ITALIAN JOURNAL OF PURE AND APPLIED MATHEMATICS N ( ) 528

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ITALIAN JOURNAL OF PURE AND APPLIED MATHEMATICS N. 40 2018 528 534 528 SOME OPERATOR α-geometric MEAN INEQUALITIES Jianing Xue Oxbridge College Kuning University of Science and Technology Kuning, Yunnan 650106 P. R. China xuejianing104@163.co Abstract. In this paper, we refine an operator α-geoetric ean inequality as follows: let Φ be a positive unital linear ap and let A and B be positive operators. If 0 < A < M B M or 0 < B < M A M, then for each α [0, 1], Φ A α Φ B 2 2 K h K 2r h Φ 2 A α B, where K h = h+12, K +1 2 h = h and r = in {α, 1 α}. Keywords: operator inequalities, α-geoetric ean, positive linear aps. 1. Introduction Throughout this paper, is the operator nor and I denotes the identity operator. A 0 A > 0 iplies that A is positive strictly positive operator. Φ is a positive unital linear ap if ΦA 0 with A 0 and ΦI = I. For A, B > 0 and α [0, 1], the α-geoetric ean A α B is defined by A α B = A 1 2 A 1 2 BA 1 α 1 2 A 2, when α = 1 2, A 1 2 B = A B is said to be the geoetric ean. Seo [1] gave the following α-geoetric ean inequality: let Φ be a positive unital linear ap. If 0 < 1 A, B M 1 for soe nubers 1 M 1. Then for α [0, 1], ΦA α ΦB K, M, α 1 ΦA α B, where = 1 M 1, M = M 1 1 and the generalized Kantorovich constant K, M, α [2, Definition 2.2] is defined by K, M, α = for any real nuber α R. M α M α α 1 M α α α 1 M α M α M α α

SOME OPERATOR α-geometric MEAN INEQUALITIES 529 Fu [3] squared operator α-geoetric ean inequality: let Φ be a positive unital linear ap. If 0 < A, B M for soe nubers M. Then for α [0, 1] 1.1 Φ A α Φ B 2 K 2 h Φ 2 A α B, where K h = h+12 with h = M is the Kantorovich constant. A great nuber of results on operator inequalities have been given in the literature, for exaple, see [4-8] and the references therein. In this paper, we will get a stronger result than 1.1 and apply it to obtain an operator α-geoetric ean inequality to the power of 2p p 2. 2. Main results In this section, the ain results of this paper will be given. To do this, the following leas are necessary. Lea 1 [9]. Let A, B > 0. Then 2.1 AB A + B2. Lea 2 [10]. Let A > 0. Then for every positive unital linear ap Φ, 2.2 ΦA 1 Φ 1 A. Lea 3 [11]. Let A, B > 0. Then for 1 r <, 2.3 A r + B r A + B r. Lea 4 [12]. Let 0 < A < M B M or 0 < B < M A M. Then for each α [0, 1], 2.4 K r h A α B A α B, where K h = h +1 2, h = M and r = in {α, 1 α}. Lea 5. Let 0 < A < M B M or 0 < B < M A M. Then for each α [0, 1], 2.5 K r h A 1 α B 1 A 1 α B 1, where K h = h +1 2, h = M and r = in {α, 1 α}.

530 JIANMING XUE Proof. If 0 < A < M B M, it follows that 0 < 1 M B 1 1 M < 1 A 1 1. By h = M = 1 1 M and 2.4, we have K r h A 1 α B 1 A 1 α B 1. If 0 < B < M A M, siilarly, 2.5 holds. This copletes the proof. Theore 1. Let Φ be a positive unital linear ap and let A and B be positive operators. If 0 < A < M B M or 0 < B < M A M, then for each α [0, 1], 2.6 Φ A α Φ B 2 K h 2 K 2r h Φ 2 A α B, where K h = h+12, K h = h +1 2 and r = in {α, 1 α}. Proof. The inequality 2.6 is equivalent to Φ A α Φ B Φ 1 A α B It is easy to see that K h K 2r h. 2.7 1 α A + MA 1 1 α M + and 2.8 α B + MB 1 α M +. Suing up inequalities 2.7 and 2.8, we get and hence A α B + M A 1 α B 1 M + 2.9 Φ A α B + MΦ A 1 α B 1 M +. Copute Φ A α Φ B MK 2r h Φ 1 A α B K r h Φ A α Φ B + MK r h Φ 1 A α B 2 K r h Φ A α Φ B + MK r h Φ A 1 α B 1 2 Φ A α Φ B + MΦ A 1 α B 1 2 by2.4, 2.5 Φ A α B + MΦ A 1 α B 1 2 by2.1 by2.2 M + 2. by2.9

SOME OPERATOR α-geometric MEAN INEQUALITIES 531 That is Φ A α Φ B Φ 1 A α B M + 2 4MK 2r h = K h K 2r h. Thus, 2.6 holds. This copletes the proof. Reark 1. Since h > 1, then K h K 2r h < K h. Thus, inequality 2.6 is tighter than 1.1. Theore 2. Let Φ be a positive unital linear ap and let A and B be positive operators. If 0 < A < M B M or 0 < B < M A M and 2 p <, then for each α [0, 1], K 2 h M 2 + 2 2 p Φ 2p A α B, 2.10 Φ A α Φ B 2p 1 16 K 4r h M 2 2 where K h = h+12, K h = h +1 2 and r = in {α, 1 α}. Proof. The inequality 2.10 is equivalent to 2.11 Φ A α Φ B p Φ p A α B 1 4 K 2 h M 2 + 2 2 K 4r h M 2 2 p 2. By the operator reverse onotonicity of inequality 2.6, we have 2.12 Φ 2 A α B Since 0 < A, B M, it follows that and hence K h 2 K 2r h Φ A α Φ B 2. Φ A α Φ B M 2.13 Φ A α Φ B 2 + M 2 2 Φ A α Φ B 2 M 2 + 2.

532 JIANMING XUE Copute Φ A α Φ B p M p p Φ p A α B p 1 p 2 K h 2 Φ A α 4 K 2r h Φ B p + M 2 2 2 Φ p A Kh α B by2.1 K 2r h 1 K h 4 K 2r h Φ A αφ B 2 + M 2 2 p Φ 2 A Kh α B by2.3 K 2r h 1 K h 4 K 2r h Φ A αφ B 2 + K h K 2r h M 2 2 ΦA α ΦB 2 p by2.12 1 Kh p 4 K 2r h M 2 + 2. by2.13 That is Φ A α Φ B p Φ p A α B K 2 h M 2 + 2 2 K 4r h M 2 2 p 2. Thus, 2.10 holds. This copletes the proof. Lea 6 [13]. For any bounded operator X, [ ti X 2.14 X ti X t X ti ] 0 t 0. Theore 3. Let Φ be a positive unital linear ap and let A and B be positive operators. If 0 < A < M B M or 0 < B < M A M and 2 p <, then for each α [0, 1], 2.15 Φ A α Φ B p Φ p A α B + Φ p A α B Φ A α Φ B p 1 K 2 h M 2 + 2 p 2 2 2 K 4r h M 2 2, where K h = h+12, K h = h +1 2 and r = in {α, 1 α}. Proof. Put t = 1 2 K2 hm 2 + 2 2 p K 4r h M 2 2 2, X 1 = Φ A α Φ B p Φ p A α B, X 2 = Φ p A α B Φ A α Φ B p and X = X 1 + X 2. By 2.11 and 2.14, we have [ ] ti X1 2.16 0 ti X 2

SOME OPERATOR α-geometric MEAN INEQUALITIES 533 and 2.17 [ ti X2 ti X 1 ] 0. Suing up 2.16 and 2.17, we have [ 2tI X X 2tI ] 0. Since X is self-adjoint, 2.15 follows fro the axial characterization of geoetric ean. This copletes the proof. Acknowledgents This work is supported by Scientific Research Fund of Yunnan Provincial Education Departent No. 2014Y645. References [1] Y. Seo, Reverses of Ando s inequality for positive linear aps, Math. Inequal. Appl., 14 2011, 905-910. [2] T. Furuta, J. Mićić, J.E. Pečarić, Y. Seo, Mond-Pečarić Method in Operator Inequalities, Monographs in Inequalities 1, Eleent, Zagreb, 2005. [3] X. Fu, An operator α-geoetric ean inequality, J. Math. Inequal., 9 2015, 947-950. [4] J. Xue, Soe refineents of operator reverse AM-GM ean inequalities, J. Inequal. Appl., 2017 2017. [5] J. Xue, On reverse weighted arithetic-geoetric ean inequalities for two positive operators, Ital. J. Pure Appl. Math., 37 2017, 113-116. [6] P. Zhang, More operator inequalities for positive linear aps, Banach J. Math. Anal., 9 2015, 166-172. [7] M. Lin, On an operator Kantorovich inequality for positive linear aps, J. Math. Anal. Appl., 402 2013, 127-132. [8] L. Zou, An arithetic-geoetric ean inequality for singular values and its applications, Linear Algebra Appl., 528 2017, 25-32. [9] R. Bhatia, F. Kittaneh, Notes on atrix arithetic-geoetric ean inequalities, Linear Algebra Appl., 308 2000, 203-211.

534 JIANMING XUE [10] R. Bhatia, Positive Definite Matrices, Princeton University Press, Princeton, 2007. [11] T. Ando, X. Zhan, Nor inequalities related to operator onotone functions, Math. Ann., 315 1999, 771-780. [12] H. Zuo, G. Shi, M. Fujii, Refined Young inequality with Kantorovich constant, J. Math. Inequal., 5 2011, 551-556. [13] R.A. Horn, C.R. Johnson, Topics in Matrix Analysis, Cabridge University Press, Cabridge, 1991. Accepted: 7.02.2018