Reverse arithmetic-harmonic mean and mixed mean operator inequalities

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:

$$\begin{aligned}& a\nabla_{\mu}b=(1-\mu)a+\mu b,\qquad a!_{\mu}b= \bigl( {(1- \mu)a^{-1}+\mu b^{-1}} \bigr)^{-1}, \\& A\nabla_{\mu}B=(1-\mu)A+\mu B, \qquad A\# _{\mu}B=A^{\frac{1}{2}} \bigl(A^{-\frac {1}{2}}BA^{-\frac{1}{2}} \bigr)^{\mu}A^{\frac{1}{2}}, \\& A!_{\mu}B= \bigl( {(1-\mu)A^{-1}+\mu B^{-1}} \bigr)^{-1}, \end{aligned}$$

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,

$$ a\nabla_{\mu}b\ge a^{1-\mu}b^{\mu}\ge a!_{\mu}b $$

(1.1)

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

$$A\nabla_{\mu}B\ge A\#_{\mu}B \ge A!_{\mu}B. $$

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:

$$ a\nabla_{\mu}b\ge\mathrm{K}(h,2)^{r}a^{1-\mu}b^{\mu}, $$

(1.2)

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:

$$\begin{aligned} &a\nabla_{\mu}b\ge a!_{\mu}b+ 2r (a\nabla b-a!b ), \end{aligned}$$

(1.3)

$$\begin{aligned} &A\nabla_{\mu}B\ge A!_{\mu}B+2r (A\nabla B-A!B), \end{aligned}$$

(1.4)

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:

$$\begin{aligned} &A\#(A\nabla B)\ge A\nabla(A\# B), \end{aligned}$$

(1.5)

$$\begin{aligned} &A\#(A!B)\le A! (A\# B). \end{aligned}$$

(1.6)

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

$$ \Biggl( {\sum_{i=1}^{n} { \lambda_{i} } x_{i} } \Biggr) \Biggl( {\sum _{i=1}^{n} {\lambda_{i} } x_{i}^{-1} } \Biggr)\le A^{2}G^{-2}, $$

(2.1)

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

$$ a\nabla_{\mu}b\le\mathrm{K}(h,2)a!_{\mu}b, $$

(2.2)

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)):

$$ a\nabla_{\mu}b\le\mathrm{K}(h,2)a!_{\mu}b\le \mathrm{K}(h,2)^{1-r}a^{1-\mu }b^{\mu}= \mathrm{K}(h,2)^{R}a^{1-\mu}b^{\mu}, $$

(2.3)

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

$$ a^{1 - \mu}b^{\mu}\le\mathrm{K}(h,2)^{R}a!_{\mu}b. $$

(2.4)

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

$$ \sum_{i=1}^{n} {p_{i} } f(x_{i} )-f \Biggl(\sum _{i=1}^{n} {p_{i} } x_{i} \Biggr)\le n\bar{p} \Biggl[ {\sum_{i=1}^{n} { \frac{1}{n}} f(x_{i} )-f \Biggl(\sum _{i=1}^{n} {\frac{1}{n}} x_{i} \Biggr)} \Biggr]. $$

(2.5)

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

$$ \sum_{i=1}^{n} {p_{i} } x_{i}^{-1} - \Biggl(\sum _{i=1}^{n} {p_{i} } x_{i} \Biggr)^{-1}\le n\bar{p} \Biggl[ {\sum_{i=1}^{n} {\frac{1}{n}} x_{i}^{-1} - \Biggl(\sum _{i=1}^{n} {\frac{1}{n}} x_{i} \Biggr)^{-1}} \Biggr], $$

(2.6)

where $\bar{p}=\max\{p_{1} ,p_{2} ,\ldots,p_{n} \}$.

In particular, when $n=2$ in the inequality (2.6), we can get

$$ a\nabla_{\mu}b\le a!_{\mu}b+2R(a\nabla b-a!b), $$

(2.7)

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

$$ a\nabla_{\mu}b\le2r(a\nabla b-a!b)+\mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K} (h,2 )^{R} a!_{\mu}b $$

(2.8)

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 $\mu\in[0,\frac{1}{2}]$,

$$\begin{aligned} a\nabla_{\mu}b-2\mu(a\nabla b-a!b) &\le a\nabla_{\mu}b-2\mu (a\nabla b-\sqrt{ab} ) \\ &= ( {1-2\mu} )a+2\mu\sqrt{ab} \\ &\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}a^{1-\mu}b^{\mu}\\ &\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K}(h,2)^{R}a!_{\mu}b. \end{aligned}$$

If $\mu\in(\frac{1}{2},1]$, then we have

$$\begin{aligned} a\nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-a!b) &\le a \nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-\sqrt{ab} ) \\ &= ( {2\mu-1} )b+2 ({1-\mu} )\sqrt{ab} \\ &\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}a^{1-\mu}b^{\mu}\\ &\le\mathrm{K} (\sqrt{h},2 )^{{R}'}\mathrm{K}(h,2)^{R}a!_{\mu}b. \end{aligned}$$

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):

$$0< a!_{\mu}b \le a\nabla_{\mu}b-2r(a\nabla b-a!b)\le \mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K} (h,2 )^{R} a!_{\mu}b. $$

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

$$\begin{aligned}[b] a\nabla_{\mu}b-2\mu(a\nabla b-a!b)&\le a\nabla_{\mu}b-2\mu (a\nabla b-\sqrt{ab} ) \\ &= ( {1-2\mu} )a+2\mu\sqrt{ab} \\ &\le\mathrm{K} (\sqrt{h} ,2 )a!_{2\mu}\sqrt{ab}. \end{aligned} $$

If $\mu\in(\frac{1}{2},1]$, then we have

$$\begin{aligned} a\nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-a!b) &\le a \nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-\sqrt{ab} ) \\ &= ( {2\mu-1} )b+2 ({1-\mu} )\sqrt{ab} \\ &\le\mathrm{K} (\sqrt{h},2 )b!_{2-2\mu} \sqrt{ab}. \end{aligned}$$

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

$$\begin{aligned} a\nabla_{\mu}b-2\mu(a\nabla b-a!b)&\le ( {1-2\mu} )a+2\mu \sqrt {ab} \\ &\le a!_{2\mu}\sqrt{ab}+2\max\{1-2\mu,2\mu\}(a\nabla\sqrt{ab}- a! \sqrt{ab}). \end{aligned}$$

If $\mu\in(\frac{1}{2},1]$, by the inequality (2.7), then we have

$$\begin{aligned} &a\nabla_{\mu}b-2 ( {1-\mu} ) (a\nabla b-a!b) \\ &\quad\le ( {2\mu-1} )b+2 ({1-\mu} )\sqrt{ab} \\ &\quad\le b!_{2-2\mu}\sqrt{ab}+2\max\{2-2\mu,2\mu-1\}(b\nabla\sqrt{ab}- b! \sqrt{ab}). \end{aligned}$$

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

$$ A\nabla_{\mu}B\le\mathrm{K}(h,2)A!_{\mu}B, $$

(3.1)

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

$$(1-\mu)+\mu x\le\mathrm{K}(x,2) \bigl( {(1-\mu)+\mu x^{-1}} \bigr)^{-1} $$

for any $x>0$, and hence

$$( {1-\mu} )I+\mu X\le\mathop{\max} _{\frac {1}{h}\le x\le h} \mathrm{K}(x,2) \bigl( {(1- \mu)I+\mu X^{-1}} \bigr)^{-1} $$

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

$$( {1-\mu} )I+\mu A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\le\mathrm{K}(h,2) \bigl( {(1- \mu)I+\mu \bigl(A^{-\frac{1}{2}}BA^{-\frac {1}{2}} \bigr)^{-1}} \bigr)^{-1}. $$

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

$$ A\nabla_{\mu}B\le A!_{\mu}B+2R(A\nabla B-A!B), $$

(3.2)

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

$$(1-\mu)+\mu x^{-1}\le \bigl((1-\mu)+\mu x \bigr)^{-1}+2R \biggl[ {\frac{1+x^{-1}}{2}- \biggl( {\frac{1+x}{2}} \biggr)^{-1}} \biggr]. $$

For a positive invertible operator T and $\mu\in[0,1]$, it follows that

$$(1-\mu)I+\mu T^{-1}\le \bigl((1-\mu)I+\mu T \bigr)^{-1}+2R \biggl[ {\frac{I+T^{-1}}{2}- \biggl( {\frac{I+T}{2}} \biggr)^{-1}} \biggr]. $$

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

$$\begin{aligned} 0&< A!_{\mu}B \\ &\le A!_{\mu}B+2r(A\nabla B-A!B) \\ &\le A\nabla_{\mu}B \\ &\le A!_{\mu}B+2R(A\nabla B-A!B). \end{aligned}$$

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 $\mu\in [0,1]$,

$$ A\nabla_{\mu}B-2r(A\nabla B-A!B)\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K}(h,2)^{R}A!_{\mu}B, $$

(3.3)

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

$$\begin{aligned} &(1-\mu)+\mu x-2r \biggl[\frac{1+x}{2}- \biggl( {\frac{1}{2}+ \frac{1}{2}x^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{x} ,2 )^{{R}'}\mathrm{K}(x,2)^{R} \bigl( {(1-\mu )+\mu x^{-1}} \bigr)^{-1} \end{aligned}$$

for any $x>0$, and hence

$$\begin{aligned} &(1-\mu)I+\mu X-2r \biggl[\frac{I+X}{2}- \biggl( {\frac{1}{2}I+ \frac {1}{2}X^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K}(h,2)^{R} \bigl( {(1-\mu )I+\mu X^{-1}} \bigr)^{-1} \end{aligned}$$

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):

$$0< A!_{\mu}B\le A\nabla_{\mu}B-2r(A\nabla B-A!B)\le\mathrm{K} (\sqrt{h} ,2 )^{{R}'}\mathrm{K}(h,2)^{R}A!_{\mu}B. $$

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

$$\begin{aligned} &(1-\mu)+\mu a-2\mu \biggl[\frac{1+a}{2}- \biggl( {\frac{1}{2}+ \frac {1}{2}a^{-1}} \biggr)^{-1} \biggr]\\ &\quad\le\mathrm{K} (\sqrt{a} ,2 ) \bigl[(1-2\mu)+2\mu a^{-\frac {1}{2}} \bigr]^{-1} \end{aligned}$$

for any $a>0$, and hence

$$\begin{aligned} &(1-\mu)I+\mu X-2\mu \biggl[\frac{I+X}{2}- \biggl( {\frac{1}{2}I+ \frac {1}{2}X^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{h} ,2 ) \bigl[(1-2\mu)I+2\mu X^{-\frac {1}{2}} \bigr]^{-1} \end{aligned}$$

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

$$\begin{aligned} &(1-\mu)I+\mu A^{-\frac{1}{2}}BA^{-\frac{1}{2}} \\ &\qquad{}-2\mu \biggl[\frac{I+A^{-\frac {1}{2}}BA^{-\frac{1}{2}}}{2}- \biggl( {\frac{1}{2}I+\frac{1}{2} \bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}} \bigr)^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{h} ,2 ) \bigl[(1-2\mu)I+2\mu \bigl(A^{-\frac {1}{2}}BA^{-\frac{1}{2}} \bigr)^{-\frac{1}{2}} \bigr]^{-1}. \end{aligned}$$

(3.6)

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

$$\begin{aligned} &(1-\mu)b+\mu-2(1-\mu) \biggl[\frac{1+b}{2}- \biggl( {\frac{1}{2}+ \frac {1}{2}b^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{b} ,2 ) \bigl[(2\mu- 1)b^{-\frac {1}{2}}+(2-2\mu) \bigr]^{-1} \end{aligned}$$

for any $b>0$, and hence

$$\begin{aligned} &(1-\mu)Y+\mu I -2(1-\mu) \biggl[\frac{I+Y}{2}- \biggl( { \frac {1}{2}I+\frac{1}{2}Y^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{h} ,2 ) \bigl[(2\mu- 1)Y^{-\frac {1}{2}}+(2-2\mu)I \bigr]^{-1} \end{aligned}$$

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

$$\begin{aligned} &\mu I+(1-\mu)B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \\ &\qquad{}-2(1-\mu) \biggl[\frac {I+B^{-\frac{1}{2}}AB^{-\frac{1}{2}}}{2}- \biggl( {\frac{1}{2}I+ \frac{1}{2}I \bigl(B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \bigr)^{-1}} \biggr)^{-1} \biggr] \\ &\quad\le\mathrm{K} (\sqrt{h} ,2 ) \bigl[(2\mu- 1) \bigl(B^{-\frac {1}{2}}AB^{-\frac{1}{2}} \bigr)^{-\frac{1}{2}}+(2-2\mu)I \bigr]^{-1}. \end{aligned}$$

(3.7)

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

$$\begin{aligned} 0&< A!_{\mu}B \\ &\le A\nabla_{\mu}B-2r(A\nabla B-A!B) \\ &\le \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \mathrm{K} (\sqrt{h} ,2 )A!_{2\mu}(A\#B), & {0\leq\mu\le \frac{1}{2},}\\ \mathrm{K} (\sqrt{h} ,2 )B!_{2(1-\mu)} (A\#B),& {\frac{1}{2}< \mu \le1.} \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned}$$

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

$$\begin{aligned} &(1-\mu)+\mu b-2\mu \biggl[\frac{1+b}{2}- \biggl( {\frac{1}{2}+ \frac {1}{2}b^{-1}} \biggr)^{-1} \biggr]- \bigl[(1-2\mu)+2 \mu b^{-\frac{1}{2}} \bigr]^{-1} \\ &\quad\le2\max\{1-2\mu,2\mu\} \biggl[\frac{1}{2}+\frac{1}{2}b^{\frac {1}{2}}- \biggl(\frac{1}{2}+\frac{1}{2}b^{-\frac{1}{2}} \biggr)^{-1} \biggr] \end{aligned}$$

for any $b>0$, and hence

$$\begin{aligned} &(1-\mu)I+\mu X-2\mu \biggl[\frac{I+X}{2}- \biggl( {\frac{1}{2}I+ \frac {1}{2}X^{-1}} \biggr)^{-1} \biggr]- \bigl[(1-2 \mu)I+2\mu X^{-\frac{1}{2}} \bigr]^{-1} \\ &\quad\le2\max\{1-2\mu,2\mu\} \biggl[\frac{1}{2}I+\frac{1}{2}X^{\frac {1}{2}}- \biggl(\frac{1}{2}I+\frac{1}{2}X^{-\frac{1}{2}} \biggr)^{-1} \biggr] \end{aligned}$$

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

$$\begin{aligned} &(1-\mu)a+\mu-2(1-\mu) \biggl[\frac{a+1}{2}- \biggl( {\frac {1}{2}a^{-1}+ \frac{1}{2}} \biggr)^{-1} \biggr] \\ &\quad\le \bigl[(2\mu-1)+(2-2\mu) a^{-\frac{1}{2}} \bigr]^{-1} +2\max\{2-2\mu,2\mu-1\} \biggl[\frac{1}{2}+\frac{1}{2}a^{\frac {1}{2}}- \biggl(\frac{1}{2}+\frac{1}{2}a^{-\frac{1}{2}} \biggr)^{-1} \biggr] \end{aligned}$$

for any $a>0$, and hence

$$\begin{aligned} &(1-\mu)Y+\mu I-2(1-\mu) \biggl[\frac{Y+I}{2}- \biggl( { \frac {1}{2}Y^{-1}+\frac{1}{2}I} \biggr)^{-1} \biggr] \\ &\quad\le \bigl[(2\mu-1)I+(2-2\mu) Y^{-\frac{1}{2}} \bigr]^{-1} \\ &\qquad{}+2\max\{2-2\mu,2\mu-1\} \biggl[\frac{1}{2}I+\frac{1}{2}Y^{\frac {1}{2}}- \biggl(\frac{1}{2}I+\frac{1}{2}Y^{-\frac{1}{2}} \biggr)^{-1} \biggr] \end{aligned}$$

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

$$\begin{aligned} 0&< A!_{\mu}B \\ &\le A\nabla_{\mu}B-2r(A\nabla B-A!B) \\ &\le \left \{ \textstyle\begin{array}{@{}l@{\quad}l} A!_{2\mu}{(A\#B)}+2R'(A\nabla{(A\#B)}- A!{(A\#B)}), & 0\leq\mu \le\frac{1}{2},\\ B!_{2-2\mu}(A\#B)+2R'(B\nabla(A\#B)- B!(A\#B)),& \frac{1}{2}< \mu\le1. \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned}$$

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

$$ f \biggl(\frac{x+y}{2} \biggr)\le\frac{1}{y-x}\int _{x}^{y} f(t)\,dt\le\frac {f(x)+f(y)}{2}. $$

(4.1)

Theorem 4.1

Let $A, B\in{\mathcal{B}}^{++}(H)$. Then

$$\begin{aligned}& A\#(A\nabla B)\ge\frac{1}{3} \bigl[4A\nabla(A\# B)-(A\#B) \bigl(A\nabla(A\# B) \bigr)^{-1}A \bigr]\ge A\nabla(A\# B), \end{aligned}$$

(4.2)

$$\begin{aligned}& A\#(A!B)\le3 \bigl[4 \bigl(A! (A\# B) \bigr)^{-1}-(A \#B)^{-1} \bigl(A! (A\# B) \bigr)A^{-1} \bigr]^{-1}\le A! (A\# B). \end{aligned}$$

(4.3)

Proof

Applying the Hermite-Hadamard’s inequalities (4.1) to the convex function $f(t)=-t^{\frac{1}{2}}$, $t>0$, we have

$$\biggl(\frac{x+y}{2} \biggr)^{1/2}\ge\frac{1}{3} \biggl[4 \biggl(\frac {x^{1/2}+y^{1/2}}{2} \biggr)-(xy)^{1/2} \biggl(\frac {x^{1/2}+y^{1/2}}{2} \biggr)^{-1} \biggr]\ge\frac{ (x^{1/2}+y^{1/2} )}{2}, $$

and hence

$$\biggl(\frac{I+C}{2} \biggr)^{1/2}\ge\frac{1}{3} \biggl[4 \biggl(\frac {I+C^{1/2}}{2} \biggr)-(C)^{1/2} \biggl(\frac{I+C^{1/2}}{2} \biggr)^{-1} \biggr]\ge\frac{ (I+C^{1/2} )}{2} $$

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:

$$ f \biggl(\frac{a+b}{2} \biggr)\le\int_{0}^{1} f \bigl((1-t)a+tb \bigr)\,dt\le\frac{f(a)+f(b)}{2}. $$

(4.4)

A natural generalization of the classical Hermite-Hadamard inequality to Hermitian matrices could be the double inequality

$$f \biggl(\frac{A+B}{2} \biggr)\le\int_{0}^{1} f \bigl((1-t)A+tB \bigr)\,dt\le\frac{f(A)+f(B)}{2}. $$

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

$$ A\#(A\nabla B)\ge\int_{0}^{1} A \#(A\nabla_{t} B)\,dt\ge A\nabla(A\# B). $$

(4.5)

Proof

Taking $f(t)=-t^{\frac{1}{2}}$, $t>0$ in (4.4), then

$$\biggl(\frac{a+b}{2} \biggr)^{1/2}\ge\int_{0}^{1} \bigl((1-t)a+tb \bigr)^{1/2}\,dt\ge \frac{ (a^{1/2}+b^{1/2} )}{2}, $$

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 $v \in[0, 1]$,

$$ f \bigl((1-v)X+vY \bigr) \le(1-v)f(X)+vf(Y). $$

(4.6)

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 $\mathbb{I}$,

$$ f \bigl((1-v)X+vY \bigr)\ge(1-v)f(X)+vf(Y). $$

(4.7)

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

$$\begin{aligned} &A\#_{s}(A\nabla_{v} B)\le A\nabla_{v} (A \#_{s} B), \end{aligned}$$

(4.8)

$$\begin{aligned} &A\#_{s}(A!_{v} B)\ge A!_{v} (A\#_{s} B). \end{aligned}$$

(4.9)

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

$$\bigl[(1-v)X+vY \bigr]^{s}\le(1-v)X^{s}+vY^{s}. $$

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. □