=,j i We use induction for m = k + n, and we assume that ) is already true for all n, k such that k + n m. We will prove that ) is also true for all n, k such that n + k = m +. Indeed, after taking the global derivative, the inequality ) becomes nn + k ) c k,n n ) k+n a i + k n a k i n j=,j i a i a j ) 0 ) According to the inductive hypothesis for n and k ), we have n ) k+n n n a i + a i a j ) 0 c k,n Moreover, because a k i j=,j i nn + k ) kc k,n c k,n n 4, the inequality ) is successfully proved. By the mixing all variables method, we only need to consider ) in case min{a, a,..., a n } = 0. WLOG, assume that a a... a n, then a n = 0 and the inequality becomes c k,n c k+,n ) k+n n a i + n a k+ i or since c k,n = c k+,n ) n ) k+n n a i + a k+ i n j=,j i n j=,j i a i a j ) 0 a i a j ) 0. This is another form of the inequality ) for n numbers a, a,..., a n and for the exponent k + instead of k). Performing this reasoning n 4 times or we can use
c GIL Publishing House. All rights reserved. 75 induction again), we can change ) to the problem of only four numbers a, a, a 3, a 4 but with the exponents k + n 4. Namely, we have to prove that a + b + c + d) k+n + M a k+n 4 a b)a c)a d) 0 ) where M = c k+n 4,4 = c k,n. Taking the global derivative of ) exactly r times r k + n 4), we obtain the following inequality 4 r k + n )k + n )...k + n r)a + b + c + d) 4 + +k + n 4)k + n 3)...k + n r 3)M a k+n 4 r a b)a c)a d) 0 [ ] We will call the inequality constructed by taking r times the global derivative of ) as the [r th ] inequality this previous inequality is the [r th ] inequality). If abcd = 0, assume d = 0, and the [r th ] inequality is true for all r {0,,,..., n + k 5} because a k+n 4 r a b)a c)a d) = a k+n 3 r a b)a c) 0. According to the principles of the mixing all variables method and global derivative, if the [r + ) th ] inequality is true for all a, b, c, d and the [r th ] inequality is true when abcd = 0 then the [r th ] inequality is true for all non-negative real numbers a, b, c, d. Because abcd = 0, the [r th ] is true for all 0 r n + k 5, so we conclude that, in order to prove the [0 th ] inequality which is exactly )), we only need to check the [n + k 4) th ] inequality. The [n + k 4) th ] inequality is as follows 4 n+k 4 k + n )k + n )...4)a + b + c + d) 4 + +k + n 4)k + n 3)...)M a b)a c)a d) 0 4 n+k 4 k+n )k+n )k+n 3)a+b+c+d) 4 +6M a b)a c)a d) 0 a + b + c + d) 4 + 7 a b)a c)a d) 0. This last inequality is clearly true by the mixing all variables method and AM-GM inequality. The inductive process is finished and the conclusion follows immediately. Consider the non-negative real numbers a, a,..., a n such that a +a +...+a n =. For k = 9 n+k 9 n + k )n + k )n + k 3), we have n n a k i a i a j ) k. j=,j i
76 c GIL Publishing House. All rights reserved. Article Looking at Familiar Expressions.3 On AM-GM Inequality Certainly, if a, b, c are positive real numbers then AM-GM inequality shows that a b + b c + c a a 3 3 b b c c a = 3. We denote Ga, b, c) = a b + b c + c 3, then Ga, b, c) 0 for all a, b, c > 0. This article a will present some nice properties regarding the function G. Example.3.. Let a, b, c, k be positive real numbers. Prove that a b + b c + c a a + k b + k + b + k c + k + c + k a + k. SOLUTION. Notice that we can transform the expression Ga, b, c) into Ga, b, c) = a b + b ) b a + c + c a b ) a a b) = + ab WLOG, assume that c = mina, b, c). Our inequality is equivalent to a b) + ab a c)b c) ac a c)b c). ac a b) a c)b c) + a + k)b + k) a + k)c + k). Because c = mina, b, c) it follows that a c)b c) 0 and the inequality is obvious. The proof is completed and equality holds for a = b = c. Example.3.. Let a, b, c be positive real numbers. If k maxa, b, c ), prove that a b + b c + c a a + k b + k + b + k c + k + c + k a + k. Pham Kim Hung)
c GIL Publishing House. All rights reserved. 77 SOLUTION. Similarly as in the preceding inequality, this one is equivalent to a b) + ab a c)b c) ac a b) a + b) a + k)b + k) WLOG, assume that c = mina, b, c). It s sufficient to prove that The first one is certainly true because The second one is equivalent to a + k)b + k) aba + b) ; a + k)c + k) aca + c)b + c). a + k)b + k) a + b ) aba + b). c k ac) + a k bc) + k abc 0 which is also obvious because k maxa, b, c ). We are done. Comment. The following inequality is stronger a c)b c)a + c)b + c) + a + k)c. + k) Let a, b, c be positive real numbers. If k maxab, bc, ca), prove that a b + b c + c a a + k b + k + b + k c + k + c + k a + k. To prove this one, we only note that if c = mina, b, c) and k maxab, bc, ca) then a + k)b + k) a + ab)b + ab) = aba + b) ; c k ac) + a k bc) + k abc c k ac) + a k bc) + ac) bc) abc = 0. The equality holds for a = b c and k = ac. Both a and c can take arbitrary values. Example.3.3. If a, b, c are the side lengths of a triangle, then a 4 b + b c + c ) 9 + a + c a c + b + c + b b + a + b + a a + c. SOLUTION. The inequality can be rewritten in the following form 4a b) + ab 4c a)c b) ac Pham Kim Hung) a b ) a + c )c + b ) + c a )c b ) a + b )a + c )
78 c GIL Publishing House. All rights reserved. WLOG, we may assume c = mina, b, c). Then it s not too difficult to show that c + a)c + b) ac a + b )c + b ) 4 ab a + b) a + c )b + c ) and the proof is completed. Equality holds for a = b = c. Example.3.4. Let a, b, c be positive real numbers. Prove that a b + b c + c 8ab + bc + ca) + a a + b + c. Nguyen Van Thach) SOLUTION. Similarly, this inequality can be rewritten in the following form a b) a + b) a b ) ) 8 a + c)b + c) 8 a + b + c +c a)c b) a c a + b + c 0. WLOG, assume that c = min{a, b, c}. We have a + b) a b 8 a + b 8 a + b + c. Moreover, a + c)b + c)a + b + c ) ca + c)a + c ) 8a c a + c)b + c) a c 8 a + b + c. Therefore we get the desired result. Equality holds for a = b = c. Example.3.5. Let a, b, c be the side lengths of a triangle. Prove that a + b a + c + c + a c + b + b + c b + a a + b a + c + a + c b + c + b + c b + a. SOLUTION. It is easy to rewrite the inequality in the following form a b) M + c a)c b)n 0, Vo Quoc Ba Can)
c GIL Publishing House. All rights reserved. 79 where M = N = a + b) a + c )b + c ) a + c)b + c) ; a + c)b + c) a + b )a + c ) a + c)a + b). WLOG, assume that c = min{a, b, c}. Clearly, M 0 and N 0 since a+c) b+c)a+b) a +b )a +c ) = a 3 b+c a)+3a bc+3abc +a c +c 3 a+b) 0. We are done. Equality holds for a = b = c. Example.3.6. For all distinct real numbers a, b, c, prove that a b) b c) c a) + + b c) c a) a b) 5. Darij Grinberg) SOLUTION. This inequality is directly deduced from the following identity a b) b c) + b c) c a) a b) = 5 + + a b b c + b c c a + c a ). a b + c a) Comment. According to this identity, we can obtain the following results Let a, b, c be distinct real numbers. Prove that G a b), b c), c a) ) 8 + 3 ) 8 Ga b, b c, c a). G a b), b c), c a) ) + G c a), b c), a b) ) 9. Example.3.7. Prove that for all positive real numbers a, b, c, we have a b + b c + c a a + c b + c + c + b a + b + b + a c + a SOLUTION. First Solution. First, we will prove that a + c a b + b a b + c + In order to prove ), we only need to prove that a b + b a a + c b + c + b + c a + c b + c a + c )
80 c GIL Publishing House. All rights reserved. Indeed, by squaring, this inequality becomes a + b ab a + b + c a + c )b + c ) or a + b ) a + c )b + c ) ab a + b + c ) or a + b ) a + c )b + c ) a b a + b + c ) or c a b ) a + b + c ) 0, which is clearly true. As a result, ) is proved. Now, returning to our problem, we assume by contradiction that the inequality is false for a certain triple a, b, c). By ), we have Combining this with ), we get that a b < a + c b + c ) a b + b a a + c b + c + b + c a + c On the other hand, from ), we have that b a > b + c a + c 3) ) a < a + c b b + c Combining with 3), we obtain a b + b a < a + c b + c + b + c a + c. a b < a + c b + c 4) This inequality contradicts the result in example.3.. Therefore, the assumption is false, or in other words, the inequality is proved successfully. Equality holds for
c GIL Publishing House. All rights reserved. 8 a = b = c. Second Solution. Recall the following result, presented in the first volume of this book. Let a, a,.., a n and b b... b n 0 be positive real numbers such that a a...a k b b...b k k {,,..., n}, then a + a +... + a n b + b +... + b n. For the case n = 3, we obtain the following result Given positive numbers a, b, c, x, y, z such that max {a, b, c} max {x, y, z}, min {a, b, c} min {x, y, z}, then a + b + c x + y + z. According to this result, we will prove a general inequality for all a, b, c, x > 0 as follows a b a x + c x ) x. b x + c x Indeed, we already have a b b c c a = a x + c x b x + c x ) x c x + b x It suffices to show that { a max b, b c, c } { a x + c x max a b x + c x { a min b, b c, c } { a x + c x min a b x + c x a x + b x ) ) ) x, c x + b x x b x + a x a x + b x x, c x + b x a x + b x c x + a x ) ) ) x =. x, b x + a x c x + a x x, b x + a x c x + a x We prove 5) and 6) can be proved similarly). WLOG, assume that a x + c x b x + c x ) { x a x + c x ) x c x + b x = max, b x + c x a x + b x ) x, b x + a x c x + a x ) } x ) } x ) } x. 5) 6) a x + c x ) x We see that, gives a b. Therefore b x + c x a x b x ax + c x b x + c x This ends the proof. Equality holds for a = b = c. a a x b + c x ) x. b x + c x Example.3.8. Let a, b, c be distinct real numbers. Prove that a b) b c) c a) + + b c) c a) a b) a + b b + c + b + c c + a + c + a a + b.
8 c GIL Publishing House. All rights reserved. Pham Kim Hung) SOLUTION. WLOG, we may assume that c = mina, b, c). Taking into account the preceding example.3., we deduce that Ga + b, b + c, c + a) Ga + b c, b + c c, c + a c). Let now x = a c, y = b c, then it remains to prove that after we consider c = 0) x y) y + y x + x x y) x + y + y y x + x x + y x y + y x + x x y) 3x y + y x + x x + y. From here, we need to consider some smaller cases i). The first case. If x y then Case y x y. The desired result is obtained by adding x y + 9 4 3x y, y x + 4 y x, x x y) 4, 3 x x + y. Case y x.3y. The desired result is obtained by adding x y + 3x y, y x + 4 y x, x.3 3., x y) 3.3 x x + y. Case.3y x.5y. The desired result is obtained by adding x y +.6 3x y, y x + 4 y x, x.5.77, x y) 3.5 x x + y. Case.5y x 3y. The desired result is obtained by adding x y +.5 3x y, y x + 4 y x, x x y).5, 3 4 x x + y. Case 3y x 4y. The desired result is obtained by adding x y 3x y, y x + 4 y x, x x y).77, 4 5 x x + y. Case x 4y. The desired result is obtained by adding x y 3x y + 4, y x + 4 y x, x x y), x x + y. ii). The second case. If x y then
c GIL Publishing House. All rights reserved. 83 Case x y.5x. The desired result is obtained by adding x y + 3x y, y x y x, x y x) 4, x x + y. Case.5x y.8x. The desired result is obtained by adding x y + 4 9 3x y, y x y x + 0.75, x y x).56,.5 x x + y. Case.8x y 3x. The desired result is obtained by adding x y +.4 3x y, y x y x +.44, x y x) 4,.8 x x + y. Case y 3x. The desired result is obtained by adding x y + 8 9 3x y, y x y x + 6, x y x) 0, x x + y. The proof has been proved completely. There s no equality case.
84 c GIL Publishing House. All rights reserved..4 On Nesbitt s Inequality The famous Nesbitt inequality has the following form If a, b, c are positive real numbers then Na, b, c) = a b + c + b c + a + c a + b 3 0. In the following pages, we will discuss some inequalities which have the same appearance as Nesbitt inequality and we will also discuss some nice properties of Na, b, c). First we give a famous generalization of Nesbitt inequality with real exponents. Example.4.. Let a, b, c be non-negative real numbers. For each real number k, find the minimum of the following expression S = ) k ) k ) k a b c + +. b + c c + a a + b SOLUTION. Certainly, Nesbitt inequality is a particular case of this inequality for k =. If k or k 0 then it s easy to deduce that ) k ) k ) k a b c + + 3 b + c c + a a + b k. If k =, we obtain a familiar result as follows a b + c + b a + c + c a + b. The most difficult case is 0 < k <. We will prove by mixing variables that fa, b, c) = ) k ) k ) k a b c + + min {, 3 } b + c c + a a + b k. WLOG, we may assume that a b c. Let t = a + b u = a b 0 then ) k ) k ) k a u + t t u c ) k = gu) = + +. b + c t u + c u + t + c t We infer that the derivative u + t g u) = k u t + c ) k t + c u t + c) + k has the same sign as the following function t u ) k t c u + t c u + t + c) hu) = k ) [ lnt + u) lnt u) ] + k + ) [ lnu + t + c) lnt u + c) ].
c GIL Publishing House. All rights reserved. 85 It s easy to check that h u) = tk ) k + )t + c) t u + t + c) u. Because t c it follows that tt+c)t c) t+c)u and therefore t+c)t u ) t t + c) u ). If k + > k) or k > /3), we have indeed that h u) > 0. Thus hu) h0) = 0 g u) 0. Therefore g is a monotonic function if u 0 and therefore ) k a gu) g0) = tk b + c t + c) k + ck t) k ) Because g0) is homogeneous, we may assume that c t =. Consider the function Since the derivative pc) = k+ + c) k + ck p c) = has the same sign as the function k.k+ + kck + c) k+ qc) = k + ) lnc + ) + k ) ln c k + ) ln and also because, as it s easy to check q c) = k + c + + k k + )c + c + )k ) = c cc + ) has no more than one real root, we obtain that p c) = 0 has no more than one real root in 0, ) because p ) = 0). Furthermore, lim pc) = +, so we obtain c 0 { } pc) min p) ; lim pc) = min { 3 ; k+} ) c 0 According to ) and ), we conclude that ) k ) k ) k { } a b c 3 + + min b + c a + c a + b k ;. The inequality has been proved in case k > /3, now we will consider the case k /. Choose three numbers α, β, γ satisfying that α = a k, β = b k, γ = c k, then b + c) k b k + c k ) k ) k b c + ak α b + c c + b b + c) k β + γ Constructing similar results and summing up, we get a k b + c) k + b k c + a) k + c k α a + b) k β + γ + β α + γ + γ α + β.
86 c GIL Publishing House. All rights reserved. Therefore the problem has been completely solved, with the conclusion that ) k ) k ) k { } a b c 3 + + min b + c a + c a + b k ;. If k = ln 3, the equality holds for a = b = c and a = b, c = 0 up to permutation. ln Otherwise, the equality only holds in the case a = b = c. In volume I we have a problem from the Vietnam TST 006, where we have already proved that if a, b, c are the side-lengths of a triangle then a + b + c) a + a b + ) a 6 c b + c + b c + a + c ), a + b or, in other words, Na, b, c) 3Na + b, b + c, c + a). Moreover, we also have some other nice results related to the expression N as follows Example.4.. Let a, b, c be the side lengths of a triangle. Prove that a b + c + b c + a + c a + b + 3 or, in other words, prove that Na, b, c) N ab ca + b) + bc ab + c) + a, b, ) c ca bc + a). = Nab, bc, ca). SOLUTION. First, we change the inequality to SOS form as follows c ab)a + b)a b) 0. Pham Kim Hung) WLOG, assume that a b c, then S a S b S c. Therefore, it s enough to prove that b S b + c S c 0 b b ac)a + c) + c c ab)a + b) 0. This last inequality is obviously true because ba + c) ca + b) and bb ac) cc ab). The equality holds for a = b = c or a, b, c),, ). Example.4.3. Let a, b, c be positive real numbers. Prove that ab ca + b) + bc ab + c) + ca bc + a) a + b c + a + b + b + c a + b + c + c + a b + c + a, or, in other words, prove that N a, b, ) Na + b, b + c, c + a). c
c GIL Publishing House. All rights reserved. 87 SOLUTION. Similarly to the preceding problem, after changing the inequality to SOS form, we only need to prove that ab 3 + c 3 ) + bcb + c ) abc a + b) a + 3b + 3c a + b + c) if a, b, c 0 and a b c. Notice that b 3 + c 3 bcb + c), so and it remains to prove that b + c) + ab + c) a + b)a + c) LHS a + b)a + c) a + b + c)a + b + c) a + b)a + c) b + c) a + b + c a + 3b + 3c a + b + c a + b + c a + b)a + c) a + b + c. This last condition is obviously true. The equality holds for a = b = c.
88 c GIL Publishing House. All rights reserved..5 On Schur Inequality Consider the following expression in three variables a, b and c F a, b, c) = a 3 + b 3 + c 3 + 3abc aba + b) bcb + c) cac + a). By the third degree-schur inequality, we have F a, b, c) 0 for all non-negative a, b, c. In this article, we will discover some interesting relations between Schur-like expressions such as F a, b, c ), F a + b, b + c, c + a), F a, b, ) and F a b, b c c, c a), etc. First, we have Example.5.. Let a, b, c be the side lengths of a triangle. Prove that F a, b, c) F a + b, b + c, c + a). SOLUTION. Notice that the expression F a, b, c) can be rewritten as F a, b, c) = aa b)a c). Pham Kim Hung) Therefore our inequality is equivalent to b + c a)a b)a b) 0 By hypothesis we have b + c a 0, c + a b 0, a + b c 0, so the above result follows from the generalized Schur inequality. We are done and the equality holds for a = b = c equilateral triangle) or a = b = c up to permutation degenerated triangle). Example.5.. Let a, b, c be the side lengths of a triangle. Prove that F a, b, c) 4a b c F a, b, ). c Pham Kim Hung) SOLUTION. Generally, the expression F a, b, c) can be represented in SOS form as F a, b, c) = a + b c)a b). Therefore we can change our inequality to the following c a + b ) ) a + b c) a b) 0, c
c GIL Publishing House. All rights reserved. 89 and therefore the coefficients S a, S b, S c can be determined from S c = c a + b ) a + b c) ; c S b = b a + c ) c + a b) ; b S a = a b + c ) b + c a). a WLOG, assume that a b c. Clearly, S a 0 and S a S b S c. It suffices to prove that S b + S c 0 or namely 4b + c ) ) b + a c + c b b c a 0. Notice that a b + c, so b + c ) b + c) a and we are done because 4b + c ) a a ; b c + c b b + c. The equality holds for a = b = c and a = b = c up to permutation. Example.5.3. Suppose that a, b, c are the side lengths of a triangle. Prove that F a, b, c ) 36F ab, bc, ca). SOLUTION. Similarly, this inequality can be transformed into a + b c )a b ) 36 or S a b c) + S b c a) + S c a b) 0, where c ac + bc ab)a b) S a = 36a ab + ac bc) b + c) b + c a ) ; S b = 36b bc + ba ca) c + a) c + a b ) ; S c = 36c ca + cb ab) a + b) a + b c ). Pham Kim Hung) WLOG, assume that a b c, then certainly S a S b S c and S c 0. Thererfore it suffices to prove that S a + S b 0, which can be reduced to 36bcb + c ) + 34ab c) b + c) b c ) a a + b + c) a 4 + a bc 0. Suppose that S is the left expression in the previous inequality. Consider the cases
90 c GIL Publishing House. All rights reserved. i). The first case. If 7b c) b + c) aa + b + c)a + b + c). Certainly, we have S 34ab c) b + c) a a + b + c) a 4 b c ) a a + b + c)a + b + c) a a + b + c) a 4 b 4 = a 3 a + b + c) a 4 b 4 0. ii). The second case. If 7b c) b + c) aa + b + c)a + b + c), we get that a a 0 where a 0 is the unique real root of the equation 7b c) b + c) = xx + b + c)x + b + c)). Clearly, S = Sa) is a decreasing function of a, if a maxa 0, b) because S a) 0), so we obtain S = Sa) Sb + c) = 36bcb + c ) + 33b c ) 5b + c) 4 + bcb + c). Everything now becomes clear. We have of course Sb + c) = 33b c ) 5b 4 + c 4 ) + 30b c + 6bcb + c ) + bcb + c) = 8b c ) + 6bcb c) + bcb c) 0. Therefore S 0 in all cases and by SOS method, we get the desired result. The equality holds for a = b = c and a = b = c up to permutation. Example.5.4. Prove that if a, b, c are the side lengths of a triangle then 9F a, b, c) F a b, b c, c a), and if a, b, c are the side lengths of an acute triangle then 3F a, b, c) F a b, b c, c a). Pham Kim Hung) SOLUTION. We will only prove the second part of this problem because the first part can be deduced similarly but simpler. Now suppose that a, b, c are side lengths of an acute triangle. Clearly, if x + y + z = 0 then x 3 + y 3 + z 3 + 3xyz xyy + x) yzy + z) zxz + x) = 9xyz. Then, the inequality is equivalent to a + b c)a b) 3a b)b c)c a).
c GIL Publishing House. All rights reserved. 9 It s possible to assume that a c b. By the mixing all variables method, we conclude that it s sufficient to prove the inequality in case a, b, c are the side lengths of a right triangle that means a = b + c ). Because of the homogeneity, we can assume that a = and b + c =. The inequality is reduced to bc3 b c) 3 b c + bc)c b) 3b + c )c b) bc5c b 3). Because b + c b + c ) =, we deduce that b + c bc = b + c b + c bc = b + c + b + c ) + =. and it remains to prove that ) 6 c b) 5c b 3 6 ) c + 6 ) 7 b 3. This last inequality is an obvious application of Cauchy-Schwarz inequality 6 ) c + 6 ) 7 b 6 ) + 6 7).9 < 3. This ends the proof. The equality holds for the equilateral triangle, a = b = c. Example.5.5. Let a, b, c be non-negative real numbers. Prove that 9F a, b, c ) 8F a b), b c), c a) ). Pham Kim Hung) SOLUTION. We use the mixing all variables method, similarly as in the preceding problem. We can assume that a b c = 0. In this case, we obtain F a b), b c), c a) ) = a 6 + b 6 + a b) 6 + 3a b a b) a + b )a b) 4 a b a + b ) a b) a 4 + b 4 ) = a b) 4aba + b ) a b ) + a b) 4 + a b ) = 8a b a b). Moreover, because F a, b, c ) = 9a b) a + b )a + b), it remains to prove that 9a + b )a + b) 7a b, which is obvious because a + b ab and a + b) 4ab. The proof is finished and the equality holds for a = b = c and a = b, c = 0 up to permutation.
9 c GIL Publishing House. All rights reserved. Article 3 Thought Brings Knowledge - Varied Ideas.6 Exponent Smash Is there anything to say about the simplest inequalities such as a + b + c ab + bc + ca or 3a + b + c ) a + b + c) 3ab + bc + ca)? In a particular situation, in an unusual situation, they become extremely complex, hard but interesting and wonderful as well. That s why I think that this kind of inequalities is very strange and exceptional. The unusual situation we have already mentioned is when each variable a, b, c stands as the exponent of another number. Putting them in places of exponents, must have broken up the simple inequalities between variables mentioned above. Let s see some problems. Example.6.. Let a, b, c be non-negative real numbers such that a + b + c = 3. Find the maximum of the following expressions a) S = ab + bc + ca. b) S 4 = 4 ab + 4 bc + 4 ca. Pham Kim Hung) SOLUTION. Don t hurry to conclude that max S = 6 and max S 4 = because the reality is different. We figure out a solution by the mixing variable method and solve a general problem that involves both a) and b). WLOG, assume that a b c and k is a positive real constant. Consider the following expression fa, b, c) = k ab + k bc + k ca. Let t = a + b, u = a b, then t, a = t + u, b = t u and fa, b, c) = k t u + k ct u) + k ct+u) = gu).
c GIL Publishing House. All rights reserved. 93 Its derivative is g u) = u ln k k t u + ln k ck ct k cu k cu ). By Lagrange theorem, there exists a real number r [ u, u] such that k cu k cu u = ck cr, and therefore k cu k cu uck cu because r u). Moreover, c and ct + u) t u)t + u) = t u so we obtain g u) 0. Thus gu) g0) = k t + k ct = k t + k t3 t) = ht). ) ) 3 We will prove that ht) max h, h) k. Since h t) = t ln k k t + 3 4t) ln k k t3 t) we infer h t) = 0 4t 3 = t k 3tt ). Consider the following function qt) = 3tt ) ln k ln4t 3) + ln t. Because q 3 t) = 6t 3) ln k is a decreasing function of t when t ), t4t 3) we deduce that the equation q t) = 0 has no more than one root t. According [ to Rolle s theorem, the equation h t) = 0 has no more than two roots t, 3 ]. Moreover, because h ) = 0, we conclude that )) 3 ht) max h), h. According to this proof, we can synthesize a general result as follows Let a, b, c be non-negative real numbers with sum 3. For all k, we have ) k ab + k bc + k ca max 3k, k 9/4 +. Example.6.. Let a, b, c be non-negative real numbers such that a + b + c = 3. Find the minimum of the expression 3 a + 3 b + 3 c. Pham Kim Hung)
94 c GIL Publishing House. All rights reserved. SOLUTION. We will again propose and solve the general problem: for each real number k > 0, find the minimum of the following expression P = k a + k b + k c. Certainly, if k then P 3k by AM-GM inequality. Therefore we only need to consider the remaining case k. WLOG, assume that a b c. Let t = a + b, u = a b, then t and a = t + u, b = t u. Let k = and consider the following k function gu) = k t u) + k t+u) + k c. Since g u) = ln k t + u)k t+u) ln k t u)k t u), we deduce that g u) = 0 lnt + u) lnt u) = 4tu ln k. Letting now we infer that hu) = lnt + u) lnt u) + 4tu ln k, h u) = t + u + t u + 4 ln k t = t t + 4t ln k. u Therefore h u) = 0 t u ) ln k = ab ln k =. Now we divide the problem into two smaller cases i) The first case. If ab, bc, ca ln k, then k a + k b + k c k 9/4 + k ln k = k 9/4 + e /. ii) The second case. If ab. From the previous result, we deduce that ln k h u) = 0 u = 0 g u) = 0 u = 0, therefore gu) g0) = k t + k 3 t) = ft). Our remaining work is to find the minimum of ft) for 3 t. Since f t) = 4 ln k t.k t 3 t).k 3 t)), we refer that f t) = 0 3 3t)3 t) ln k = ln3 t) ln t. Denote qt) = 3 3t)3 t) ln k ln3 t) ln t then q t) = 6t ) ln k + 3 t + t = 6t ) ln 3 k + t3 t).
c GIL Publishing House. All rights reserved. 95 [ In the range, 3 ], the function t3 t) t) = t 3 7t +6t is decreasing, hence the equation q t) = 0 has no more than one real root. By Rolle s theorem, the equation f t) has no more than two roots in ft) min f), f [, 3 ]. It s then easy to get that )) 3 = min 3k, + k 9/4). According to the previous solution, the following inequality holds k a + k b + k c min 3k, + k 9/4, e / + k 9/4). Notice that if k /3 then min 3k, + k 9/4, e / + k 9/4) = 3k. Otherwise, if k /3 then + k 9/4 e / + k 9/4 min 3k, + k 9/4, e / + k 9/4) = min 3k, + k 9/4). Therefore, we can conclude that k a + k b + k c min 3k, + k 9/4, e / + k 9/4) = min 3k, + k 9/4). The initial problem is a special case for k =. In this case, we have 3 3 a + 3 b + 3 c. However, if k = then the following stranger inequality holds a + b + c + 5/4. Example.6.3. Let a, b, c be non-negative real numbers such that a + b + c = 3. Prove that 4ab + 4bc + 4ca 3abc 53. Pham Kim Hung) SOLUTION. In fact, this problem reminds us of Schur inequality only that we know have exponents notice that if a + b + c = 3 then Schur inequality is equivalent to 4ab + bc + ca) 9 + 3abc). According to the example.6., we deduce that 4ab + 4bc + 4ca = 6 ab + 6 bc + 6 ca 6 9/4 + = 9 +,
96 c GIL Publishing House. All rights reserved. moreover, we also have the obvious inequality 3abc, so 4ab + 4bc + 4ca 3abc 9 + = 53. Transforming an usual inequality into one with exponents, you can obtain a new one. This simple idea leads to plenty inequalities, some nice, hard but also interesting. As a matter of fact, you will rarely encounter this kind of inequalities, however, I strongly believe that a lot of enjoyable, enigmatic matters acan be found here. So why don t you try it yourself?.7 Unexpected Equalities Some people often make mistakes when they believe that all symmetric inequalities of three variables in fraction forms) have their equality just in one of two standard cases: a = b = c or a = b, c = 0 and permutations of course). Sure almost all inequalities belong to this kind, but some are stranger. These inequalities, very few of them in comparison, make up a different and interesting area, where the usual SOS method is nearly impossible. Here are some examples. Example.7.. Let x, y, z be non-negative real numbers. Prove that x y + z + y z + x + z 5xy + yz + zx) + x + y x + y + z) 8. SOLUTION. WLOG, assume that x y z. Denote We infer that fx, y, z) = x y + z + y z + x + z 5xy + yz + zx) + x + y x + y + z). Pham Kim Hung) fx, y, z) fx, y + z, 0) = y z + x + z x + y y + z 5yz x x + y + z) ) 5 = yz x + y + z) xx + y). xx + z) Notice that x y z, so we have z x + y + y x + z z y + z + y y + z = x + y + x + z 3 x + y + z
c GIL Publishing House. All rights reserved. 97 xx + y) + xx + z) 3 xx + y + z) 5 x + y + z). This shows that fx, y, z) fx, y + z, 0). Denote t = y + z, then we have fx, y + z, 0) = x t + t x + 5xt x + t) = + + 5xt x + t) xt x + t) 8, and the conclusion follows. The equality holds for x + t) = 5xt or x t = 3 ± 5. 3 ± ) 5 In the initial inequality, the equality holds for x, y, z),, 0. Comment. In general, the following inequality can be proved by the same method Given non-negative real numbers x, y, z. For all k 6, prove that x y + z + y z + x + z kxy + yz + zx) ) + x + y x + y + z) k. Example.7.. Let x, y, z be non-negative real numbers. Prove that x y + z + y z + x + z 6xy + yz + zx) + x + y x + y + z 8. Tan Pham Van) SOLUTION. We use mixing variables to solve this problem. Denote x = max{x, y, z} and fx, y, z) = x y + z + y z + x + z 6xy + yz + zx) + x + y x + y + z. It is easy to see that the statament fx, y, z) fx, y + z, 0) is equivalent to 6x + y + z) xx + z)x + y) x + y + z) x + y + z ) x + y + z) ) Notice that x = max{x, y, z}, so x + y)x + z) x + y + z ) ; 6x + y + z) x x + y + z) x + y + z) ) ; These two results show that fx, y, z) fx, y + z, 0). Normalizing x + y + z =, we get fx, y, z) fx, y + z, 0) = fx, x, 0) = gx). We have g x) = x )x x )6x 6x + ) x x ) x x + )
98 c GIL Publishing House. All rights reserved. notice that 3 x, and it follows that the equation g x) = 0 has two roots { x ; 3 + } 3. 6 It s then easy to infer that gx) g 3 + ) 3 = 4, 6 as desired. The equality holds for x, y, z) 3 + 3 ; 3 3 ; 0 ). Example.7.3. Let a, b, c be non-negative real numbers with sum. Prove that a b + c + b c + a + SOLUTION. WLOG, assume that a b c. Denote fa, b, c) = a b + c + c 7a + b + c) a + + b a + b + c 5. b c + a + We will prove that fa, b, c) fa, b + c, 0). Indeed fa, b, c) f a, ) b + c, 0 = b c + a + Moreover, because c a + b b + c b c a a + b ) + a a + c ) b c a a + b ) + a a + c ) c 7a + b + c) a + + b a + b + c. Pham Kim Hung) a + 7a + b + c) 7 a + b + c ) a + b + c ) + 54a b + c b + c ) ) + a + b + c 54b c 4bca + b + c ) 3 a 54 4bc ; a a + b ) + a a + c ) 3 a + b + c ; We infer that fa, b, c) f a, b + c, 0 ). Denote t = b + c, then we have fa, t, 0) = a t + t 7a + t) + a a + t = + a + t ) a t + 54at a + t + 7
c GIL Publishing House. All rights reserved. 99 = a + t ) a t + 7at a + t + 7at a + t + 5 3 3 7 7 + 5 = 5. 3 ± ) 5 This ends the proof. The equality holds for a, b, c),, 0. Comment. In a similar way, we can prove the following general inequality Given non-negative real numbers x, y, z. For all k 8, prove that a b + c + b c + a + c ka + b + c) a + + b a + b + c k + 3 k. Example.7.4. Let a, b, c be non-negative real numbers. Prove that a + b) + b + c) + c + a) + 4 a + b + c) 8 ab + bc + ca. SOLUTION. WLOG, assume that a b c. Denote fa, b, c) = We get that Pham Kim Hung) a + b) + b + c) + c + a) + 4 a + b + c) 8 ab + bc + ca. fa, b, c) fa, b + c, 0) = a + b) a + b + c) + c + a) a 8 ab + bc + ca + 8 ab + c) 8bc ca + c) ab + c)ab + bc + ca) a a + c). Because a b c, we get that 8ab a + c)b + c) ; a + c) ab + bc + ca) ; These two results show that fa, b, c) fa, b + c, 0) 0. Denote t = b + c, then we get fa, b + c, 0) = fa, t, 0) = a + t + 5 a + t) 8 at. By AM-GM inequality, we have at a + t + 5 ) a + t) a + t) = + + 5 at a + t) + 5 = 8.
300 c GIL Publishing House. All rights reserved. This ends the proof. The equality holds for a, b, c) 3 ± ) 5,, 0. Comment. By a similar method, we can prove the following generalization Given non-negative real numbers a, b, c. For all k 5, prove that a + b) + b + c) + c + a) + k a + b + c) k + ab + bc + ca. Moreover, the following result can also be proved similarly. Given non-negative real numbers a, b, c. Prove that a + b + b + c + c + a + 8 a + b + c 6 ab + bc + ca. Example.7.5. Let a, b, c be non-negative real numbers. Prove that a b + c + b c + a + c 4a + b + c)ab + bc + ca) + a + b a 3 + b 3 + c 3 5. SOLUTION. WLOG, assume that a b c. Denote t = b + c and We have fa, b, c) = fa, b, c) fa, b + c, 0) = b c + a + = because a b + c + b c + a + c 4a + b + c)ab + bc + ca) + a + b a 3 + b 3 + c 3. c a + b b + c 4a + b + c)ab + bc + ca) + a a 3 + b 3 + c 3 4a 3 + 3ab + c) + b + c) 3 )bc a + b + c)bc a ab + c) + b + c) )a 3 + b 3 + c 3 ) aa + b)a + c) 0 a + b)a + c) a ab + c) + b + c) ; Tan Pham Van) 4ab + c) a ab + c) + b + c) 4aa 3 + 3ab + c) + b + c) 3 ) a + b + c)a 3 + b 3 + c 3 ). Moreover, by AM-GM inequality, we have fa, b + c, 0) = fa, t, 0) = a t + t a + = a at + t at 4at a at + t 4at + a at + t + 5. This is the end of the proof. The equality holds for a, b, c) 3 ± ) 5,, 0.