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www.batdangthuc.net 5 Inequalities From 2007 Mathematical Competition Over The World  Example 1 (Iran National Mathematical Olympiad 2007). Assume that a, b, c are three different positive real numbers. Prove that     a + b a − b + b + c b − c + c + a c − a     > 1. Example 2 (Iran National Mathematical Olympiad 2007). Find the largest real T such that for each non-negative real numbers a, b, c, d, e such that a + b = c + d + e, then  a 2 + b 2 + c 2 + d 2 + e 2 ≥ T ( √ a + √ b + √ c + √ d + √ e) 2 . Example 3 (Middle European Mathematical Olympiad 2007). Let a, b, c, d be positive real numbers with a + b + c + d =4. Prove that a 2 bc + b 2 cd + c 2 da + d 2 ab ≤ 4. Example 4 (Middle European Mathematical Olympiad 2007). Let a, b, c, d be real num- bers which satisfy 1 2 ≤ a, b, c, d ≤ 2 and abcd =1. Find the maximum value of  a + 1 b  b + 1 c  c + 1 d  d + 1 a  . Example 5 (China Northern Mathematical Olympiad 2007). Let a, b, c be side lengths of a triangle and a + b + c =3. Find the minimum of a 2 + b 2 + c 2 + 4abc 3 . Example 6 (China Northern Mathematical Olympiad 2007). Let α, β be acute angles. Find the maximum value of  1 − √ tan α tan β  2 cot α + cot β . Example 7 (China Northern Mathematical Olympiad 2007). Let a, b, c be positive real numbers such that abc =1. Prove that a k a + b + b k b + c + c k c + a ≥ 3 2 , for any positive integer k ≥ 2. 6 www.batdangthuc.net Example 8 (Croatia Team Selection Test 2007). Let a, b, c > 0 such that a + b + c =1. Prove that a 2 b + b 2 c + c 2 a ≥ 3(a 2 + b 2 + c 2 ). Example 9 (Romania Junior Balkan Team Selection Tests 2007). Let a, b, c three pos- itive reals such that 1 a + b +1 + 1 b + c +1 + 1 c + a +1 ≥ 1. Show that a + b + c ≥ ab + bc + ca. Example 10 (Romania Junior Balkan Team Selection Tests 2007). Let x, y, z ≥ 0 be real numbers. Prove that x 3 + y 3 + z 3 3 ≥ xyz + 3 4 |(x − y)(y − z)(z − x)|. Example 11 (Yugoslavia National Olympiad 2007). Let k be a given natural number. Prove that for any positive numbers x, y, z with the sum 1 the following inequality holds x k+2 x k+1 + y k + z k + y k+2 y k+1 + z k + x k + z k+2 z k+1 + x k + y k ≥ 1 7 . Example 12 (Cezar Lupu & Tudorel Lupu, Romania TST 2007). For n ∈ N,n ≥ 2,a i ,b i ∈ R, 1 ≤ i ≤ n, such that n  i=1 a 2 i = n  i=1 b 2 i =1,  n i=1 a i b i =0. Prove that  n  i=1 a i  2 +  n  i=1 b i  2 ≤ n. Example 13 (Macedonia Team Selection Test 2007). Let a, b, c be positive real numbers. Prove that 1+ 3 ab + bc + ca ≥ 6 a + b + c . Example 14 (Italian National Olympiad 2007). a) For each n ≥ 2, find the maximum constant c n such that 1 a 1 +1 + 1 a 2 +1 + .+ 1 a n +1 ≥ c n , for all positive reals a 1 ,a 2 , .,a n such that a 1 a 2 ···a n =1. b) For each n ≥ 2, find the maximum constant d n such that 1 2a 1 +1 + 1 2a 2 +1 + .+ 1 2a n +1 ≥ d n for all positive reals a 1 ,a 2 , .,a n such that a 1 a 2 ···a n =1. www.batdangthuc.net 7 Example 15 (France Team Selection Test 2007). Let a, b, c, d be positive reals such taht a + b + c + d =1. Prove that 6(a 3 + b 3 + c 3 + d 3 ) ≥ a 2 + b 2 + c 2 + d 2 + 1 8 . Example 16 (Irish National Mathematical Olympiad 2007). Suppose a, b and c are positive real numbers. Prove that a + b + c 3 ≤  a 2 + b 2 + c 2 3 ≤ 1 3  ab c + bc a + ca b  . For each of the inequalities, find conditions on a, b and c such that equality holds. Example 17 (Vietnam Team Selection Test 2007). Given a triangle ABC. Find the minimum of cos 2 A 2 cos 2 B 2 cos 2 C 2 + cos 2 B 2 cos 2 C 2 cos 2 A 2 + cos 2 C 2 cos 2 A 2 cos 2 B 2 . Example 18 (Greece National Olympiad 2007). Let a,b,c be sides of a triangle, show that (c + a − b) 4 a(a + b − c) + (a + b − c) 4 b(b + c − a) + (b + c − a) 4 c(c + a− b) ≥ ab + bc + ca. Example 19 (Bulgaria Team Selection Tests 2007). Let n ≥ 2 is positive integer. Find the best constant C(n) such that n  i=1 x i ≥ C(n)  1≤j<i≤n (2x i x j + √ x i x j ) is true for all real numbers x i ∈ (0, 1),i =1, ., n for which (1 − x i )(1 − x j ) ≥ 1 4 , 1 ≤ j<i≤ n. Example 20 (Poland Second Round 2007). Let a, b, c, d be positive real numbers satisfying the following condition: 1 a + 1 b + 1 c + 1 d =4. Prove that: 3  a 3 + b 3 2 + 3  b 3 + c 3 2 + 3  c 3 + d 3 2 + 3  d 3 + a 3 2 ≤ 2(a + b + c + d) − 4. Example 21 (Turkey Team Selection Tests 2007). Let a, b, c be positive reals such that their sum is 1. Prove that 1 ab +2c 2 +2c + 1 bc +2a 2 +2a + 1 ac +2b 2 +2b ≥ 1 ab + bc + ac . 8 www.batdangthuc.net Example 22 (Moldova National Mathematical Olympiad 2007). Real numbers a 1 ,a 2 , .,a n satisfy a i ≥ 1 i , for all i = 1,n. Prove the inequality (a 1 +1)  a 2 + 1 2  ·····  a n + 1 n  ≥ 2 n (n + 1)! (1 + a 1 +2a 2 + ···+ na n ). Example 23 (Moldova Team Selection Test 2007). Let a 1 ,a 2 , .,a n ∈ [0, 1]. Denote S = a 3 1 + a 3 2 + .+ a 3 n , prove that a 1 2n +1+S − a 3 1 + a 2 2n +1+S − a 3 2 + .+ a n 2n +1+S − a 3 n ≤ 1 3 . Example 24 (Peru Team Selection Test 2007). Let a, b, c be positive real numbers, such that a + b + c ≥ 1 a + 1 b + 1 c . Prove that a + b + c ≥ 3 a + b + c + 2 abc . Example 25 (Peru Team Selection Test 2007). Let a, b and c be sides of a triangle. Prove that √ b + c − a √ b + √ c − √ a + √ c + a − b √ c + √ a − √ b + √ a + b − c √ a + √ b − √ c ≤ 3.  Example 26 (Romania Team Selection Tests 2007). If a 1 ,a 2 , .,a n ≥ 0 satisfy a 2 1 + ···+ a 2 n =1, find the maximum value of the product (1 − a 1 )···(1 − a n ). Example 27 (Romania Team Selection Tests 2007). Prove that for n, p integers, n ≥ 4 and p ≥ 4, the proposition P(n, p) n  i=1 1 x i p ≥ n  i=1 x i p for x i ∈ R,x i > 0,i=1, .,n , n  i=1 x i = n, is false. Example 28 (Ukraine Mathematical Festival 2007). Let a, b, c be positive real numbers and abc ≥ 1. Prove that (a).  a + 1 a +1  b + 1 b +1  c + 1 c +1  ≥ 27 8 . (b). 27(a 3 +a 2 +a+1)(b 3 +b 2 +b+1)(c 3 +c 2 +c+1) ≥≥ 64(a 2 +a+1)(b 2 +b+1)(c 2 +c+1). Example 29 (Asian Pacific Mathematical Olympiad 2007). Let x, y and z be positive real numbers such that √ x + √ y + √ z =1. Prove that x 2 + yz  2x 2 (y + z) + y 2 + zx  2y 2 (z + x) + z 2 + xy  2z 2 (x + y) ≥ 1. www.batdangthuc.net 9 Example 30 (Brazilian Olympiad Revenge 2007). Let a, b, c ∈ R with abc =1. Prove that a 2 +b 2 +c 2 + 1 a 2 + 1 b 2 + 1 c 2 +2  a + b + c + 1 a + 1 b + 1 c  ≥ 6+2  b a + c b + a c + c a + c b + b c  . Example 31 (India National Mathematical Olympiad 2007). If x, y, z are positive real numbers, prove that (x + y + z) 2 (yz + zx + xy) 2 ≤ 3(y 2 + yz + z 2 )(z 2 + zx + x 2 )(x 2 + xy + y 2 ). Example 32 (British National Mathematical Olympiad 2007). Show that for all positive reals a, b, c, (a 2 + b 2 ) 2 ≥ (a + b + c)(a + b− c)(b + c− a)(c + a − b). Example 33 (Korean National Mathematical Olympiad 2007). For all positive reals a, b, and c, what is the value of positive constant k satisfies the following inequality? a c + kb + b a + kc + c b + ka ≥ 1 2007 . Example 34 (Hungary-Isarel National Mathematical Olympiad 2007). Let a, b, c, d be real numbers, such that a 2 ≤ 1,a 2 + b 2 ≤ 5,a 2 + b 2 + c 2 ≤ 14,a 2 + b 2 + c 2 + d 2 ≤ 30. Prove that a + b + c + d ≤ 10. 10 www.batdangthuc.net SOLUTION  Please visit the following links to get the original discussion of the ebook, the problems and solution. We are appreciating every other contribution from you! http://www.batdangthuc.net/forum/showthread.php?t=26 http://www.batdangthuc.net/forum/showthread.php?t=26&page=2 http://www.batdangthuc.net/forum/showthread.php?t=26&page=3 http://www.batdangthuc.net/forum/showthread.php?t=26&page=4 http://www.batdangthuc.net/forum/showthread.php?t=26&page=5 http://www.batdangthuc.net/forum/showthread.php?t=26&page=6  For Further Reading, Please Review:  UpComing Vietnam Inequality Forum's Magazine  Secrets in Inequalities (2 volumes), Pham Kim Hung (hungkhtn)  Old And New Inequalities, T. Adreescu, V. Cirtoaje, M. Lascu, G. Dospinescu  Inequalities and Related Issues, Nguyen Van Mau  We thank a lot to Mathlinks Forum and their member for the reference to problems and some nice solutions from them! www.batdangthuc.net 11 Problem 1 (1, Iran National Mathematical Olympiad 2007). Assume that a, b, c are three different positive real numbers. Prove that     a + b a − b + b + c b − c + c + a c − a     > 1. Solution 1 (pi3.14). Due to the symmetry, we can assume a>b>c. Let a = c + x; b = c + y, then x>y>0. We have     a + b a − b + b + c b − c + c + a c − a     = 2c + x + y x − y + 2c + y y − 2c + x x =2c  1 x − y + 1 y − 1 x  + x + y x − y . We have 2c  1 x − y + 1 y − 1 x  =2c  1 x − y + x − y xy  > 0. x + y x − y > 1. Thus     a + b a − b + b + c b − c + c + a c − a     > 1. Solution 2 (2, Mathlinks, posted by NguyenDungTN). Let a + b a − b = x; b + c b − c = y; a + c c − a = z; Then xy + yz + xz =1. By Cauchy-Schwarz Inequality (x + y + z) 2 ≥ 3(xy + yz + zx)=3⇒|x + y + z|≥ √ 3 > 1. We are done. ∇ Problem 2 (2, Iran National Mathematical Olympiad 2007). Find the largest real T such that for each non-negative real numbers a, b, c, d, e such that a + b = c + d + e, then  a 2 + b 2 + c 2 + d 2 + e 2 ≥ T ( √ a + √ b + √ c + √ d + √ e) 2 12 www.batdangthuc.net Solution 3 (NguyenDungTN). Let a = b =3,c= d = e =2, we find √ 30 6( √ 3+ √ 2) 2 ≥ T. With this value of T , we will prove the inequality. Indeed, let a + b = c + d + e = X.By Cauchy-Schwarz Inequality a 2 + b 2 ≥ (a + b) 2 2 = X 2 2 c 2 + d 2 + e 2 ≥ (c + d + e) 2 3 = X 2 3 ⇒  a 2 + b 2 + c 2 + d 2 + e 2 ≥ 5X 2 6 (1) By Cauchy-Schwarz Inequality, we also have √ a + √ b ≤  2(a + b)= √ 2X √ c + √ d + √ e ≤  3(c + d + e)=3X ⇒ ( √ a + √ b + √ c + √ d + √ e) 2 ≤ ( √ 2+ √ 3) 2 X 2 (2) From (1) and (2), we have √ a 2 + b 2 + c 2 + d 2 + e 2 ( √ a + √ b + √ c + √ d + √ e) 2 ≥ √ 30 6( √ 3+ √ 2) 2 . Equality holds for 2a 3 = 2b 3 = c = d = e. ∇ Problem 3 (3, Middle European Mathematical Olympiad 2007). Let a, b, c, d non- negative such that a + b + c + d =4. Prove that a 2 bc + b 2 cd + c 2 da + d 2 ab ≤ 4. Solution 4 (mathlinks, reposted by pi3.14). Let {p,q,r,s} = {a, b, c, d} and p ≥ q ≥ r ≥ s. By rearrangement Inequality, we have a 2 bc + b 2 cd + c 2 da + d 2 ab = a(abc)+b(bcd)+c(cda)+d(dab) ≤ p(pqr)+q(pqs)+r(prs)+s(qrs)=(pq + rs)(pr + qs) ≤  pq + rs + pr + qs 2  2 = 1 4 (p + s) 2 (q + r) 2 ≤ 1 4   p + q + r + s 2  2  2 =4. Equality holds for q = r =1vp + s =2. Easy to refer (a, b, c, d)=(1, 1, 1, 1), (2, 1, 1, 0) or permutations. www.batdangthuc.net 13 ∇ Problem 4 ( 5- Revised by VanDHKH). Let a, b, c be three side-lengths of a triangle such that a + b + c =3. Find the minimum of a 2 + b 2 + c 2 + 4abc 3 Solution 5. Let a = x + y, b = y + z, c = z + x, we have x + y + z = 3 2 . Consider a 2 + b 2 + c 2 + 4abc 3 = (a 2 + b 2 + c 2 )(a + b + c)+4abc 3 = 2((x + y) 2 +(y + z) 2 +(z + x) 2 )(x + y + z)+4(x + y)(y + z)(z + x) 3 = 4(x 3 + y 3 + z 3 +3x 2 y +3xy 2 +3y 2 z +3yz 2 +3z 2 x +3zx 2 +5xyz) 3 = 4((x + y + z) 3 − xyz) 3 = 4( 26 27 (x + y + z) 3 +( x+y+z 3 ) 3 − xyz) 3 ≥ 4( 26 27 (x + y + z) 3 ) 3 = 13 3 . Solution 6 (2, DDucLam). Using the familiar Inequality (equivalent to Schur) abc ≥ (b + c − a)(c + a − b)(a + b − c) ⇒ abc ≥ 4 3 (ab + bc + ca) − 3. Therefore P ≥ a 2 + b 2 + c 2 + 16 9 (ab + bc + ca) − 4 =(a + b + c) 2 − 2 9 (ab + bc + ca) − 4 ≥ 5 − 2 27 (a + b + c) 2 =4+ 1 3 . Equality holds when a = b = c =1. Solution 7 (3, pi3.14). With the conventional denotion in triangle, we have abc =4pRr , a 2 + b 2 + c 2 =2p 2 − 8Rr − 2r 2 . Therefore a 2 + b 2 + c 2 + 4 3 abc = 9 2 − 2r 2 . Moreover, p ≥ 3 √ 3r ⇒ r 2 ≤ 1 6 . Thus a 2 + b 2 + c 2 + 4 3 abc ≥ 4 1 3 . 14 www.batdangthuc.net ∇ Problem 5 (7, China Northern Mathematical Olympiad 2007). Let a, b, c be positive real numbers such that abc =1. Prove that a k a + b + b k b + c + c k c + a ≥ 3 2 . for any positive integer k ≥ 2. Solution 8 (Secrets In Inequalities, hungkhtn). We have a k a + b + b k b + c + c k c + a ≥ 3 2 ⇔ a k−1 + b k−1 + c k−1 ≥ 3 2 + a k−1 b a + b + b k−1 c b + c + c k−1 a c + a By AM-GM Inequality, we have a + b ≥ 2 √ ab, b + c ≥ 2 √ bc, c + a ≥ 2 √ ca. So, it remains to prove that a k− 3 2 b 1 2 + b k− 3 2 c 1 2 + c k− 3 2 a 1 2 +3≤ 2  a k−1 + b k−1 + c k−1  . This follows directly by AM-GM inequality, since a k−1 + b k−1 + c k−1 ≥ 3 3 √ a k−1 b k−1 c k−1 =3 and (2k − 3)a k−1 + b k−1 ≥ (2k − 2)a k− 3 2 b 1 2 (2k − 3)b k−1 + c k−1 ≥ (2k − 2)b k− 3 2 c 1 2 (2k − 3)c k−1 + a k−1 ≥ (2k − 2)c k− 3 2 a 1 2 Adding up these inequalities, we have the desired result. ∇ Problem 6 (8, Revised by NguyenDungTN). Let a, b, c > 0 such that a + b + c =1. Prove that: a 2 b + b 2 c + c 2 a ≥ 3(a 2 + b 2 + c 2 ). [...]... c+a+1 a+b b+c c+a + + ≥2 a+b+1 b+c+1 c+a+1 By Cauchy-Schwarz Inequality, ⇔ LHS ≥ (a + b + b + c + c + a)2 = 2 cyc (a + b)(a + b + 1) We are done Comment This second very beautiful solution uses Contradiction method If you can't understand the principal of this method, have a look at Sang Tao Bat Dang Thuc, or Secrets In Inequalities, written by Pham Kim Hung Problem 8 (10, Romanian JBTST V 2007) Let x,... |xy(x − y)| 4 This last inequality can be checked easily Problem 9 (11, Yugoslavia National Olympiad 2007) Let k be a given natural number Prove that for any positive numbers x, y, z with the sum 1, the following inequality holds xk+2 yk+2 z k+2 1 + k+1 + k+1 ≥ xk+1 + yk + z k y + z k + xk z + xk + y k 7 When does equality occur? Solution 15 (NguyenDungTN) We can assume that x ≥ y ≥ z By this assumption,... NguyenDungTN) a) Let a1 = n−1 , ak = 1 ∀k = 1, then let → 0, we easily get cn ≤ 1 We will prove the inequality with this value of cn Without loss of generality, assume that a1 ≤ a2 ≤ · · · ≤ an Since a1 a2 ≤ 1, we have n k=1 1 1 1 1 a1 a1 1 ≥ + = + + = 1 ≥ ak + 1 a1 + 1 a2 + 1 a1 + 1 a2 + a1a2 a1 + 1 a1 + 1 This ends the proof b) Consider n = 2, it is easy to get d2 = becomes 2 3 Indeed, let a1 = a, a2 = 1 a The... ≥2 42 1 2( a + 1 + b 1 c + 1 ) d = 32 =4 8 This ends the proof Problem 17 (21, Turkey Team Selection Tests 2007) Let a, b, c be positive reals such that their sum is 1 Prove that: 1 1 1 1 + + ≥ ab + 2c2 + 2c bc + 2a2 + 2a ac + 2b2 + 2b ab + bc + ac Solution 25 (NguyenDungTN) First, we will prove that ab ab + ac + bc ≥ 2 + 2c ab + 2c ab + ac + bc Indeed, this is equivalent to a2b2 + b2c2 + c2 a2 +... www.batdangthuc.net 26 So it is enough to prove that 2 (a + b + c)2 ≥ 3 + (a + b + c)2 ⇔ (a + b + c)2 ≥ 9 3 This inequality is true due to a + b + c ≥ 3 Solution 30 (2, DDucLam) We have a+b+c ≥ 2 1 1 1 1 2 3 (a + b + c) + ( + + ) ≥ (a + b + c) + 3 3 a b c 3 a+b+c We only need to prove that a+b+c ≥ 3 , abc but this inequality is always true since (a + b + c)2 ≥ 1 1 1 + + a b c 2 ≥3 1 1 1 + + ab bc ca = 3 (a... Solution 10 (2, By Zaizai) a2 b2 c2 + + ≥ 3(a2 + b2 + c2 ) b c a a2 − 2a + b b ⇔ ≥ 3(a2 + b2 + c2 ) − (a + b + c)2 (a − b)2 ≥ (a − b)2 + (b − c)2 + (c − a)2 b ⇔ ⇔ ⇔ (a − b)2 1 −1 b ≥0 (a − b)2(a + c) ≥ 0 b This ends the solution, too Problem 7 (9, Romania Junior Balkan Team Selection Tests 2007) Let a, b, c be three positive reals such that 1 1 1 + + ≥ 1 a+b+1 b+c+1 c+a+1 Show that a + b + c ≥ ab + bc +... 2007) Real numbers a1 , a2, · · · , an satisfy ai ≥ 1 , for all i = 1, n Prove the inequality i (a1 + 1) a2 + 1 2 · · · · · an + 1 n ≥ 2n (1 + a1 + 2a2 + · · · + nan ) (n + 1)! Solution 26 (NguyenDungTN) This inequality is equivalent to (a1 + 1)(2a2 + 1) · · (nan + 1) ≥ 2n (1 + a1 + 2a2 + + nan ) n+1 It is clearly true when n = 1 Assume that it si true for n = k, we have to prove it for n = k + 1 Indeed,... xyz z3 3 k=1 9 a1 a3 , z= a2 2 9 a1 a2 a2 3 = 1 Thus x3 y3 z3 1 ≥ 3 + 3 + 3 ak + 1 x +2 y +2 z +2 x2 y2 z2 + 2 + 2 x2 + 2yz y + 2xz z + 2xy x2 y2 z2 + 2 + 2 = 1 x2 + y 2 + z 2 x + y 2 + z 2 x + y2 + z 2 This ends the proof Problem 12 (15, France Team Selection Test 2007) Let a, b, c, d be positive reals such that a + b + c + d = 1 Prove that: 1 6(a3 + b3 + c3 + d3) ≥ a2 + b2 + c2 + d2 + 8 www.batdangthuc.net... − y)(x − z) 2 By AM-GM inequality, we have √ √ b+c−a √ √ = b+ c− a 1− (x − y)(x − z) (x − y)(x − z) ≤ 1− 2 2x 4x2 We will prove that x−2(x − y)(x − z) + y−2 (y − z)(y − z) + z −2(z − x)(z − y) ≥ 0 But this immediately follows the general Schur inequality, with the assumption that x ≥ y ≥ z ⇒ x−2 ≤ y−2 ≤ z −2 We are done! www.batdangthuc.net 27 Problem 22 (26, Romania Team Selection Tests 2007) If a1... that x1 ≤ x2 ≤ thus x2 + x3 + x2 + x3 ≥ 2k2/3 + k4/3 = 1.07 ≥ 1 Similarly, we have f(x1 , x2, , xn) ≤ f(x1 , x2x3, 1, x4, , xn) ≤ f(x1 , x2x3 x4, 1, 1, x5, , xn) ≤ ≤ f(x1 , x2x3 xn, 1, 1, , 1), From this, easy to get the final result Problem 23 (28, Ukraine Mathematic Festival 2007) Let a, b, c > 0 và abc ≥ 1 Prove that a) a+ 1 a+1 b+ 1 b+1 c+ 1 c+1 ≥ 27 8 b) 27(a3 + a2 + a + 1)(b3 + b2 + b + 1)(c3 . We are done Comment. This second very beautiful solution uses Contradiction method. If you can't understand the principal of this method, have a look. 2 b 1 2 + b k− 3 2 c 1 2 + c k− 3 2 a 1 2 +3≤ 2  a k−1 + b k−1 + c k−1  . This follows directly by AM-GM inequality, since a k−1 + b k−1 + c k−1 ≥ 3 3