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www.batdangthuc.net Dien Dan Bat Dang Thuc Viet Nam www.batdangthuc.net Editors Dien Dan Bat Dang Thuc Viet Nam B i Viet Nay (cung voi file PDF di kem) duoc tao ra vi muc dich giao duc. Khong duoc su dung ban EBOOK nay duoi bat ky muc dich thuong mai nao, tru khi duoc su dong y cua tac gia. Moi chi tiet xin lien he: www.batdangthuc.net.
4 www.batdangthuc.net Dien Dan Bat Dang Thuc Viet Nam www.batdangthuc.net Contributors Of The Book Editor. Pham Kim Hung (hungkhtn) Admin, VIF Forum, Student, Stanford University Editor. Nguyen Manh Dung (NguyenDungTN) Super Mod, VIF Forum, Student, Hanoi National University Editor. Vu Thanh Van (VanDHKH) Moderator, VIF Forum, Student, Hue National School Editor. Duong Duc Lam (dduclam) Super Moderator, VIF Forum, Student, Civil Engineering University Editor. Le Thuc Trinh (pi.4) Moderator, VIF Forum, Student, High School Editor. Nguyen Thuc Vu Hoang (zaizai) Super Moderator, VIF Forum, Student, High School Editors. And Other VIF members who help us a lot to complete this verion
www.batdangthuc.net 5 Inequalities From 007 Mathematical Competition Over The World Example (Iran National Mathematical Olympiad 007). 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 >. Example (Iran National Mathematical Olympiad 007). 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 + b + c + d + e T ( a + b + c + d + e). Example (Middle European Mathematical Olympiad 007). Let a, b, c, d be positive real numbers with a + b + c + d =4. Prove that a bc + b cd + c da + d ab 4. Example 4 (Middle European Mathematical Olympiad 007). Let a, b, c, d be real numbers which satisfy a, b, c, d and abcd =. Find the maximum value of ( a + )( b + )( c + )( d + ). b c d a Example 5 (China Northern Mathematical Olympiad 007). Let a, b, c be side lengths of a triangle and a + b + c =. Find the minimum of a + b + c + 4abc. Example 6 (China Northern Mathematical Olympiad 007). Let α, β be acute angles. Find the maximum value of ( ) tan α tan β. cot α + cot β Example 7 (China Northern Mathematical Olympiad 007). Let a, b, c be positive real numbers such that abc =. Prove that for any positive integer k. a k a + b + bk b + c + ck c + a,
6 www.batdangthuc.net Example 8 (Croatia Team Selection Test 007). Let a, b, c > 0 such that a + b + c =. Prove that a b + b c + c a (a + b + c ). Example 9 (Romania Junior Balkan Team Selection Tests 007). Let a, b, c three positive reals such that a + b + + b + c + + c + a +. Show that a + b + c ab + bc + ca. Example 0 (Romania Junior Balkan Team Selection Tests 007). Let x, y, z 0 be real numbers. Prove that x + y + z xyz + (x y)(y z)(z x). 4 Example (Yugoslavia National Olympiad 007). Let k be a given natural number. Prove that for any positive numbers x, y, z with the sum the following inequality holds x k+ x k+ + y k + z + y k+ k y k+ + z k + x + z k+ k z k+ + x k + y k 7. Example (Cezar Lupu & Tudorel Lupu, Romania TST 007). For n N,n,a i,b i R, i n, such that n a i = n i= i= ( n ) ( n ) a i + b i n. i= b i =, n i= a ib i =0. Prove that Example (Macedonia Team Selection Test 007). Let a, b, c be positive real numbers. Prove that + ab + bc + ca 6 a + b + c. Example 4 (Italian National Olympiad 007). a) For each n, find the maximum constant c n such that a + + a + +...+ a n + c n, for all positive reals a,a,...,a n such that a a a n =. i= b) For each n, find the maximum constant d n such that a + + a + +...+ a n + d n for all positive reals a,a,...,a n such that a a a n =.
www.batdangthuc.net 7 Example 5 (France Team Selection Test 007). Let a, b, c, d be positive reals such taht a + b + c + d =. Prove that 6(a + b + c + d ) a + b + c + d + 8. Example 6 (Irish National Mathematical Olympiad 007). Suppose a, b and c are positive real numbers. Prove that a + b + c a + b + c ( ab c + bc a + ca ). b For each of the inequalities, find conditions on a, b and c such that equality holds. Example 7 (Vietnam Team Selection Test 007). Given a triangle ABC. Find the minimum of cos A cos B cos C + cos B cos C cos A + cos C cos A cos B. Example 8 (Greece National Olympiad 007). 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) Example 9 (Bulgaria Team Selection Tests 007). Let n is positive integer. Find the best constant C(n) such that n x i C(n) i= j<i n (x i x j + x i x j ) is true for all real numbers x i (0, ),i =,..., n for which ( x i )( x j ) 4, j<i n. Example 0 (Poland Second Round 007). Let a, b, c, d be positive real numbers satisfying the following condition: a + b + c + d =4. Prove that: a + b + b + c + c + d + d + a (a + b + c + d) 4. Example (Turkey Team Selection Tests 007). Let a, b, c be positive reals such that their sum is. Prove that ab +c +c + bc +a +a + ac +b +b ab + bc + ac.
8 www.batdangthuc.net Example (Moldova National Mathematical Olympiad 007). Real numbers a,a,...,a n satisfy a i, for all i =,n. Prove the inequality i ( (a +) a + ) ( a n + ) n n (n + )! ( + a +a + + na n ). Example (Moldova Team Selection Test 007). Let a,a,...,a n [0, ]. Denote S = a + a +...+ a n, prove that a a n ++S a + n ++S a +...+ n ++S a n. Example 4 (Peru Team Selection Test 007). Let a, b, c be positive real numbers, such that a + b + c a + b + c. Prove that a + b + c a + b + c + abc. Example 5 (Peru Team Selection Test 007). 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 n a + b c a + b c. Example 6 (Romania Team Selection Tests 007). If a,a,...,a n 0 satisfy a + + a n =, find the maximum value of the product ( a ) ( a n ). Example 7 (Romania Team Selection Tests 007). Prove that for n, p integers, n 4 and p 4, the proposition P(n, p) n x n n p x p i for x i R, x i > 0, i =,...,n, x i = n, i is false. i= i= Example 8 (Ukraine Mathematical Festival 007). Let a, b, c be positive real numbers and abc. Prove that (a). ( a + )( b + )( c + ) 7 a + b + c + 8. (b). 7(a +a +a+)(b +b +b+)(c +c +c+) 64(a +a+)(b +b+)(c +c+). Example 9 (Asian Pacific Mathematical Olympiad 007). Let x, y and 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). i=
www.batdangthuc.net 9 Example 0 (Brazilian Olympiad Revenge 007). Let a, b, c R with abc =. Prove that a +b +c + a + b + ( c + a + b + c + a + b + ) ( b 6+ c a + c b + a c + c a + c b + b ). c Example (India National Mathematical Olympiad 007). If x, y, z are positive real numbers, prove that (x + y + z) (yz + zx + xy) (y + yz + z )(z + zx + x )(x + xy + y ). Example (British National Mathematical Olympiad 007). Show that for all positive reals a, b, c, (a + b ) (a + b + c)(a + b c)(b + c a)(c + a b). Example (Korean National Mathematical Olympiad 007). 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 007. Example 4 (Hungary-Isarel National Mathematical Olympiad 007). Let a, b, c, d be real numbers, such that a,a + b 5,a + b + c 4,a + b + c + d 0. Prove that a + b + c + d 0.
0 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=6 http://www.batdangthuc.net/forum/showthread.php?t=6&page= http://www.batdangthuc.net/forum/showthread.php?t=6&page= http://www.batdangthuc.net/forum/showthread.php?t=6&page=4 http://www.batdangthuc.net/forum/showthread.php?t=6&page=5 http://www.batdangthuc.net/forum/showthread.php?t=6&page=6 For Further Reading, Please Review: UpComing Vietnam Inequality Forum's Magazine Secrets in Inequalities ( 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 Problem (, Iran National Mathematical Olympiad 007). 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 >. Solution (pi.4). 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 We have = c + x + y + c + y c + x x y y x ( =c x y + y ) + x + y x x y. ( c x y + y ) ( =c x x y + x y ) > 0. xy x + y x y >. Thus a + b a b + b + c b c + c + a c a >. Solution (, Mathlinks, posted by NguyenDungTN). Let a + b a b = x; b + c b c = y; a + c c a = z; Then xy + yz + xz =. By Cauchy-Schwarz Inequality We are done. (x + y + z) (xy + yz + zx) = x + y + z >. Problem (, Iran National Mathematical Olympiad 007). 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 + b + c + d + e T ( a + b + c + d + e)
www.batdangthuc.net Solution (NguyenDungTN). Let a = b =,c= d = e =, we find 0 6( + ) T. With this value of T, we will prove the inequality. Indeed, let a + b = c + d + e = X. By Cauchy-Schwarz Inequality a + b (a + b) = X c + d + e (c + d + e) = X a + b + c + d + e 5X 6 By Cauchy-Schwarz Inequality, we also have () a + b (a + b) = X c + d + e (c + d + e) =X ( a + b + c + d + e) ( + ) X () From () and (), we have a + b + c + d + e ( a + b + c + d + e) Equality holds for a = b = c = d = e. 0 6( + ). Problem (, Middle European Mathematical Olympiad 007). Let a, b, c, d nonnegative such that a + b + c + d =4. Prove that a bc + b cd + c da + d ab 4. Solution 4 (mathlinks, reposted by pi.4). Let {p,q,r,s} = {a, b, c, d} and p q r s. By rearrangement Inequality, we have a bc + b cd + c da + d 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 = 4 (p + s) (q + r) ( (p ) ) + q + r + s =4. 4 Equality holds for q = r =vp + s =. Easy to refer (a, b, c, d) =(,,, ), (,,, 0) or permutations.
www.batdangthuc.net Problem 4 ( 5- Revised by VanDHKH). Let a, b, c be three side-lengths of a triangle such that a + b + c =. Find the minimum of a + b + c + 4abc Solution 5. Let a = x + y, b = y + z,c = z + x, we have Consider x + y + z =. a + b + c + 4abc = (a + b + c )(a + b + c)+4abc = ((x + y) +(y + z) +(z + x) )(x + y + z)+4(x + y)(y + z)(z + x) = 4(x + y + z +x y +xy +y z +yz +z x +zx +5xyz) = 4((x + y + z) xyz) 6 4( 7 = (x + y + z) +( x+y+z ) xyz) 6 4( 7 (x + y + z) ) =. Solution 6 (, DDucLam). Using the familiar Inequality (equivalent to Schur) abc (b + c a)(c + a b)(a + b c) abc 4 (ab + bc + ca). Therefore P a + b + c + 6 (ab + bc + ca) 4 9 =(a + b + c) 9 (ab + bc + ca) 4 5 7 (a + b + c) =4+. Equality holds when a = b = c =. Solution 7 (, pi.4). With the conventional denotion in triangle, we have abc =4pRr, a + b + c =p 8Rr r. Therefore Moreover, Thus a + b + c + 4 abc = 9 r. p r r 6. a + b + c + 4 abc 4.
4 www.batdangthuc.net Problem 5 (7, China Northern Mathematical Olympiad 007). Let a, b, c be positive real numbers such that abc =. Prove that for any positive integer k. a k a + b + bk b + c + ck c + a. Solution 8 (Secrets In Inequalities, hungkhtn). We have a k a + b + bk b + c + ck c + a a k + b k + c k + ak b a + b + bk c b + c + ck a c + a By AM-GM Inequality, we have So, it remains to prove that a + b ab, b + c bc, c + a ca. a k b + b k c + c k a + ( a k + b k + c k ). This follows directly by AM-GM inequality, since a k + b k + c k a k b k c k = and (k )a k + b k (k )a k b (k )b k + c k (k )b k c (k )c k + a k (k )c k a 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 =. Prove that: a b + b c + c a (a + b + c ).
www.batdangthuc.net 5 Solution 9. By Cauchy-Schwarz Inequality: It remains to prove that So we are done! Solution 0 (, By Zaizai). a b + b c + c a (a + b + c ) a b + b c + c a. (a + b + c ) a b + b c + c a (a + b + c ) (a + b + c )(a + b + c) (a b + b c + c a) a + b + c + ab + bc + ca (a b + b c + c a) a(a b) + b(b c) + c(c a) 0. a b + b c + c a (a + b + c ) ( a ) b a + b (a + b + c ) (a + b + c) (a b) b (a b) +(b c) +(c a) (a b) ( b ) 0 This ends the solution, too. (a b) (a + c) b 0. Problem 7 (9, Romania Junior Balkan Team Selection Tests 007).. Let a, b, c be three positive reals such that a + b + + b + c + + c + a +. Show that a + b + c ab + bc + ca. Solution (Mathlinks, Reposted by NguyenDungTN). By Cauchy-Schwarz Inequality, we have (a + b + )(a + b + c ) (a + b + c).
6 www.batdangthuc.net Therefore or a + b + c + a + b (a + b + c), a + b + + b + c + + c + a + a + b + c +(a + b + c) (a + b + c) a + b + c +(a + b + c) (a + b + c) a + b + c ab + bc + ca. Solution (DDucLam). Assume that a + b + c = ab + bc + ca, we have to prove that a + b + + b + c + + c + a + a + b a + b + + b + c b + c + + c + a c + a + By Cauchy-Schwarz Inequality, We are done LHS (a + b + b + c + c + a) =. cyc (a + b)(a + b +) 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 (0, Romanian JBTST V 007). Let x, y, z be non-negative real numbers. Prove that x + y + z xyz + (x y)(y z)(z x). 4 Solution (vandhkh). We have x + y + z x + y + z xyz + (x y)(y z)(z x) 4 xyz (x y)(y z)(z x) 4 ((x y) +(y z) +(z x) (((x+y)+(y + z)+(z +x)) 9 (x y)(y z)(z x). Notice that and by AM-GM Inequality, x + y x y ; y + z y z ; z + x z x, ((x y) +(y z) +(z x) )( x y + y z + z x ) 9 (x y)(y z)(z x). So we are done. Equality holds for x = y = z.
www.batdangthuc.net 7 Solution 4 (Secrets In Inequalities, hungkhtn). The inequality is equivalent to (x + y + z) (x y) 9 (x y)(y z)(z x). By the entirely mixing variable method, it is enough to prove when z =0 This last inequality can be checked easily. x + y 9 xy(x y). 4 Problem 9 (, Yugoslavia National Olympiad 007). Let k be a given natural number. Prove that for any positive numbers x, y, z with the sum, the following inequality holds When does equality occur? x k+ x k+ + y k + z + y k+ k y k+ + z k + x + z k+ k z k+ + x k + y k 7. Solution 5 (NguyenDungTN). We can assume that x y z. By this assumption, easy to refer that x k+ x k+ + y k + z k y k+ y k+ + z k + x k z k+ z k+ + x k + y k ; z k+ + y k + x k y k+ + x k + z k x k+ + z k + y k ; and x k y k z k. By Chebyshev Inequality, we have x k+ x k+ + y k + z + y k+ k y k+ + z k + x + z k+ k z k+ + x k + y k x + y + z ( x k+ x k+ + y k + z k + y k+ y k+ + z k + x k + z k+ ) z k+ + x k + y k = ( x k+ x k+ + y k + z k + y k+ y k+ + z k + x k + z k+ ) cyc (xk+ + y k + z k ) z k+ + x k + y k cyc (xk+ + y k + z k ) ( ( )) = x k+ x k+ + y k + z k (x k+ + y k + z k ) cyc (xk+ + y k + z k ) cyc cyc (xk+ +y k+ +z k+ ). cyc (xk+ + y k + z k ) = xk+ + yk+ + zk+ x k+ + y k+ + z k+ +(x k + y k + z k )
8 www.batdangthuc.net Also by Chebyshev Inequality, Thus (x k+ + y k+ + z k+ ) x + y + z (x k + y k + z k )=x k + y k + z k. x k+ + y k+ + z k+ x k+ + y k+ + z k+ +(x k + y k + z k ) x k+ + y k+ + z k+ x k+ + y k+ + z k+ +6(x k+ + y k+ + z k+ ) = 7. So we are done. Equality holds for a = b = c =. Problem 0 (Macedonia Team Selection Test 007). Let a, b, c be positive real numbers. Prove that + ab + bc + ca 6 a + b + c. Solution 6 (VoDanh). The inequality is equivalent to a + b + c + (a + b + c) ab + bc + ca 6. By AM-GM Inequality, (a + b + c) (a + b + c) a + b + c + ab + bc + ca ab + bc + ca. It is obvious that (a + b + c) (ab + bc + ca), so we are done! Problem (4, Italian National Olympiad 007). a). For each n, find the maximum constant c n such that: a + + a + +...+ a n + c n, for all positive reals a,a,...,a n such that a a a n =. b). For each n, find the maximum constant d n such that a + + a + +...+ a n + d n, for all positive reals a,a,...,a n such that a a a n =.
www.batdangthuc.net 9 Solution 7 (Mathlinks, reposted by NguyenDungTN). a). Let a = ɛ n,a k = k, ɛ then let ɛ 0, we easily get c n. We will prove the inequality with this value of c n. Without loss of generality, assume that a a a n. Since a a, we have n k= a k + a + + a + = a + + a a + a a a + + a a + =. This ends the proof. b). Consider n =, it is easy to get d =. Indeed, let a = a, a = a. The inequality becomes a + + a a + (a +)+a(a +) (a + )(a +) (a ) 0. When n, similar to (a), we will show that d n =. Indeed, without loss of generality, we may assume that a a a n a a a. Let x = 9 a a a a a a a,y= 9 a, z = 9 a then a x, a y,a z,xyz =. Thus n k= a k + = k= a k + x x + + y y + + x x +yz + y y +xz + z z +xy z z + x x + y + z + y x + y + z + z x + y + z =. This ends the proof. Problem (5, France Team Selection Test 007).. Let a, b, c, d be positive reals such that a + b + c + d =. Prove that: 6(a + b + c + d ) a + b + c + d + 8.
0 www.batdangthuc.net Solution 8 (NguyenDungTN). By AM-GM Inequality a + 4 a 4 a + 4 a. Therefore 6(a + b + c + d )+ 6 9(a + b + c + d ) 4 5(a + b + c + d ) + 5 4 6 5(a + b + c + d = 5 8 8 Adding up two of them, we get Solution 9 (Zaizai). We known that 6(a + b + c + d ) a + b + c + d + 8 6a a + 5a 8 8 (4a ) (a +) 8 Adding up four similar inequalities, we are done! 0 Problem (6, Revised by NguyenDungTN). Suppose a, b and c are positive real numbers. Prove that a + b + c a + b + c ( bc a + ca b + ab ). c Solution 0. The left-hand inequality is just Cauchy-Schwarz Inequality. We will prove the right one. Let The inequality becomes bc a = x, ca b xy + yz + zx Squaring both sides, the inequality becomes = y, ab c = z. x + y + z. (x + y + z) (xy + yz + zx) (x y) +(y z) +(z x) 0, which is obviously true. Problem 4 (7, Vietnam Team Selection Test 007). Given a triangle ABC. Find the minimum of: (cos ( A )(cos ( B ) cos ( C )
www.batdangthuc.net Solution (pi.4). We have T = (cos ( A )(cos ( B ) (cos ( C ) = ( + cosa)( + cosb). ( + cosc) Let a = tan A ; b = tan B ; c = tan C. We have ab + bc + ca =.So T = ( + a ) ( + b )( + c ) = (+b )(+c ) +a By Iran96 Inequality, we have = (ab+bc+ca+b )(ab+bc+ca+c ) (ab+bc+ca+a ) = (a+b)(c+b)(a+c)(b+c) (b+a)(b+c) = (b + c) (b + c) + (c + a) + (a + b) 9 4(ab + bc + ca). Thus F 9 4 Equality holds when ABC is equilateral. Problem 5 (8, Greece National Olympiad 007).. Let a, b, c be sides of a triangle, show that (b + c a) 4 (c + a b)4 (b + c a)4 + + ab + bc + ca. a(a + b c) b(b + c a) a(c + a b) Solution (NguyenDungTN). Since a, b, c are three sides of a triangle, we can substitute The inequality becomes a = y + z,b = z + x, c = x + y. 8x 4 (x + y)y + 8y4 (y + z)z + 8z4 (z + x)x x + y + z +(xy + yz + zx). By Cauchy-Schwarz Inequality, we have 8x 4 (x + y)y + 8y4 (y + z)z + 8z4 (z + x)x 8(x + y + z ) x + y + z + xy + yz + zx.
www.batdangthuc.net We will prove that 8(x + y + z ) x + y + z + xy + yz + zx x + y + z +(xy + yz + zx) 8(x + y + z ) (x + y + z + xy + yz + zx)(x + y + z +(xy + yz + zx)) 8 x 4 +6 x y x 4 + x y + +4 x (y + z)+xyz(x + y + z)+ x y +6xyz(x + y + z) 7 x 4 + x y 4 x (y + z)+0xyz(x + y + z). By AM-GM and Schur Inequality Adding up two inequalities, we are done! x 4 + x y 4xyz(x + y + z); ( ) 4 x 4 + xyz(x + y + z) 4 x (y + z) Solution (, DDucLam). By AM-GM Inequality, we have (b + c a) 4 a(a + b c) + a(a + b c) (b + c a). Construct two similar inequalities, then adding up, we have We are done! (b + c a) 4 (c + a b)4 (b + c a)4 + + a(a + b c) b(b + c a) a(c + a b) [(a + b + c ) (ab + bc + ca)] (a + b + c ) =5(a + b + c ) 4(ab + bc + ca) ab + bc + ca. Problem 6 (0, Poland Second Round 007).. Let a, b, c, d be positive real numbers satisfying the following condition a + b + c + d =4Prove that: a + b + b + c + c + d + d + a (a + b + c + d) 4.
www.batdangthuc.net Solution 4 (Mathlinks, reposted by NguyenDungTN). First, we show that which is equivalent to a + b a + b a + b, (a b) 4 (a + ab + b ) 0. Therefore, we refer that a + b + b + c + c + d + d + a a + b a + b + b + c b + c + c + d c + d + d + a d + a It remains to prove that a + b a + b + b + c b + c + c + d c + d + d + a (a + b + c + d) 4. d + a However, a + b a + b a + b = ab a + b = a +, b So, due to Cauchy-Schwarz Inequality, we get ( a + b (a + b + c + d) a + b + b + c b + c + c + d c + d + d + a ) d + a This ends the proof. = a + b 4 ( a + b + c + d ) = 8 =4 Problem 7 (, Turkey Team Selection Tests 007).. Let a, b, c be positive reals such that their sum is. Prove that: ab +c +c + bc +a +a + ac +b +b ab + bc + ac. Solution 5 (NguyenDungTN). First, we will prove that Indeed, this is equivalent to ab + ac + bc ab +c +c ab ab + ac + bc. a b + b c + c a +abc(a + b + c) a b +abc +abc, which is always true since abc(a + b + c) =abc and due to AM-GM Inequality a c + b c abc.
4 www.batdangthuc.net Similarly, we have ab + ac + bc bc +a +a ab + ac + bc ac +b +b Adding up three inequalities, we are done! bc ab + ac + bc. ca ab + ac + bc. Problem 8 (, Moldova National Mathematical Olympiad 007). Real numbers a,a,,a n satisfy a i i, for all i =,n. Prove the inequality ( (a +) a + ) ( a n + ) n n (n + )! ( + a +a + + na n ). Solution 6 (NguyenDungTN). This inequality is equivalent to (a + )(a +).. (na n +) n n + ( + a +a +... + na n ). It is clearly true when n =. Assume that it si true for n = k, we have to prove it for n = k +. Indeed, (a +)(a +).. (ka k +)((k+)a k+ +) k k + (+a +a +...+ka k )((k+)a k+ +) Let a =(k +)a k+ s = a +a +... + ka k s k. We need to show that k k+ ( + s)( + a) ( + s + a) k + k + (as k)+k(a )(s ) 0. Since a k, the above one is true for n = k +. The proof ends! Equality holds for a i =,i=,n. i Solution 7 (NguyenDungTN). The inequality is equivalent to ( )( ) ( ) +a +a +nan... +a +a +... + na n. n + Let x i = iai 0, it becomes ( + x )( + x )...( + x n ) + n + (x + x +... + x n ). But ( + x )( + x )...( + x n ) +x + x +... + x n + n + (x + x +... + x n ). So we have the desired result.
www.batdangthuc.net 5 Problem 9 (, Moldova Team Selection Test 007). Let a,a,..., a n [0, ]. Denote S = a + a +... + a n. Prove that a a n ++S a + n ++S a + + n ++S a n a n. Solution 8 (NguyenDungTN). By AM-GM Inequality, we have S a +(n )=(a +)+(a +)+ +(a n +) (a + a + + a n ). Thus a n ++S a a ( + a + a + + a n ) a (a + a + + a n ). Similar for a,a,..., a n, we have a a n ++S a + n ++S a + + n ++S a n a + a + + a n = a + a + + a n. The equality holds for a = a =... = a n =. a n Problem 0 (4, Peru Team Selection Test 007). Let a, b, c be positive real numbers, such that a + b + c a + b + c. Prove that: a + b + c a + b + c + abc. Solution 9 (NguyenDungTN). By Cauchy-Schwarz Inequality, we have a + b + c a + b + c 9 a + b + c a + b + c. Our inequality is equivalent to ( (a + b + c) + ab + bc + ). ca By AM-GM Inequality ( ab + bc + ) ( ca a + b + ) (a + b + c) c
6 www.batdangthuc.net So it is enough to prove that (a + b + c) + (a + b + c) (a + b + c) 9. This inequality is true due to a + b + c. Solution 0 (, DDucLam). We have a + b + c (a + b + c)+ ( a + b + c ) (a + b + c)+ a + b + c. We only need to prove that a + b + c abc, but this inequality is always true since (a + b + c) ( a + b + ) ( c ab + bc + ) = (a + b + c). ca abc Problem (5, Revised by NguyenDungTN). Let a, b and c be sides of a triangle. Prove that b + c a c + a b a + b c + +. b + c a c + a b a + b c Solution. Let x = b + c a, y = c + a b, z = a + b c, then b + c a = x (x y)(x z) By AM-GM inequality, we have b + c a (x y)(x z) (x y)(x z) = b + c a x 4x We will prove that x (x y)(x z)+y (y z)(y z)+z (z x)(z y) 0. But this immediately follows the general Schur inequality, with the assumption that We are done! x y z x y z.
www.batdangthuc.net 7 Problem (6, Romania Team Selection Tests 007). If a,a,...,a n 0 are such that a + + a n =, find the maximum value of the product ( a ) ( a n ). Solution (hungkhtn, reposted by NguyenDungTN). We use contradiction method. Assume that x,x,..., x n [0, ] such that x x...x n =( ). We will prove f(x,x,..., x n )=( x ) +( x ) +... +( x n ) () Indeed, first, we prove that: Lemma: If x, y [0, ],x+ y + xy then Proof. Notice that ( x) +( y) ( xy). ( x) +( y) ( xy) =(x + y ) x y =(x )(y )(x + y + xy ) 0. The lemma is asserted. Return to the problem, let k =. Assume that x x... x n, then x x x k x x k 4/, thus x + x + x + x k / + k 4/ =.07. Similarly, we have f(x,x,..., x n ) f(x,x x,,x 4,..., x n ) f(x,x x x 4,,,x 5,..., x n )... f(x,x x...x n,,,..., ), From this, easy to get the final result. Problem (8, Ukraine Mathematic Festival 007). Let a, b, c > 0 và abc. Prove that ( a). a + )( b + )( c + ) 7 a + b + c + 8. b). 7(a + a + a + )(b + b + b + )(c + c + c +) 64(a + a + )(b + b + )(c + c +).
8 www.batdangthuc.net Solution (pi.4). Consider the case abc =. Let a = x y, b = y z, c = z. The inequality x becomes x y + x y + x y + 7 8 or 8(x + xy + y )(y + yz + z )(x + zx + z ) 7xyz(x + y)(y + z)(z + x) () We have since (x + xy + y ) xy(x + y), (x + xy + y ) (x +xy + y ) xy(x + y). Write two similar inequalities, then multiply all of them, we get () immediately. If abc >, we let a = ka ; b = kb ; c = kc ; with k = abc. We have k>and a b c =. Then a + a + a + a +. a + a + Since the inequality is proved for a,b,c, this is true for a, b, c immediately. b). By AM-GM inequality Therefore a + a (a +) (a + a +). (a + a + a + ) = (a + )(a +) 6 a. (a + a +)=4 a(a + a +). Constructing similar inequalities, then multiply all of them, we get 7(a +a +a+)(b +b +b+)(c +c +c+) 64(a +a+)(b +b+)(c +c+). Solution 4 (, NguyenDungTN). By AM-GM inequality Adding up two inequalities, we get Similar for b, c, and finally we have ( a + a + a + 4 )( b + b + Equality holds for a = b = c =. + a + ; a 4 + 4 a ; a + a + a. )( c + ) c + 7 7 abc 8 8.
www.batdangthuc.net 9 Problem 4 (9, Asian Pacific Mathematical Olympiad 007). Let x, y and 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). Solution 5 (NguyenDungTN). We have the transformation cyc x + yz x (y + z) = (x y)(x z) cyc x (y + z) + cyc Moreover, by Cauchy-Schwarz Inequality y + z. y + z cyc cyc y + z =. So it is enough to prove that cyc (x y)(x z) x (y + z) 0 Without loss of generality, assume that x y z, then x (y + z) y (z + x) z (x + y). Using the general Schur Inequality, we have the desired result. Problem 5 ( 0, Brazilian Olympiad Revenge 007). Let a, b, c R with abc =. Prove that a +b +c + a + b + ( c + a + b + c + a + b + ) ( a 6+ c b + b c + c a + b a + c b + a ). c Solution 6 (NguyenDungTN). Since abc =, we have ( a + b + c + a + b + ) = a + b + c +(ab + bc + ca) =(a + b + c). c a + b + c +(a + b + c) =a b + b c + c a +abc(a + b + c) =(ab + bc + ca). ( a b + b c + c a + b a + c b + a ) c + (ab(a + b)+bc(b + c)+ca(c + a)+abc) = abc
0 www.batdangthuc.net =(a + b + c)(ab + bc + ca). By AM-GM Inequality, (a + b + c) +(ab + bc + ca) (a + b + c)(ab + bc + ca) (a + b + c)(ab + bc + ca). This ends the proof. The equality holds for a = b = c =. Problem 6 (, Revised by NguyenDungTN). If x, y, z are positive real numbers, prove that (x + y + z) (yz + zx + xy) (y + yz + z )(z + zx + x )(x + xy + y ). Solution 7. Using the inequality We have 4(a + b + ab) (a + b) a, b( (a b) 0) (y + yz + z )(z + zx + x )(x + xy + y ) 4 (x + y) (y + z) (z + x). By AM-GM inequality, we get 9(x + y)(y + z)(z + x) =9(xy(x + y)+yz(y + z)+zx(z + x)+xyz) =8(xy(x + y)+yz(y + z)+zx(z + x)+xyz)+xy(x + y)+yz(y + z)+zx(z + x) 6xyz 8(x + y + z)(xy + yz + zx). So we have the desired result. Problem 7 (, British National Mathematical Olympiad 007). Show that for all positive reals a, b, c (a + b ) (a + b + c)(a + b c)(b + c a)(c + a b). Solution 8 (NguyenDungTN). Using the familiar inequality we have xy (x + y) x, y R, 4 (a + b + c)(a + b c)(b + c a)(c + a b) = ( (a + b) c )( c (a b) ) ( (a + b) c + c (a b) ) =(a + b ). 4 Equality holds when (a + b) c = c (a b) c = a + b.
www.batdangthuc.net Problem 8 (4, Mathlinks, Revised by VanDHKH). Let a, b, c, d be real numbers such that a,a +b 5,a +b +c 4,a +b +c +d 0 Prove that a+b+c+d 0. Solution 9. By hypothesis, we have By Cauchy-Schwarz Inequality, we have 00 = (a +6b +4c +d ) a +6b +4c +d 0. Therefore a + b + c + d a + b + c + d 0. ( + 6 + 4 + ) (a + b + c + d)