Sharp Bounds for a Generalized Logarithmic Operator Mean and Heinz Operator Mean by Weighted Ones of Classical Operator Ones

: In this paper, using a criteria for the monotonicity of the quotient of two power series, we present some sharp bounds for a generalized logarithmic operator mean and Heinz operator mean by weighted ones of classical operator ones.


Introduction
Since scholars ( [1][2][3]) found that the original weighted arithmetic-geometric mean inequality of two positive numbers can be extended to positive invertible operators, this research field has been developing in the direction of prosperity.In addition to establishing the relationships between various classical means, researchers have also introduced new means and generalized classical means, created new inequalities, and finally tried to extend these conclusions to positive invertible operators.In recent years, progress and achievements in this field have been very rich, as can be seen in the related body of literature .
The Heinz mean (see Bhatia and Davis [1]), denoted as H ν (a, b), is introduced and defined by (2) Bhatia [3] defined the weighted mean of the arithmetic mean and geometric mean as the Heron means by and obtained the relationship between the Heinz mean and the Heron mean.
Kittaneh, Moslehian and Sababheh [20] (Theorem 2.1) also obtained the above result in a very flexible way.Recently, in [32], the author of this paper sharpened the above inequality and obtained the following results.
Ref. [32] also gave a lower bound and a better upper bound for H ν (a, b), as follows.
In fact, the left-hand side of the inequality ( 7) is a mean of two positive numbers, a and b, and it is a generalization of the logarithmic mean, which was first introduced by Shi in [28].For the sake of narration, we give it the sign L ν ; that is, Recently, Alsaafin and Burqan [41] produced richer results for the relationship between the Heinz mean H ν (a, b) and the above generalization of the logarithmic mean L ν (a, b).
Considering the particular mean L ν (a, b) has the following properties we come to the first conclusion of this paper, which is about its monotonicity.
The second topic of this paper is to introduce the weighted mean of the geometric mean G(a, b) and the logarithm mean L(a, b) and the weighted mean of the geometric mean A(a, b) and the logarithm mean L(a, b) by holds, where β 1 cannot be replaced by any larger number and β 2 cannot be replaced by any smaller number.
holds, where λ 1 cannot be replaced by any larger number and λ 2 cannot be replaced by any smaller number.
holds, where σ 2 cannot be replaced by any larger number and σ 1 cannot be replaced by any smaller number.
Letting θ = 1/2 in the above Theorems gives Corollary 1.Let a, b > 0, a = b.Then, the following three double inequalities hold: The double inequality ( 16) is equivalent to We know that Proposition 2 reveals the relationship between H ν (a, b) and F α (a, b).The third goal of this paper is to determine the relationships between holds with the best constants 0 and 3θ 2 .
Letting θ = 3/4 and θ = 2/5 in the above two Theorems, respectively, gives Corollary 2. Let a, b > 0, a = b.Then, the following two double inequalities hold: The right-hand side of ( 20) and the left-hand side inequality of (21) are equivalent to The last section of this paper gives the operator conclusions of the above results.

Proof.
Let where Then, Since {κ n } n≥1 is decreasing for 0 < |θ| < 1, from Lemma 1, we know that the function the proof of Lemma 2 is complete.
Lemma 4. Let t = 0, and 0 < |θ| < 1.Then, the double inequality holds with the best constants 1 and 3 − θ 2 /2. where Then, We can prove that ε n > ε n+1 is based on the following facts Let θ 2 = x.Then, 0 < x < 1, and the last inequality above is equivalent to which is true for Since {ε n } n≥1 is decreasing for 0 < |θ| < 1, from Lemma 1, we know that the function the proof of Lemma 4 is complete.
In a word, when 0 < x = θ 2 ≤ 1/5, the sequence {η n } n≥1 is decreasing.From Lemma 1, we know that the function A 5 (t)/B 5 (t) is decreasing on (0, ∞).Note that lim so the proof of Lemma 6 is complete.
Remark 1.The difficulty of this paper lies in the proofs of Lemmas 2-6.The difficulty of the proofs can be divided into three levels.The proofs of Lemmas 2 and 3 are relatively concise, while the proof of Lemma 4 is concise and ingenious.The proofs of Lemmas 5 and 6 are difficult.We know that the limiting condition comes from the problem itself.From the proof processes of Lemmas 5 and 6, it can be seen that these proof methods are of great benefit to future research in this area.At the same time, the author encourages readers and scholars to do some work to relax the conditions of Lemmas 5 and 6.

Proofs of the Main Results
Proof of Theorem 1.Since L ν (a, b) has the properties (10), we just need to prove that L ν (a, b) decreases as ν increases on (0, 1/2).Through calculations, we can get The proof of Theorem 1 is complete when proving Let s(1 − 2ν)/2 = y.Then, y > 0, and the last inequality above is equivalent to tanh y < y, which is true for all y > 0.  Through the above relationships, we know that the double inequalities ( 24)-( 28) are equivalent to (11)-(19).This completes the proofs of Theorems 2-6.
respectively and to obtain the following results about the relationships between L ν (a, b) and E β (a, b), L ν (a, b) and F α (a, b), L ν (a, b) and J σ (a, b).
we let b/a = e s , then s > 0 and (30) is equivalent to