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More results on weighted means

We give a refined Young inequality which generalizes the inequality by Zou--Jiang. We also show the upper bound for the logarithmic mean by the use of the weighted geometric mean and the weighted arithmetic mean. Furthermore, we show some inequalities among the weighted means. Based on the obtained essential scalar inequalities, we give some operator inequalities. In particular, we give a generalization of the result by Zou--Jiang, that is, we show the operator inequalities with the operator relative entropy with the weighted parameter. Finally, we give the further generalized inequalities by the Tsallis operator relative entropy.


Introduction
The logarithmic mean is defined by for two positive numbers a and b. (We usually define L(a, b) = a, if a = b.) It is known the inequality: which is called the classical Pólya inequality in [14], where A(a, b) := 1 2 (a + b), G(a, b) := √ ab.
It is also known the inequality [10]: We have the following relation [5, Lemma 1.1]: which is a refinement of the Pólya inequality.
where A v (a, b) := (1 − v)a + vb is the weighted arithmetic mean and G v (a, b) := a 1−v b v is the weighted geometric mean. However, the following inequality does not hold in general.
since we have counter-examples. See [11] for example. We have the following relation for a, b > 0 and 0 ≤ v ≤ 1, where H v (a, b) := (1 − v)a −1 + vb −1 −1 is the weighted harmonic mean. The inequality ) is often called the Young inequality. Many refinements and reverses for this inequality have been studied. See [3,Chapter 2] for example. See also [7,8] for the recent advanced results. In the paper [4], one of authors studied some inequalities on the weighted means, especially the weighted logarithmic mean. In this paper, we give the further results on the weighted mean and obtain some new inequalities for them.

Main results
We firstly give a new refinement of the Young inequality which is a generalization for the known result by using the weighted parameter v ∈ [0, 1].
Proof. The first inequality of (9) is trivial. To prove the second inequality of (9), we assume Then we have We also have For the case 1/2 ≤ v ≤ 1 can be proven similarly. Finally, putting t := a/b and then multiplying b > 0 to both sides, we obtain the desired result.
Remark 2.2. It is remarkable that Theorem 2.1 recovers the following inequality [15]: when v = 1/2. In addition, the following reverse of the second inequality in (9): does not hold in general, because of (10). The authors have been unaware the paper [8] until the review. Although our generalization given in Theorem 2.1 for (10) is natural and simple, we have to point out that the inequality given in [8, (2.1)] is better than our one.
Considering r-logarithmic function which is defined by ln r x := x r − 1 r for x > 0 and r = 0, we find the following corollary of Theorem 2.1. We note that lim r→0 ln r x = log x. Corollary 2.3. Let 0 ≤ v ≤ 1, r = 0 and a, b > 0. If we have the conditions (i)r > 0 and 0 < a ≤ b, or (ii)r < 0 and a ≥ b > 0, then we have the following inequalities.
Proof. The first inequality of (11) is trivial. We consider the function .
t r log t r ≥ 0 for t > 0 and r ∈ R. Thus we have df v,r (t) dr ≥ 0 for r > 0 and 0 < t ≤ 1, The last inequality is thanks to the second inequality of (9) with the fact lim r→0 ln r x = log x. We also have df v,r (t) dr ≤ 0 for r < 0 and t ≥ 1, which implies f v,r (t) ≥ f v,0 (t) ≥ 0, similarly. Therefore, if we have the conditions (i)r > 0 and 0 < t ≤ 1, or (ii)r < 0 and t ≥ 1, then we have the following inequalities: Putting t := a/b in the above and multiplying b > 0 to both sides, we get the second inequality of (11).
It is well known that where is the Heinz mean. Then we have the following inequalities from Theorem 2.1.

Proof.
Replacing v by 1− v in (9) and adding it to (9) and then dividing 2, we get the result.
The inequality does not hold in general, since we have counter-examples such as and However, we have the following inequality for the weighted mean.
Then we have Therefore we have We here prove f v min (t) ≥ 0 for t > 0. Then we set for t > 0 the function g(t) by Then we have (16) with same procedure in the Corollary 2.4, we get the following inequality.
This inequality is also proven by the use of (6) with v = 1/2 and (12) as (ii) From g(t) ≥ 0 in the proof of Theorem 2.5 and [2, Lemma 3], we have the bounds of t We study the properties on the function which is a representing function of the weighted logarithmic mean.
We easily see bL L v (t) = 1 and lim t→0 L v (t) = 0. We have the following property.
Proposition 2.7. The function L v (t) given in (18) is increasing with respect to v when 0 < t ≤ 1 and decreasing with respect to v when t ≥ 1.
Proof. We calculate We also calculate we easily see that the representing functions for the weighted arithmetic mean, the weighted geometric mean and the weighted harmonic mean have similar properties. Proposition 2.7 will be applied to the proof of Proposition 3.1.
We further study the inequalities among means. In the sequel, we consider the bounds of the nested means for the weighted means. From the simple calculations and numerical computations, we see A(a, b), G (G v (a, b), G(a, b), Because we have L(L 1/4 (10, 1), L 3/4 (10, 1)) − L(10, 1) ≃ 0.0173327 as an example. As given in (8), the distance between the geometric mean and the arithmetic mean is not so tight. Therefore the proof for the following relations are not so difficult. and Proof. The inequalities (20) can be proven by The inequalities (20) are just same to the inequalities given in (12).
From Proposition 2.8, we have the following.
Proposition 2.9. For 0 ≤ v ≤ 1 and a, b > 0, we have and Proof. Replacing a and b by 1/a and 1/b in Proposition 2.8 respectively, and taking inverse we have (22) and (23), G(a, b), and (a, b)) .
We have the following relation on the arithmetic mean and the logarithmic mean. Their proofs are not so easy, since the distance between the logarithmic mean and the arithmetic mean is relatively tight.
and v (a, b), we have the second inequality of (24). As for the first inequality of (24) , it is sufficient to prove f v (t) ≥ 0 for t > 0 and 0 L v (t) = 1, for the special cases such as which is known. So we assume 0 < v < 1. By the simple calculations, we have Since The second inequality of (25) can be easily proven by As for the first inequality of (25), it is sufficient to prove the following inequality for t > 0 and 0 ≤ v ≤ 1. Since the equality holds when v = 0, 1/2, 1, we assume 0 < v < 1 with v = 1/2.
(ii) For the case t ≥ 1 and 1/2 < v < 1, the inequality (26) is equivalent to So we set the function l v (t) : Therefore we have (26) for t ≥ 1 and 1/2 < v < 1. By the similar way to the last part of (i), we have the inequality (26) for t > 0 and 1/2 < v < 1.
Theorem 2.11. For 0 ≤ v ≤ 1 and a, b > 0, we have and Proof. The first inequality of (28) is easily proven by To prove the second inequality of (28), we set the function (since L (G v (a, b), By the similar way, we can prove Putting t := 1/s ≥ 1, we obtain Thus we have Putting t := a/b in (31) and multiplying b > 0 to the both sides, we obtain the second inequality of (28). The first inequality of (29) is easily proven by a, b)) ≥ G (G v (a, b), G(a, b). 2 because we put t := a/b > 0 and multiply b 2 > 0 to both sides. Since (log t) 2 > 0 for all t > 0, we set the function Since g v (t) = g 1−v (t) and g v (t) ≥ 0, (t ≥ 1) implies g v (s) ≥ 0, (0 < s ≤ 1) by putting t := 1/s ≥ 1, we have only to prove g v (t) ≥ 0 for t ≥ 1 and 0 ≤ v ≤ 1/2. Then we calculate We further calculate

Thus we have
In the end of this section, we state some operator inequalities for the essential scalar inequalities which were obtained above. To this end, we give a notation for a self-adjoint operator A. If a self-adjoint operator A satisfies Ax, x ≥ 0 for all vectors x = 0, then A is called a positive operator, and we use the notation A ≥ 0. If Ax, x > 0 for all vectors x = 0, then A is called strictly positive operator, and we use the notation A > 0. It is known that the scalar order is equivalent to the operator partial order by Kubo-Ando theory [9]. Therefore it is often important to obtain a new scalar inequality. To express the logarithmic mean, we use the integral form such as L(t, by putting t := A −1/2 BA −1/2 and multiplying A 1/2 to both sides, where is the weighted operator geometric mean for A, B > 0. In the proof of Theorem 2.11, we proved L v (t)L 1−v (t) ≤ L 1/2 (t) 2 which is also rewritten as Thus we similarly have for A, B > 0 where the weighted operator logarithmic mean is defined by and the operator logarithmic mean is written by From (2.5), we also have for A, B > 0 known as the operator relative entropy [1]. We see that the second inequality in (32) gives a generalization of [15,Theorem 4.1]. Furthermore, the inequalities (11) is equivalent to the inequalities: under the conditions (i)r > 0 and 0 < t ≤ 1, or (ii)r < 0 and t ≥ 1. Therefore we have the following proposition.
Proposition 2.12. Under the conditions (i)r > 0 and 0 < B ≤ A, or (ii)r < 0 and 0 < A ≤ B, we have the following operator inequalities: known as the Tsallis operator relative entropy [13].
The other obtained scalar inequalities give the corresponding operator inequalities. However, we omit them.
It is quite natural to consider the maximum (optimal) value p such that By the numerical computation shows that We close this paper with the following conjecture. From (19) with several numerical computations indicate that the inequality seems to be true L(a, b) ≤ L(L v (a, b), L 1−v (a, b)), a, b > 0, 0 ≤ v ≤ 1. (36) However we have not proven this inequality due to its complicated computations, and we also have not found any counter-examples. If the conjectured inequality (36) will be shown, then it will give a tight inequality for the first inequalities in (24) and (25).