Đang chuẩn bị nút TẢI XUỐNG, xin hãy chờ
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Tham khảo tài liệu 'tuyển tập đề thi toán quốc tế p4', khoa học tự nhiên, toán học phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 4.36 Shortlisted Problems 1995 601 28. Let F x f x 95 for x 1. Writing k for m 95 the given condition becomes F k F n F k n k 96 n 1. 1 Thus for x z 96 and an arbitrary y we have F x y z F x y F z F x F F y z F x F y z and consequently F x y F x F y whenever x 96. Moreover since then F x y F 96 F x y 96 F x F y 96 F x F y F 96 for any x y we obtain F x y F x F y x y G N. 2 It follows by induction that F n nc for all n where F 1 c. Equation 1 becomes ck c2n ck n and yields c 1. Hence F n n and f n n 95 for all n. Finally 1 1 f k 96 97 114 1995. Second solution. First we show that f n 95 for all n. If to the contrary f n 95 we have f m n f m 95 f n so by induction f m kn f m k 95 f n kn for all k which is impossible. Now for m 95 we have f m f n 95 n f m and again by induction f m k f n 95 kn f m for all m n k. It follows that with n fixed f m k f n 95 n V m lim _ k- m k f n 95 f n 95 hence f s lim ------- s s n f n - 95. Hence f w _95 does not depend on n i.e. f n cn 95 for some constant c. It is easily checked that only c 1 is possible. 602 4 Solutions 4.37 Solutions to the Shortlisted Problems of IMO 1996 1. We have a5 b5 a2b2 a b a3 b3 a2 b2 0 i.e. a5 b5 a2b2 a b . Hence ab ab abc2 c a5 b5 ab a2b2 a b ab a2b2c2 a b abc2 a b c Now the left side of the inequality to be proved does not exceed a b c a b c a b c 1. Equality holds if and only if a b c. 2. Clearly ai 0 and if p ai we must have an 0 an ai and p an. But then for sufficiently large odd k an an k n 1 a1 k so that ak an n 1 a1 k an k 0 a contradiction. Hence p ai. Now let x a1. From a1 an 0 we deduce 2 2 x j n 1 x n 1 so by the AM-GM inequality x n-1 x a2 x an x - xn-1 xn-2ai an-1. 1 n 1 i The last inequality holds because n- n 1 r for all r 0. Multiplying 1 by x ai yields the desired inequality. 3. Since a1 2 it can be written as a1 b b-1 for some b 0. Furthermore a2 2 b2 b-2 and hence a2 b2 b-2 b b-1 . We prove that an b b-1 b2 b-2 b4 b-4 b b by induction. Indeed Onii p-V 2 A2 b
