tailieunhanh - Tuyển tập đề thi vô địch bất đẳng thức thế giới P3

Tuyển tập đề thi vô địch bất đẳng thức thế giới P3 , tài liệu tham khảo, tài liệu gồm các bài toán bất đẳng thức cực khó, các bạn có đào sâu kiến thức toán về mảng này, Tai liệu được viết bằng tiêng anh. Chúc các bạn học tốt. | Old and New Inequalities 61 First solution Using the identity a c c a a b c a be ca 1 we reduce the problem to the following one 3 ab be ca 4-------- 4. a b c Now we can apply the AM-GM Inequality in the following form ab be ca 4--------- a b c J ab bc ca 3 y 9 a -I- b 4- c And so it is enough to prove that ab be ca 3 9 a b c . But this is easy because we clearly have ab be ca 3 and ab be 4- ca 2 3abc a b c 3 a b c . Second solution We will use the fact that a b b c c a 2 1 So it is enough to prove that - ab be ca 4--- 9 a 4- b - Inequality we can write g - a b c ab be ca . 1. Using the AM-GM - ab bc ca 4----- --- 9V a b c A ab bc ca 2 y 81 a 4 b 4 c because ab bc ca 2 3abc a b c 3 a b c . 57. Prove that for any a b c 0 a2 b2 c2 a b c b c a c a b abc ab be ca . Solution Clearly if one of the factors in the left-hand side is negative we are done. So we may assume that a c are the side lenghts of a triangle ABC. With the usual notations in a triangle the inequality becomes IGA 2 a2 b2 c2 ---------- abc ab bc ca O a b c ab bc ca R2 abc a2 b2 c2 . a b c But this follows from the fact that a 4- b c ab bc ca 9abc and 0 OH2 9R2 -a2 - b2 - c2. 62 Solutions 58. Let a b c 0. Prove that 1 1 1 a b c 3 iz c 7 7 a b c b c a o a 1 l c 1 O 1 abc Kvant 1988 Solution The inequality is equivalent to the following one __ i n ab a Yr 3 --------t- a b abc 1 or u c a i- 2c 2 a 3a But this follows from the inequalities 9 b . 9 c a a be - 2ab b ca 4 2bc c ab 4 2ca cab and a2c 2a b2a - 2b c2b 7 2c. c a b 59. Gabriel Dospinescu Prove that for any positive real numbers Xj x xn with product 1 we have the inequality n n n n n f D e . e- 1 i l 1 1 V Solution Using the AM-GM Inequality we deduce that Old and New Inequalities 63 Of course this is true for any other variable so we can add all these inequalities to obtain that which is the desired inequality. 60. Let a b c d 0 such that a b c 1. Prove that a3 b3 c3 abed min t t d 1 4 9 27 J Kvant 1993 the inequality d Solution .