Tuyển tập báo cáo các nghiên cứu khoa học quốc tế ngành hóa học dành cho các bạn yêu hóa học tham khảo đề tài: Research Article On Shafer-Fink-Type Inequality Ling Zhu | Hindawi Publishing Corporation Journal ofInequalities and Applications Volume 2007 Article ID 67430 4 pages doi 2007 67430 Research Article On Shafer-Fink-Type Inequality Ling Zhu Received 5 January 2007 Accepted 14 April 2007 Recommended by Laszlo I. Losonczi A new simple proof of Shafer-Fink-type inequality proposed by Malesevic is given. Copyright 2007 Ling Zhu. This is an open access article distributed under the Creative Commons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited. 1. Introduction R. E. Shafer see Mitrinovic 1 page 247 gives us a result as follows. Theorem . Let x 0. Then arcsin x 6 V1 x - V1 - x 4 V1 x V1 -x 3x 2 V1 - x2 The theorem is generalized by Fink 2 as follows. Theorem . Let 0 x 1. Then 3x 2 V1 - x2 arcsin x nx 2 V1 - x2 Furthermore 3 and n are the best constants in . From the theorems above it is possible to improve the upper bound of inverse sine and deduce the following property see 3 4 . Theorem . Let 0 x 1. Then 3x ---- 2 V1 - x2 6 V1 x - V1 - x 4 V1 x ựĩ -x arcsin x n V2 1 2 V1 x - yJ1 - x nx 4 41 x - 1 x 2 - I x2 2 Journal of Inequalities and Applications Furthermore 3 and n 6 and n ff2 1 2 are the best constants in . MaleSeviC 5 6 obtained the following theorem by using A-method and computer separately. Theorem . For all x e 0 1 the following inequality is valid arcsinx n 2 V2 n 2V2 V1 x V1 x V2 n 4 n 2 V2 V1 x V1 x Recently Malesevic 7 obtains the inequality by using further method on computer. In this paper we show a new simple proof of inequality and obtain the following further result. Theorem . Let 0 x 1. Then 6 V1 x V1 x . n 2 V2 n 2V2 V1 x V1 x . arcsinx ------ - . . 4 V1 x a 1 x 2 n 4 ft 2 V2 ặ 1 x ặ 1 x Furthermore 4 and ff2 4 n n 2a 2 are the best constants in . 2. One lemma L Hospital s rule for monotonicity Lemma 8-10 . Let f g a b R be two continuous functions .
