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Inequalities Among Complementary Means of Heron Mean and Classical Means

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Inequalities Among Complementary Means of Heron Mean and Classical Means Advances and Applications in Mathematical Sciences Volume 20, Issue 7, May 2021, Pages 1249-1258 © 2021 Mili Publications INEQUALITIES AMONG COMPLEMENTARY MEANS OF HERON MEAN AND CLASSICAL MEANS AMBIKA M. HEJIB, K. M. NAGARAJA, R. SAMPATHKUMAR and B. S. VENKATARAMANA Department of Mathematics R N S Institute of Technology Uttarahalli-Kengeri Main Road R Rnagar post, Bengaluru-98, Karnataka, India E-mail: [email protected] [email protected] Department of Mathematics J.S.S. Academy of Technical Education Uttarahalli-Kengeri Main Road Bengaluru-60, Karnataka, India E-mail: [email protected] Department of Mathematics K S Institute of Technology Kannakapura Main Road Bengaluru - 560 109, Karnataka, India E-mail: [email protected] Abstract In this paper, the complementary means of arithmetic, geometric, harmonic and contra harmonic with respect to Heron mean are defined and verified them as means. Further, inequalities among them and classical means are established. I. Introduction In Pythagorean school ten Greek means are defined based on proportion, the following are the familiar means in literature and are given as follows: 2010 Mathematics Subject Classification: Primary 26D10, secondary 26D15. Keywords: Complementary mean, Heron mean, Classical means. Received September 7, 2020; Accepted February 18, 2021 1250 HEJIB, NAGARAJA, SAMPATHKUMAR and VENKATARAMANA u v For two real numbers u and v which are positive, Au , v 1 1 ; 1 1 1 1 2 2 2 2u1v1 u1 v1 Gu1, v1 u1v1 ; Hu1, v1 and Cu1, v1 . These are u1 v1 u1 v1 called Arithmetic, Geometric, Harmonic and Contra harmonic mean respectively. The Hand book of Means and their Inequalities, by Bullen [1], gave the tremendous work on Mathematical means and the corresponding inequalities involving huge number of means. The authors in [2, 3, 4] discussed about the relations between the well-known means and series. The generalization of the means is discussed in [5, 6, 18, 19]. Relevant to this paper the authors in [7- 17] established the good number of inequalities, double inequalities, introduced new means, studied homogeneous functions as an application inequalities are obtained. In 1958, C. Gini introduced Complementary means and G. Toader in 1991 proposed a generalization of complementariness and inversion ([20], [21]). This work motivates us to introduce the following definitions and to establish some double inequalities in this paper. Definition 1.1 [1]. For two real numbers u1, v1 0, the Heron mean is u u v v given by H 1 1 1 1 . e 3 Definition 1.2 [20, 21]. A mean N is called complementary to M with respect to P (or P-complementary to M) if it verifies PM, N P, it is denoted by M P N. (1) Complementary of Arithmetic mean with respect to Heron mean is 1 denoted by AHe and is given by AHe 3A 2G 4AG 5A2 2 1 AHe 3u 3v 4 u v u v 5u 5v 8 u v . 4 1 1 1 1 1 1 1 1 1 1 (2) Complementary of Geometric mean with respect to Heron mean is 1 denoted by GHe and is given by GHe 4A G 8AG G2 2 Advances and Applications in Mathematical Sciences, Volume 20, Issue 7, May 2021 INEQUALITIES AMONG COMPLEMENTARY MEANS OF 1251 1 GHe 2u 2v u v 4u v u v u v . 2 1 1 1 1 1 1 1 1 1 1 (3) Complementary of Harmonic mean with respect to Heron mean is denoted by H He and is given by 1 H He 4A2 G2 2AG G 8A2 3G2 4AG 2A 2 He u1v12u1 v1 3u1v1 2 u1v1 u1 v1 H u1 v1 u1v1 . u1 v1 (4) Complementary of Contra harmonic mean with respect to Heron mean is denoted by 1 1 CHe 2A2 G2 2AG 2A2 G2 2A2 3G2 4AG 2A 2A u v u v u2 v2 CHe 1 1 1 1 1 1 2u1 v1 u2 v2 2 8u v u2 v2 4 u v u2 v2 u v 1 1 1 1 1 1 1 1 1 1 1 1 . 2u1 v1 II. Main Results In this section, it can be verified that above definitions are means. Inequalities involving these and classical means are established. Theorem 2.1. For u1, v1 0, complementary of geometric mean and arithmetic mean with respect to Heron mean satisfy the inequality AHe GHe . Proof. Consider, 1 GHe AHe A G 4AG 5A2 8AG G2 2 G A A A2 A 1 4 5 8 1. 2 G G G G Advances and Applications in Mathematical Sciences, Volume 20, Issue 7, May 2021 1252 HEJIB, NAGARAJA, SAMPATHKUMAR and VENKATARAMANA A G Let k 1, k 1 4k 5k2 8k 1. G 2 Put k 1 z 0 k z 1, 4k 5k2 5z2 14z 9, 8k 1 8z 9 G GHe AHe f z 2 where f z z 5z2 14z 9 8z 9. To verify f z 0. Clearly, z2 6z 9 2z 5z2 14z 9 0 z2 5z2 14z 9 2z 5z2 14z 9 8z 9. Therefore, z 5z2 14z 9 8z 9. Hence f z 0, GHe AHe 0. Hence proved that AHe GHe. Theorem 2.2. For u1, v1 0, complementary of harmonic mean and geometric mean with respect to Heron mean satisfy the inequality H He GHe. Proof. Consider, 1 H He GHe 4A2 G2 2AG G 8A2 3G2 4AG 2A 1 4A G 8AG G2 2 1 AG G2 A 8AG G2 G 8A2 3G2 4AG 2A G2 A A A A2 A 1 8 1 8 3 4 . 2A G G G 2 G G A Let k 1 G G2 k 1 k 8k 1 8k2 3 4k. 2A Advances and Applications in Mathematical Sciences, Volume 20, Issue 7, May 2021 INEQUALITIES AMONG COMPLEMENTARY MEANS OF 1253 Put k 1 z k z 1, 8k2 4k 3 8z2 20z 9, 8k 1 8z G2 9 z z 1 8z 9 8z2 20z 9. 2A To prove z z 1 8z 9 8z2 20z 9 0. Clearly, 8z3 18z2 6z 2zz 1 8z 9 0. Therefore, z z 1 8z 9 8z2 20z 9, thus H He GHe 0. Hence proved that H He GHe. Theorem 2.3. For u1, v1 0, complementary of arithmetic mean and contra harmonic mean with respect to Heron mean satisfy the inequality AHe CHe. Proof. Consider 1 AHe CHe A2 G2 2A2 G2 2A2 3G2 4AG 2A 1 A 4AG 5A2 2A Α dividing by G2 and let k 1, then G G2 AHe CHe k2 1 2k2 1 2k2 3 4k k 4k 5k2 put 2A k 1 z or k z 1, then G2 z2 2z 2z2 4z 1 2z2 8z 9 z 1 5z2 14z 9 2A G2 f z, 2A To verify f z 0. Clearly, 4z3 14z2 12z 2zz 2 2z2 4z 1 2z2 8z 9 0 Advances and Applications in Mathematical Sciences, Volume 20, Issue 7, May 2021 1254 HEJIB, NAGARAJA, SAMPATHKUMAR and VENKATARAMANA 5z4 28z3 56z2 44z 9 2z2 4z 2z2 4z 1 2z2 8z 9 5z4 24z3 42z2 32z 9 z2 2z 2z2 4z 1 2z2 8z 9 z 1 5z2 14z 9. Therefore, f z 0. Hence AHe CHe 0. Hence proved that CHe AHe. Theorem 2.4. For u1, v1 0, complementary of arithmetic mean with respect to Heron mean and arithmetic mean satisfy the inequality AHe A. Proof. Consider 1 A AHe 2u 2v 3u 3v 4 v v u v 5u 5v 8 u v 4 1 1 1 1 1 1 1 1 1 1 1 1 1 u v 5u 5v 8 u v u v 4 u v 4 1 1 1 1 1 1 1 1 1 1 To verify whether, u1 v1 5u1 5v1 8 u1v1 u1 v1 4 u1v1 0. 2 2 2 Clearly v1 u1 0, u1 v1 16u1v1 8u1 v1 u1v1 5u1 v1 8u1 v1 u1v1 u1 v1 5u1 5v1 8 u1v1 u1 v1 4 u1v1 . Hence proved that AHe A. Theorem 2.5. For u1, v1 0, complementary of geometric mean with respect to Heron mean and arithmetic mean satisfy the inequality GHe A. 1 Proof. Consider GHe A u v u v 4u v u v u v . 2 1 1 1 1 1 1 1 1 1 1 To verify u1 v1 u1v1 4u1 v1 u1v1 u1v1 0. Clearly u1 v1 2 u1 v1 0 2 u1 b 2 u1v1 u1 v1. Advances and Applications in Mathematical Sciences, Volume 20, Issue 7, May 2021 INEQUALITIES AMONG COMPLEMENTARY MEANS OF 1255 Therefore u1 v1 u1v1 4u1 v1 u1v1 u1v1 . Hence proved that GHe A. Theorem 2.6. For u1, v1 0, complementary of contraharmonic mean with respect to Heron mean and geometric mean satisfy the inequality CHe G. Proof. Consider u2 v2 4u v CHe G 1 1 1 1 2u1 v1 u2 v2 2 8u v u2 v2 4 u v u2 v2 u v 1 1 1 1 1 1 1 1 1 1 1 1 . 2u1 v1 Clearly 2 2 2 2 2 2 2 2 2 u1 v1 4u1v1 u1 v1 8u1v1u1 v1 4 u1v1 u1 v1 u1 v1 . 2 2 2 2 2 2 2 2 2 2 As u1 v1 4u1v1 u1 v1 8u1v1u1 v1 4 u1v1 u1 v1 u1 v1. 2 2 2 2 4 2 Simplifying gives 4u1v1 u1v1 u1 v1 u1 v1 which is 4G 4GA C, which is always true as G A and G C. Therefore CHe G 0. Hence proved that CHe G. Theorem 2.7. For u1, v1 0, complementary of contraharmonic mean with respect to Heron mean and harmonic mean satisfy the inequality CHe H.
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