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Reverse arithmetic-harmonic mean and mixed mean operator inequalities
Journal of Inequalities and Applications volume 2015, Article number: 215 (2015)
Abstract
This note aims to present some scalar inequalities and operator inequalities on a Hilbert space. Firstly, the direct reverse weighted arithmetic-harmonic mean inequalities for scalars are obtained. Secondly, based on these scalar inequalities, the corresponding operator inequalities are established. Finally, we present the mixed arithmetic-geometric and geometric-harmonic means inequalities for two positive operators.
1 Introduction
Let \(\mathcal{B}(H)\) be the \(C^{\ast}\)-algebra of all bounded linear operators on a complex separable Hilbert space H. I stands for the identity operator. \({\mathcal{B}}^{++}(H)\) denotes the cone of all positive invertible operators on H. As a matter of convenience, we use the following notations to define the μ-weighted arithmetic mean (AM), geometric mean (GM), and harmonic mean (HM) for scalars and operators:
where \(a,b>0\), \(A,B\in{\mathcal{B}}^{++}(H)\), and \(\mu\in[0,1]\). When \(\mu=\frac{1}{2}\), we write \(a\nabla b\), \(a!b\), \(A\nabla B\), \(A\# B\) and \(A!B\) for brevity, respectively.
It is well known that the famous μ-weighted A-G-H mean inequalities hold,
for \(a,b>0\) and \(\mu\in[0,1]\) with equalities if and only if \(a=b\). The first inequality of (1.1) is the classical Young inequality.
An operator version of (1.1) proved in [1] says that if \(A, B \in{\mathcal{B}}^{++}(H) \) and \(\mu\in[0,1]\), then
In recent years, the study of the A-G-H mean inequalities has received increasing attention in the literature (see [2–5]).
Zuo et al. (see [5]) refined the Young inequality with the Kantorovich constant and obtained the following results:
where \(a,b>0\), \(\mu\in[0,1]\), \(r=\min \{ {\mu,1-\mu} \} \), and \(h=\frac{b}{a}\). In addition, they also refined the μ-weighted arithmetic-harmonic mean inequality and extended it to an operator version as follows:
where \(a,b>0\), \(A,B\in\mathcal{B}^{++}(H) \), \(\mu\in[0,1]\), and \(r=\min \{ {\mu,1-\mu} \}\). By (1.3) and (1.4), we are encouraged to investigate whether there exist reverse forms of the μ-weighted arithmetic-harmonic mean inequality, so we give an affirmative answer to this question in our paper.
Moreover, mixed mean inequalities are also extremely attractive. Sagae and Tanabe [6] establish a mixed A-G mean inequality for a finite number of strictly positive operators. Mond and Pečarić [7] establish a mixed A-G and G-H mean inequalities for two noncommutative strictly positive operators as follows: Let A and B be positive invertible operators. The mixed arithmetic-geometric and geometric-harmonic mean inequalities are valid:
In this paper, we are concerned with the weighted arithmetic-harmonic, mixed arithmetic-geometric and mixed geometric-harmonic mean inequalities. In Section 2, we present direct reverse weighted arithmetic-harmonic mean inequalities by the Kantorovich constant and deduce some auxiliary results. In Section 3, we extend inequalities proved in Section 2 from the scalars setting to a Hilbert space operator setting. In Section 4, we establish mixed weighted arithmetic-geometric and geometric-harmonic means inequalities for two positive operators which are the refinements of (1.5) and (1.6).
2 Reverse arithmetic-harmonic mean inequalities
In this section, we mainly present the direct reverse forms of the μ-weighted arithmetic-harmonic mean inequality for two positive numbers a, b.
First of all, we recall the classical Kantorovich inequality in [8].
Lemma 2.1
Let \(0< a=x_{1} < x_{2} <\cdots<x_{n} =b\) be given positive numbers, \(\lambda_{1} ,\lambda_{2} ,\ldots,\lambda_{n} \ge0\) and \(\sum_{i=1}^{n} {\lambda_{i} } =1\). Then
where \(A=\frac{1}{2}(a+b)\), \(G=\sqrt{ab}\) are the arithmetic and geometric means, respectively.
The inequality (2.1) is the Kantorovich inequality and the number \(\frac{(a+b)^{2}}{4ab}\) is called the Kantorovich constant. For convenience, we write the Kantorovich constant as \(\mathrm{K}(t,2)=\frac{(t+1)^{2}}{4t}\) \((t=\frac{b}{a} )\), which has the properties \(\mathrm{K}(1,2)=1\), \(\mathrm{K}(t,2)=\mathrm{K} ( {\frac{1}{t},2} )\ge1\) (\(t>0\)), and \(\mathrm{K}(t,2)\) is monotone increasing on \([1,\infty)\), and monotone decreasing on \((0,1]\).
When \(n=2\), we can get a special form for the inequality (2.1) as follows.
Proposition 2.1
Let \(a,b>0\) and \(\mu\in[0,1]\). Then
where \(h=\frac{b}{a}\). Equality holds if and only if \(a=b\).
Remark 2.1
The inequality (2.2) is a direct reverse of the μ-weighted arithmetic-harmonic mean inequality. It is very interesting that by the inequality (2.2) and (1.2) we can get a reverse of Young inequality (it is also a reverse of the inequality (1.2)):
where \(a,b>0\), \(\mu\in[0,1]\), \(r=\min \{ \mu,1-\mu \}\), and \(R = \max\{1 - \mu,\mu\}\). Replacing a, b by \(a^{ - 1}\), \(b^{ - 1}\)in the above inequalities, respectively, we have
Next, we deduce a direct reverse of the inequality (1.3) by the following lemma.
Lemma 2.2
[9]
Let \(x_{i}\) (\(i=1,2,\ldots,n\)) belong to a fixed closed interval \(\mathbb{I}\), \(p_{i} \ge0\) with \(\sum_{i=1}^{n} {p_{i} } =1\) and \(\bar{p}=\max\{p_{1} ,p_{2} ,\ldots ,p_{n} \}\). If f is a convex function on \(\mathbb{I}\), then
It is easy to see that if we take \(f(x)=x^{-1}\) in the inequality (2.5), then we have the following.
Proposition 2.2
If \(x_{1} ,x_{2} ,\ldots,x_{n} >0\) and \(p_{i} \ge0 \) (\(i=1,2,\ldots,n\)) with \(\sum_{i=1}^{n} {p_{i} } =1\), then
where \(\bar{p}=\max\{p_{1} ,p_{2} ,\ldots,p_{n} \}\).
In particular, when \(n=2\) in the inequality (2.6), we can get
where \(a, b>0\), \(\mu\in[0,1]\), \(R = \max\{1 - \mu,\mu\}\). Equality holds if and only if \(a=b\).
Note that (2.7) is a reverse of the inequality (1.3) with a similar form.
The following three theorems are our main reverse forms of the μ-weighted arithmetic-harmonic mean inequality for scalars.
Theorem 2.1
Let \(a,b>0\), and \(\mu\in[0,1]\). Then the inequality
holds, where \(h=\frac{b}{a}\), \(r=\min \{ {\mu,1-\mu} \}\), \(R=\max \{ \mu, 1-\mu \}\), and \({R}'=\max\{ 2r,1-2r\}\). Equality holds if and only if \(a=b\).
Proof
By the inequalities (2.3) and (2.4), firstly, we consider the case \(\mu\in[0,\frac{1}{2}]\),
If \(\mu\in(\frac{1}{2},1]\), then we have
By the above discussion, for any \(\mu\in[0,1]\), the inequality (2.8) always holds. □
Note that (2.8) can also be considered as a reverse ratio inequality of (1.3):
Theorem 2.2
Let \(a,b>0\), \(\mu\in[0,1]\), and \(h=\frac{b}{a}\).
-
(I)
If \(0\leq\mu\le\frac{1}{2}\), then
$$a\nabla_{\mu}b-2\mu(a\nabla b-a!b)\le\mathrm{K} (\sqrt{h} ,2 )a!_{2\mu}\sqrt{ab}. $$Equality holds if and only if \(a=b\).
-
(II)
If \(\frac{1}{2}<\mu\le1 \), then
$$a\nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-a!b) \le\mathrm{K} (\sqrt{h},2 )b!_{2-2\mu} \sqrt{ab}. $$Equality holds if and only if \(a=b\).
Proof
By the inequality (2.2), firstly, we consider the case \(\mu\in[0,\frac{1}{2}]\), then we have
If \(\mu\in(\frac{1}{2},1]\), then we have
The proof is completed. □
Theorem 2.3
Let \(a,b>0\) and \(\mu\in[0,1]\).
-
(I)
If \(0\leq\mu\le\frac{1}{2}\), then
$$a\nabla_{\mu}b-2\mu(a\nabla b-a!b)\le a!_{2\mu} \sqrt{ab}+2R'(a\nabla \sqrt{ab}- a!\sqrt{ab}). $$Equality holds if and only if \(a=b\).
-
(II)
If \(\frac{1}{2}<\mu\le1 \), then
$$a\nabla_{\mu}b-2(1-\mu) (a\nabla b-a!b)\le b!_{2-2\mu}\sqrt {ab}+2R'(b\nabla\sqrt{ab}- b!\sqrt{ab}). $$Equality holds if and only if \(a=b\).
Proof
Firstly, we consider the case \(\mu\in[0,\frac{1}{2}]\), by the inequality (2.7), then we have
If \(\mu\in(\frac{1}{2},1]\), by the inequality (2.7), then we have
The proof is completed. □
Note that the inequalities proved in Theorem 2.1-2.3 are all the reverses of the inequality (1.3).
3 Reverse arithmetic-harmonic mean operator inequalities
In this section, we present the operator versions of these reverse arithmetic-harmonic mean inequalities proved in Section 2. The techniques are based on the monotonicity property of operator functions described in the following lemma (for more details, see [1, 10]).
Lemma 3.1
Let \(X\in{\mathcal{B}}(H)\) be self-adjoint operator and if f and g are both continuous functions with \(f(t)\ge g(t)\) for \(t\in \operatorname{Sp}(X)\) (the spectrum of X), then \(f(X)\ge g(X)\) with equality if and only if \(f(t)=g(t)\) for all \(t\in\operatorname{Sp}(X)\).
Based on Proposition 2.1, we can deduce the operator inequality which is also the noncommutative Kantorovich inequality proved by Furuta et al. [1], but our method is different.
Proposition 3.1
Let \(A,B\in{\mathcal{B}}^{++}(H)\) and positive real numbers m, M satisfy \(0< mI\le A,B\le MI\). Then
where \(\mu\in[0,1]\) and \(h=\frac{M}{m}\). Equality holds if and only if \(A=B\) and \(m=M\).
Proof
By the inequality (2.2), we have
for any \(x>0\), and hence
for the positive operator X such that \(0<\frac{1}{h}I\le X\le hI\).
Since \(0<\frac{1}{h}I\le A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\le hI\) and the Kantorovich constant \(\mathrm{K}(t,2)\) is an increasing function for \(t>1\) and \(\mathrm{K}(\frac{1}{h},2)=\mathrm{K}(h,2)\), substituting \(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) for X in the above inequality, we have
Multiplying both sides of the above inequality by \(A^{\frac{1}{2}}\), we can deduce the required inequality (3.1). □
We prove an operator inequality obtained by Krnić et al. in [11] by a different method. But our method is more transparent and simpler than the one given in [11].
Proposition 3.2
Let \(A,B\in{\mathcal{B}}^{++}(H)\) and \(\mu\in [0,1]\). Then
where \(R=\max\{1-\mu,\mu\}\). Equality holds if and only if \(A=B\).
Proof
By the inequality (2.7), for \(x>0\) and \(\mu\in[0,1]\), we have
For a positive invertible operator T and \(\mu\in[0,1]\), it follows that
Putting \(T=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) in the above inequality and multiplying both sides by \(A^{\frac{1}{2}}\), we deduce the inequality (3.2). □
Note that, by (1.4) and (3.2), we have
Based on Theorem 2.1, we have the following.
Theorem 3.1
Let \(A,B\in{\mathcal{B}}^{++}(H)\) and positive real numbers m, M satisfy \(0< mI\le A,B\le MI\). Then for \(\mu\in [0,1]\),
where \(h=\frac{M}{m}\), \(r=\min \{ {\mu,1-\mu} \}\), \(R=\max \{ {\mu,1-\mu} \}\), and \({R}'=\max \{ {2r,1-2r} \}\). Equality holds if and only if \(A=B\) and \(m=M\).
Proof
By the inequality (2.8), we have
for any \(x>0\), and hence
for the positive operator X such that \(0<\frac{1}{h}I\le X\le hI\).
By a similar process to Proposition 3.1, we can deduce the required inequality (3.3). □
Note that (3.3) is a reverse of (1.4):
Now, we exhibit the operator inequalities based on Theorem 2.2 and Theorem 2.3.
Theorem 3.2
Let \(A,B\in{\mathcal{B}}^{++}(H)\) and \(\mu\in [0,1]\). The positive real numbers m, M satisfy \(0< mI\le A,B\le MI\) and \(h=\frac{M}{m}\).
-
(I)
If \(0\leq\mu\le\frac{1}{2}\), then
$$ A\nabla_{\mu}B-2\mu(A\nabla B-A!B)\le\mathrm{K} (\sqrt{h} ,2 )A!_{2\mu}(A\#B). $$(3.4)Equality holds if and only if \(A=B\) and \(m=M\).
-
(II)
If \(\frac{1}{2}<\mu\le1 \), then
$$ A\nabla_{\mu}B-2(1-\mu) (A\nabla B-A!B)\le\mathrm{K} ( \sqrt{h} ,2 )B!_{2(1-\mu)} (A\#B). $$(3.5)Equality holds if and only if \(A=B\) and \(m=M\).
Proof
If \(0\leq\mu\le\frac{1}{2}\), by (I) of Theorem 2.2, we have
for any \(a>0\), and hence
for the positive invertible operator X such that \(0<\frac{1}{h}I\le X\le hI\).
Substituting \(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) for X in the above inequality, we have
Multiplying both sides of (3.6) by \(A^{\frac{1}{2}}\), we can deduce the required inequality (3.4).
Likewise, if \(\frac{1}{2}<\mu\le1 \), by (II) of Theorem 2.2, we have
for any \(b>0\), and hence
for the positive invertible operator Y such that \(0<\frac{1}{h}I\le Y\le hI\).
Substituting \(B^{-\frac{1}{2}}AB^{-\frac{1}{2}}\) for Y in the above inequality, we have
Multiplying both sides of (3.7) by \(B^{\frac{1}{2}}\), we can deduce the required inequality (3.5). □
Note that, by (1.4), (3.4), and (3.5), we have
Theorem 3.3
Let \(A, B\in{\mathcal{B}}^{++}(H)\) and \(\mu\in[0,1]\).
-
(I)
If \(0\leq\mu\le\frac{1}{2}\), then
$$ A\nabla_{\mu}B-2\mu(A\nabla B-A!B)\le A!_{2\mu}{(A\#B)}+2R' \bigl(A\nabla {(A\#B)}- A!{(A\#B)} \bigr). $$(3.8)Equality holds if and only if \(A=B\).
-
(II)
If \(\frac{1}{2}<\mu\le1 \), then
$$ A\nabla_{\mu}B-2(1-\mu) (A\nabla B-A!B)\le B!_{2-2\mu}(A\#B)+2R' \bigl(B\nabla (A\#B)- B!(A\#B) \bigr). $$(3.9)Equality holds if and only if \(A=B\).
Proof
If \(0\leq\mu\le\frac{1}{2}\), by (I) of Theorem 2.3, we have
for any \(b>0\), and hence
for the positive invertible operator X.
Substituting \(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) for X in the above inequality and then multiplying both sides by \(A^{\frac{1}{2}}\), we can deduce the required inequality (3.8).
Likewise, if \(\frac{1}{2}<\mu\le1 \), by (II) of Theorem 2.3, we have
for any \(a>0\), and hence
for the positive invertible operator Y.
Substituting \(B^{-\frac{1}{2}}AB^{-\frac{1}{2}}\) for Y in the above inequality and then multiplying both sides by \(B^{\frac{1}{2}}\), we can deduce the required inequality (3.9). □
Note that, by (1.4), (3.8), and (3.9), we have
Remark 3.1
These inequalities proved in Theorem 3.1-3.3 are all the reverse forms of the inequality (1.4). It is easy to see that the right-hand side of these inequalities can not be compared with each other, but they are indeed new versions of reverse ratio arithmetic-harmonic mean inequality.
4 The mixed mean inequalities
In this section, we obtain refinements of the inequalities (1.5) and (1.6) and deduce some mixed weighted arithmetic-geometric and geometric-harmonic means inequalities.
First, we need the following lemma.
Lemma 4.1
(Hermite-Hadamard’s inequality [12])
If \(f:\mathbb{I}\rightarrow\mathbb{R}\) is a convex function on the interval \(\mathbb{I}\subset\mathbb{R}\), then for any \(x,y \in\mathbb{I}\) with \(x\neq y\), we have
Theorem 4.1
Let \(A, B\in{\mathcal{B}}^{++}(H)\). Then
Proof
Applying the Hermite-Hadamard’s inequalities (4.1) to the convex function \(f(t)=-t^{\frac{1}{2}}\), \(t>0\), we have
and hence
for a positive invertible operator C.
Putting \(C=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) in the above inequality and multiplying both sides by \(A^{\frac{1}{2}}\), we can obtain (4.2).
Substituting \(A^{-1}\) for A and \(B^{-1}\) for B and then taking the inverse in (4.2), we get (4.3). □
Note that the inequalities (4.2) and (4.3) are the refinements of (1.5) and (1.6), respectively.
If f is convex on a segment \([a, b]\) of a linear space, one can easily observe that (4.1) is equivalent to the following double inequality:
A natural generalization of the classical Hermite-Hadamard inequality to Hermitian matrices could be the double inequality
However, Moslehian [13] pointed out that this was not true. We will show that the following result is valid.
Theorem 4.2
Let \(A, B\in{\mathcal{B}}^{++}(H)\). Then
Proof
Taking \(f(t)=-t^{\frac{1}{2}}\), \(t>0\) in (4.4), then
and by a method resembling Theorem 4.1, we obtain (4.5). □
In the next theorems, mixed weighted arithmetic-geometric and geometric-harmonic mean inequalities are established. First, we show the definition of operator convex (see [10], p.113).
Definition 4.1
A continuous function \(f:\mathbb{I}\rightarrow \mathbb{R}\) on the interval \(\mathbb{I}\subset\mathbb{R}\) is said to be operator convex if for every pair of self-adjoint operators X, Y on a Hilbert space H with spectrum in \(\mathbb{I}\) and each \(v \in[0, 1]\),
A function f is called operator concave if the function −f is operator convex. The function \(f(X)=X^{s}\) is operator convex on a self-adjoint operator space for \(-1\le s\le0\) or \(1\le s\le2\) and is operator concave for \(0\le s\le1\).
For an operator convex function \(f:\mathbb{I}\rightarrow\mathbb{R}\) on the interval \(\mathbb{I}\subset\mathbb{R}\), we have the following property (see [14], p.71): For each \(v\notin[0,1]\), self-adjoint operators X, Y and \((1-v)X+vY\) with spectra in \(\mathbb{I}\),
In the next theorems, we still use the notations \(A\nabla_{v} B\), \(v\notin[0,1] \), and \(A\# _{s} B\), \(s\in[-1,0]\cup[1,2]\).
Theorem 4.3
Let \(A, B\in{\mathcal{B}}^{++}(H)\), and \(v\in[0,1]\). If \(s\in [-1,0]\cup[1,2]\), then
Proof
By the inequality (4.6), for every pair of self-adjoint operators X, Y and the operator convex function \(f(x)=x^{s} \) (\(x>0\)), \(s\in[-1,0]\cup [1,2]\), we have
Putting \(X=I\) and \(Y=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) in the above inequality and multiplying both sides by \(A^{\frac{1}{2}}\), we can get (4.8).
Substituting \(A^{-1}\) for A and \(B^{-1}\) for B in (4.8) and then taking the inverse of both sides, we get (4.9). □
Theorem 4.4
Let \(A, B\in{\mathcal{B}}^{++}(H)\) and \((1-v)A+vB\in{\mathcal {B}}^{++}(H)\) for \(v\notin[0,1]\).
-
(I)
If \(s \in[0,1]\), then
$$\begin{aligned} &A\#_{s}(A\nabla_{v} B)\le A\nabla_{v} (A \#_{s} B), \\ &A\#_{s}(A!_{v} B)\ge A!_{v} (A\#_{s} B). \end{aligned}$$ -
(II)
If \(s\in[-1,0]\cup[1,2]\), then
$$\begin{aligned} &A\#_{s}(A\nabla_{v} B)\ge A\nabla_{v} (A \#_{s} B), \\ &A\#_{s}(A!_{v} B)\le A!_{v} (A\#_{s} B). \end{aligned}$$
Proof
By the inequality (4.7) and using the same ideas as in the proof of Theorem 4.2, we can deduce this theorem. □
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Acknowledgements
The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript. The authors are grateful to Dr. Pujun Long for fruitful discussion and revising the manuscript.
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Liao, W., Wu, J. Reverse arithmetic-harmonic mean and mixed mean operator inequalities. J Inequal Appl 2015, 215 (2015). https://doi.org/10.1186/s13660-015-0735-z
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DOI: https://doi.org/10.1186/s13660-015-0735-z