Tuyển tập đề thi vô địch bất đẳng thức thế giới P4

Tuyển tập đề thi vô địch bất đẳng thức thế giới P4 , 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 91 n is trivial because Xi-i 2 n l xi Xj i 2 n 1 x for all i. This shows fe i that mn ----------Taking xi xl the expression becomes 2 n 1 1 n 2 x xn 1 x a -1 2 n 1 1 2 n l o x2 1 2 n l xn-1 xn 2 and taking the limit when x approaches 0 we find that mn 1 n Ö7----Tv 2 n 1 ------r and thus 2 n - 1 Now we will prove that Mn Xi X2 xn 0 we have Of course it suffices to 2 prove that for any But it is clear that 2x Xi-i 2 n - l xi xi 1 1 - 2 2x x 2 n - l xi xi 1 2 n l xt . 1 1 n - 1 ------------- Xi Taking n di we have to prove that if JJ ai i l 1 then E i Mn Mn --------- 1. But this has already been proved in the problem 84. Thus n 1 ai 1 - and because for Xi x- which solves the problem. xn we have equality we deduce that 96. Vasile Cirtoaje If x y z are positive real numbers then 1 1 19 x2 xy y2 y2 yz z2 z2 zx z2 x y z 2 Gazeta Matematicä Solution Considering the relation x2 xy y2 x y z 2 xy yz zx x y z z 92 Solutions x y z 2 x2 xy y2 or x y x2 xy - we get 1 xy yz zx z x y z 2 x y z 02 1 y2 1 ab be cd c where a -------- b ------- c ----------. The inequality can be rewritten x y z x y z x y z as 1 1 1 I I 9 1-d-c-----------------1-d-b-1-d-a where a b c are positive reals with a b c 1 and d ab be ca. After making some computations the inequality becomes 9d3 - 6d2 - 3d 1 9abc 0 or d 3d - l 2 1 - 4d 9abc 0. which is Schur s Inequality. 97. Vasile Cirtoaje For any a b c d 0 prove that 2 a3 1 3 l c3 l d3 1 1 abcd l a2 l 2 1 c2 l d2 . Gazeta Matematica Solution Using Huygens Inequality JJ 1 a4 1 abed 4 we notice that it is enough to show that that 24 J a3 l 4 JJ 1 a4 l a2 4. Of course it suffices to prove that 2 a3 l 4 a4 l a2 I 4 for any positive reala. But a2 l 4 a l 2 a3 l 2 and we are left with the inequality 2 a3 1 2 a l 2 a4 l 44 2 a2 - a l 2 a4 1 44 a - l 4 0 which follows. 98. Prove that for any real numbers a b c a b 4 b c 4 c a 4 a4 b4 c4 . Vietnam TST 1996 Old and New Inequalities 93 Solution Let us make the substitution a b 2z b c 2x c a .