# Some reverse mean inequalities for operators and matrices

## Abstract

In this paper, we present some new reverse arithmetic–geometric mean inequalities for operators and matrices due to Lin (Stud. Math. 215:187–194, 2013). Among other inequalities, we prove that if $$A, B\in B(\mathcal{H})$$ are accretive and $$0< {mI}\le \Re (A), \Re (B)\le {MI}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \biggl(\Re \biggl(\frac{A+B}{2} \biggr) \biggr)\le \bigl(K(h) \bigr)^{2}\varPhi ^{2} \bigl(\Re (A\sharp B) \bigr), \end{aligned}

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$. Moreover, some reverse harmonic–geometric mean inequalities are also presented.

## Introduction

Throughout this paper, $$B(\mathcal{H})$$ stands for all bounded linear operators on a complex Hilbert space $$\mathcal{H}$$. In the finite-dimensional setting, $$\mathbb{M}_{n}$$ denotes the set of all $$n\times n$$ complex matrices. For $$A, B\in B(\mathcal{H})$$, we use $$A\ge B$$ ($$B\le A$$) to mean that $$A-B$$ is positive. The operator norm is denoted by $$\Vert \cdot \Vert$$. An operator $$A\in B(\mathcal{H})$$ is called accretive if in its Cartesian (or Toeptliz) decomposition, $$A=\Re A+i \Im A$$, A is positive, where $$\Re A=\frac{A+A^{*}}{2}$$, $$\Im A=\frac{A-A^{*}}{2i}$$. A linear map Φ: $$B(\mathcal{H}) \rightarrow B(\mathcal{H})$$ is called positive if $$\varPhi (A)\ge 0$$ whenever $$A \ge 0$$. If $$\varPhi (I)=I$$, where I denotes the identity operator, then we say that Φ is unital. We reserve M, m for scalars. In the finite-dimensional setting, we use $$I_{n}$$ for the identity.

The numerical range of $$A\in \mathbb{M}_{n}$$ is defined by

\begin{aligned}& W(A)= \bigl\{ x^{*}Ax : x\in \mathbb{C}^{n}, x^{*}x=1 \bigr\} . \end{aligned}

For $$\alpha \in [0,\frac{\pi }{2})$$, $$S_{\alpha }$$ denote the sector regions in the complex plane as follows:

\begin{aligned}& S_{\alpha }= \bigl\{ z\in \mathbb{C} : \Re z\ge 0, \vert \Im z \vert \le (\Re z) \tan \alpha \bigr\} . \end{aligned}

Clearly, A is positive semidefinite if and only if $$W(A)\subset S _{0}$$, and if $$W(A), W(B)\subset S_{\alpha }$$ for some $$\alpha \in [0,\frac{ \pi }{2})$$, then $$W(A+B)\subset S_{\alpha }$$. As $$0\notin S_{\alpha }$$, if $$W(A)\subset S_{\alpha }$$, then A is nonsingular. Moreover, $$W(A)\subset S_{\alpha }$$ implies $$W(X^{*}AX)\subset S_{\alpha }$$ for any nonzero $$n\times m$$ matrix X, thus $$W(A^{-1})\subset S_{\alpha }$$.

Recent studies on matrices with numerical ranges in a sector can be found in [3,4,5, 8,9,10, 14] and the references therein.

In , Tominaga presented an operator inequality as follows: Let A, B be positive operators on a Hilbert space with $$0< {mI}\le A,B \le {MI}$$, then

\begin{aligned} \frac{A+B}{2}\le S(h)A\sharp B, \end{aligned}
(1)

where $$S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}$$ is called Specht’s ratio and $$h=\frac{M}{m}$$.

Lin  found that, for a positive unital linear map Φ between $$C^{*}$$-algebra,

\begin{aligned} \varPhi \biggl( \frac{A+B}{2} \biggr)\le K(h)\varPhi (A\sharp B) \end{aligned}
(2)

due to (1) and the following observation :

\begin{aligned} S(h)\le K(h)\le S^{2}(h)\quad (h\ge 1), \end{aligned}

where $$K(h)=\frac{(h+1)^{2}}{4h}$$.

It is well known that, for two general positive operators (or positive definite matrices) AB,

\begin{aligned} A\ge B\quad \nRightarrow \quad A^{2}\ge B^{2}. \end{aligned}

However, Lin  showed that (2) can be squared as follows:

\begin{aligned} \varPhi ^{2} \biggl( \frac{A+B}{2} \biggr)\le K^{2}(h)\varPhi ^{2}(A\sharp B). \end{aligned}
(3)

Zhang  generalized (3) when $$p\ge 2$$ as follows:

\begin{aligned} \varPhi ^{2p} \biggl( \frac{A+B}{2} \biggr)\le \frac{(K(h)(M^{2}+m^{2}))^{2p}}{16M ^{2p}m^{2p}}\varPhi ^{2p}(A\sharp B). \end{aligned}
(4)

For two accretive operators $$A, B\in B(\mathcal{H})$$, Drury  defined the geometric mean of A and B as follows:

\begin{aligned} A\sharp B= \biggl( {\frac{2}{\pi } \int _{0}^{\infty } \bigl(tA+t^{-1}B \bigr)^{-1} \frac{dt}{t}} \biggr) ^{-1}. \end{aligned}
(5)

This new geometric mean defined by (5) possesses some similar properties compared to the geometric mean of two positive operators. For instance, $$A\sharp B=B\sharp A$$, $$(A\sharp B)^{-1}=A^{-1}\sharp B^{-1}$$. For more information about the geometric mean of two accretive operators, see . Moreover, if $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B)\subset S _{\alpha }$$, then $$W(A\sharp B)\subset S_{\alpha }$$.

Following an idea of Lin , we shall give some new reverse arithmetic–geometric mean inequalities for operators and matrices which can be seen as a complementary of (3) and (4). Moreover, some reverse harmonic–geometric mean inequalities are also presented.

## Main results

To reach our goal, we need the following lemmas.

### Lemma 2.1

()

If $$A, B\in B(\mathcal{H})$$ are accretive, then

\begin{aligned} \Re (A)\sharp \Re (B)\le \Re (A\sharp B). \end{aligned}

### Lemma 2.2

()

If $$A, B\in B(\mathcal{H})$$ are accretive, then

\begin{aligned} \Re \biggl( \biggl(\frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr)\ge \biggl( \frac{{\Re (A)}^{-1}+{\Re (B)}^{-1}}{2} \biggr)^{-1}. \end{aligned}

### Lemma 2.3

()

If $$A\in \mathbb{M}_{n}$$ has a positive definite real part, then

\begin{aligned} \Re \bigl(A^{-1} \bigr)\le \Re (A)^{-1}. \end{aligned}

### Lemma 2.4

()

If $$A\in \mathbb{M}_{n}$$ with $$W(A)\subset S _{\alpha }$$, then

\begin{aligned} \sec ^{2}(\alpha )\Re \bigl(A^{-1} \bigr)\ge \Re (A)^{-1}. \end{aligned}

It is easy to verify that $$\Re ( (\frac{A^{-1}+B^{-1}}{2} ) ^{-1} )\le \Re (A\sharp B)\le \Re (\frac{A+B}{2} )$$ does not persist for two accretive operators A and B. However, Lin presented the following extension of the arithmetic–geometric mean inequality.

### Lemma 2.5

()

Let $$A ,B\in \mathbb{M}_{n}$$ be such that $$W(A), W(B)\subset S_{\alpha }$$. Then

\begin{aligned} \Re (A\sharp B)\le \sec ^{2}(\alpha )\Re \biggl( \frac{A+B}{2} \biggr). \end{aligned}
(6)

### Lemma 2.6

()

Let $$A, B\in B(\mathcal{H})$$ be positive. Then

\begin{aligned} \Vert AB \Vert \le \frac{1}{4} \Vert A+B \Vert ^{2}. \end{aligned}

### Lemma 2.7

()

Let $$A\in B(\mathcal{H})$$ be positive. Then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{-1}(A)\le \varPhi \bigl(A^{-1} \bigr). \end{aligned}

### Lemma 2.8

()

Let $$A, B\in B(\mathcal{H})$$ be positive. Then, for $$1\le r<+\infty$$,

\begin{aligned} \bigl\Vert A^{r}+B^{r} \bigr\Vert \le \bigl\Vert (A+B)^{r} \bigr\Vert . \end{aligned}

An operator Kantorovich inequality obtained by Marshall and Olkin  reads as follows.

Let $$0< {mI}\le A\le {MI}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi \bigl(A^{-1} \bigr)\le K(h)\varPhi (A)^{-1}, \end{aligned}
(7)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

Lin  showed that (7) can be squared as follows:

\begin{aligned} \varPhi ^{2} \bigl(A^{-1} \bigr)\le \bigl(K(h) \bigr)^{2}\varPhi (A)^{-2}, \end{aligned}
(8)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

Let $$A\in \mathbb{M}_{n}$$ have a positive definite real part, $$0< {mI}_{n}\le \Re (A)\le {MI}_{n}$$ and Φ be a unital positive linear map. By (7) and Lemma 2.3, we can obtain the following inequality:

\begin{aligned} \varPhi \bigl(\Re \bigl(A^{-1} \bigr) \bigr)\le K(h) \varPhi \bigl(\Re (A) \bigr)^{-1}, \end{aligned}
(9)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

As an analog of inequality (8), we show that inequality (9) can be squared nicely as follows.

### Theorem 2.9

If $$A\in \mathbb{M}_{n}$$ has a positive definite real part and $$0< {mI}_{n}\le \Re (A)\le {MI}_{n}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \bigl(\Re \bigl(A^{-1} \bigr) \bigr) \le \bigl(K(h) \bigr)^{2}\varPhi \bigl(\Re (A) \bigr)^{-2}, \end{aligned}
(10)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

Since

\begin{aligned} {mI}_{n}\le \Re (A)\le {MI}_{n}, \end{aligned}

we have

\begin{aligned} \bigl( {MI}_{n}-\Re (A) \bigr) \bigl( {mI}_{n}- \Re (A) \bigr){\Re (A)}^{-1}\le 0, \end{aligned}

which is equivalent to

\begin{aligned} \Re (A)+Mm\Re (A)^{-1}\le (M+m)I_{n}. \end{aligned}
(11)

By Lemma 2.3 and (11), we get

\begin{aligned} &\Re (A)+Mm\Re \bigl(A^{-1} \bigr) \\ &\quad \le \Re (A)+Mm\Re (A)^{-1} \\ &\quad \le (M+m)I_{n}. \end{aligned}
(12)

Inequality (10) is equivalent to

\begin{aligned} \bigl\Vert \varPhi \bigl(\Re \bigl(A^{-1} \bigr) \bigr)\varPhi \bigl( \Re (A) \bigr) \bigr\Vert \le K(h). \end{aligned}

By computation, we have

\begin{aligned} & \bigl\Vert Mm\varPhi \bigl(\Re \bigl(A^{-1} \bigr) \bigr)\varPhi \bigl(\Re (A) \bigr) \bigr\Vert \\ &\quad \le \frac{1}{4} \bigl\Vert Mm\varPhi \bigl(\Re \bigl(A^{-1} \bigr) \bigr)+\varPhi \bigl(\Re (A) \bigr) \bigr\Vert ^{2}\quad \text{(by Lemma 2.6)} \\ &\quad \le \frac{1}{4}(M+m)^{2}\quad \bigl(\text{by (12)}\bigr). \end{aligned}

That is,

\begin{aligned} \bigl\Vert \varPhi \bigl(\Re \bigl(A^{-1} \bigr) \bigr)\varPhi \bigl( \Re (A) \bigr) \bigr\Vert \le K(h). \end{aligned}

This completes the proof. □

Let $$A, B\in B(\mathcal{H})$$ be accretive, $$0< {mI}\le \Re (A), \Re (B) \le {MI}$$ and Φ be a unital positive linear map. By inequality (2) and Lemma 2.1, we can obtain the following inequality:

\begin{aligned} \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\le K(h)\varPhi \bigl( \Re (A\sharp B) \bigr), \end{aligned}
(13)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

Following an idea of Lin , we give a squaring version of inequality (13) below.

### Theorem 2.10

If $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B) \subset S_{\alpha }$$ and $$0< {mI}_{n}\le \Re (A), \Re (B)\le {MI}_{n}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\le \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{2} \varPhi ^{2} \bigl(\Re (A\sharp B) \bigr), \end{aligned}
(14)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

From Theorem 2.9 we have

\begin{aligned} \frac{1}{2}\Re (A)+\frac{1}{2}Mm\Re (A)^{-1}\le \frac{1}{2}(M+m)I_{n} \end{aligned}
(15)

and

\begin{aligned} \frac{1}{2}\Re (B)+\frac{1}{2}Mm\Re (B)^{-1}\le \frac{1}{2}(M+m)I_{n}. \end{aligned}
(16)

Summing up inequalities (15) and (16), we get

\begin{aligned} \Re \biggl( \frac{A+B}{2} \biggr)+Mm \biggl( \frac{\Re (A)^{-1}+ \Re (B)^{-1}}{2} \biggr)\le (M+m)I_{n}. \end{aligned}
(17)

Inequality (14) is equivalent to

\begin{aligned} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le \sec ^{4}(\alpha )K(h). \end{aligned}

By computation, we have

\begin{aligned} & \biggl\Vert \sec ^{4}(\alpha )Mm\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm\varPhi ^{-1} \bigl(\Re (A \sharp B) \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.6)} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm\varPhi \bigl( \bigl(\Re (A\sharp B) \bigr)^{-1} \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.7)} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+\sec ^{2}(\alpha )Mm\varPhi \bigl(\Re \bigl(A^{-1} \sharp B^{-1} \bigr) \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.4)} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+\sec ^{4}(\alpha )Mm\varPhi \biggl(\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2}\quad \bigl(\text{by (6)}\bigr) \\ &\quad =\frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr)+Mm\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{4}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr)+Mm \biggl( \frac{\Re (A)^{-1}+\Re (B)^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.3)} \\ &\quad \le \frac{1}{4}\sec ^{8}(\alpha ) (M+m)^{2} \quad \bigl(\text{by (17)}\bigr). \end{aligned}

That is,

\begin{aligned} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le \sec ^{4}(\alpha )K(h). \end{aligned}

This completes the proof. □

Next we give a pth ($$p\ge 2$$) powering of inequality (14).

### Theorem 2.11

If $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B) \subset S_{\alpha }$$, $$0< {mI}_{n}\le \Re (A), \Re (B)\le {MI}_{n}$$, $$1<\beta \le 2$$ and $$p\ge 2\beta$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{p} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\le \frac{( \sec ^{2\beta }(\alpha )K(h)^{\frac{\beta }{2}}(M^{\beta }+m^{\beta }))^{\frac{2p}{ \beta }}}{16M^{p}m^{p}}\varPhi ^{p} \bigl(\Re (A\sharp B) \bigr), \end{aligned}
(18)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

Since

\begin{aligned} {mI}_{n}\le \varPhi \biggl(\Re \biggl(\frac{A+B}{2} \biggr) \biggr)\le {MI} _{n}, \end{aligned}

we have

\begin{aligned} M^{\beta }m^{\beta }\varPhi ^{-{\beta }} \biggl( \Re \biggl(\frac{A+B}{2} \biggr) \biggr)+ \varPhi ^{\beta } \biggl(\Re \biggl(\frac{A+B}{2} \biggr) \biggr)\le M^{ \beta }+m^{\beta }. \end{aligned}
(19)

By (14) and the L-H inequality , we obtain

\begin{aligned} \varPhi ^{-{\beta }} \bigl(\Re (A\sharp B) \bigr)\le \bigl( \sec ^{4}(\alpha )K(h) \bigr)^{\beta } \varPhi ^{-{\beta }} \biggl( \Re \biggl( \frac{A+B}{2} \biggr) \biggr). \end{aligned}
(20)

Inequality (18) is equivalent to

\begin{aligned} \biggl\Vert \varPhi ^{\frac{p}{2}} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-{\frac{p}{2}}} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le \frac{( \sec ^{2\beta }(\alpha )K(h)^{\frac{\beta }{2}}(M^{\beta }+m^{\beta }))^{\frac{p}{ \beta }}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}. \end{aligned}

By computation, we have

\begin{aligned} & \biggl\Vert M^{\frac{p}{2}}m^{\frac{p}{2}}\varPhi ^{\frac{p}{2}} \biggl( \Re \biggl( \frac{A+B}{2} \biggr) \biggr)\varPhi ^{-{\frac{p}{2}}} \bigl(\Re (A \sharp B) \bigr) \biggr\Vert \\ &\quad \le \frac{1}{4} \biggl\Vert \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{\frac{p}{4}} \varPhi ^{\frac{p}{2}} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+ \biggl(\frac{M^{2}m^{2}}{\sec ^{4}(\alpha )K(h)} \biggr)^{\frac{p}{4}} \varPhi ^{-{\frac{p}{2}}} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{ \frac{\beta }{2}}\varPhi ^{\beta } \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+ \biggl(\frac{M^{2}m^{2}}{\sec ^{4}(\alpha )K(h)} \biggr)^{\frac{\beta }{2}} \varPhi ^{-{\beta }} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert ^{\frac{p}{\beta }} \\ &\quad \le \frac{1}{4} \biggl\Vert \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{ \frac{\beta }{2}}\varPhi ^{\beta } \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \\ &\quad\quad{} + \bigl( \sec ^{4}(\alpha )K(h) \bigr)^{\frac{\beta }{2}}M^{\beta }m^{\beta } \varPhi ^{-{\beta }} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \biggr\Vert ^{\frac{p}{\beta }} \\ &\quad =\frac{1}{4} \biggl\Vert \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{\frac{\beta }{2}} \biggl(\varPhi ^{\beta } \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+M ^{\beta }m^{\beta }\varPhi ^{-{\beta }} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \biggr) \biggr\Vert ^{\frac{p}{\beta }} \\ &\quad \le \frac{1}{4} \bigl(\sec ^{2\beta }(\alpha )K(h)^{\frac{\beta }{2}} \bigl(M ^{\beta }+m^{\beta } \bigr) \bigr)^{\frac{p}{\beta }}, \end{aligned}

where the first inequality is by Lemma 2.6, the second one is by Lemma 2.8, the third one is by (20) and the last one is by (19).

That is,

\begin{aligned} \biggl\Vert \varPhi ^{\frac{p}{2}} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-{\frac{p}{2}}} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le \frac{( \sec ^{2\beta }(\alpha )K(h)^{\frac{\beta }{2}}(M^{\beta }+m^{\beta }))^{\frac{p}{ \beta }}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}. \end{aligned}

This completes the proof. □

We are not satisfied with the factor $$(\sec ^{4}(\alpha )K(h))^{2}$$ in Theorem 2.10, the ideal factor should be $$(K(h))^{2}$$. We shall prove it in the following theorem.

### Theorem 2.12

If $$A, B\in B(\mathcal{H})$$ are accretive and $$0< {mI}\le \Re (A), \Re (B)\le {MI}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\le \bigl(K(h) \bigr)^{2} \varPhi ^{2} \bigl( \Re (A\sharp B) \bigr), \end{aligned}
(21)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

From Theorem 2.10 one can get

\begin{aligned} \Re \biggl( \frac{A+B}{2} \biggr)+Mm \biggl( \frac{\Re (A)^{-1}+ \Re (B)^{-1}}{2} \biggr)\le (M+m)I. \end{aligned}
(22)

Inequality (21) is equivalent to

\begin{aligned} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le K(h). \end{aligned}

By computation, we have

\begin{aligned} & \biggl\Vert Mm\varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \\ &\quad \le \frac{1}{4} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.6)} \\ &\quad \le \frac{1}{4} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm \varPhi \bigl( \bigl(\Re (A\sharp B) \bigr)^{-1} \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.7)} \\ &\quad \le \frac{1}{4} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm \varPhi \bigl( \bigl(\Re (A)\sharp \Re (B) \bigr)^{-1} \bigr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.1)} \\ &\quad =\frac{1}{4} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm \varPhi \bigl(\Re (A)^{-1}\sharp \Re (B)^{-1} \bigr) \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+Mm \varPhi \biggl( \frac{\Re (A)^{-1}+\Re (B)^{-1}}{2} \biggr) \biggr\Vert ^{2}\quad \text{(by AM-GM inequality)} \\ &\quad \le \frac{1}{4}(M+m)^{2}\quad\bigl(\text{by (22)}\bigr). \end{aligned}

That is,

\begin{aligned} \biggl\Vert \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr) \varPhi ^{-1} \bigl(\Re (A\sharp B) \bigr) \biggr\Vert \le K(h). \end{aligned}

This completes the proof. □

### Remark 2.13

Letting $$A, B\ge 0$$ in Theorem 2.12, inequality (21) coincides with inequality (3).

Next we give a pth ($$p\ge 2$$) powering of inequality (21) along the same line as in Theorem 2.11.

### Theorem 2.14

If $$A, B\in B(\mathcal{H})$$ are accretive and $$0< {mI}\le \Re (A), \Re (B)\le {MI}$$ $$1<\beta \le 2$$ and $$p\ge 2\beta$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{p} \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)\le \frac{(K(h)^{\frac{ \beta }{2}}(M^{\beta }+m^{\beta }))^{\frac{2p}{\beta }}}{16M^{p}m^{p}} \varPhi ^{p} \bigl(\Re (A\sharp B) \bigr), \end{aligned}
(23)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Remark 2.15

Letting $$A, B\ge 0$$ and $$\beta =2$$ in Theorem 2.14, inequality (23) coincides with inequality (4).

The following theorem corrects Theorem 1.2 of Liu et al. .

### Theorem 2.16

Let $$A, B\in \mathbb{M}_{n}$$ be such that $$W(A),W(B)\subset S_{\alpha }$$, then

\begin{aligned} \Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \le \sec ^{4}(\alpha )\Re (A\sharp B). \end{aligned}
(24)

### Proof

We can get

\begin{aligned} \biggl( \Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr)^{-1} \le \sec ^{2}(\alpha ) \bigl( \Re \bigl(A^{-1} \bigr)\sharp \Re \bigl(B^{-1} \bigr) \bigr) ^{-1} \end{aligned}
(25)

along the same line as Liu et al. did in  by Lemma 2.1 and Lemma 2.5.

Thus we have

\begin{aligned} \Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) &\le \biggl( \Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr)^{-1} \quad \text{(by Lemma 2.3)} \\ &\le \sec ^{2}(\alpha ) \bigl( \Re \bigl(A^{-1} \bigr)\sharp \Re \bigl(B^{-1} \bigr) \bigr) ^{-1} \quad \bigl(\text{by (25)}\bigr) \\ &=\sec ^{2}(\alpha ) \bigl(\Re \bigl(A^{-1} \bigr) \bigr)^{-1}\sharp \bigl( \Re \bigl(B^{-1} \bigr) \bigr)^{-1} \\ &\le \sec ^{4}(\alpha ) \bigl(\Re (A)\sharp \Re (B) \bigr)\quad \text{(by Lemma 2.4)} \\ &\le \sec ^{4}(\alpha )\Re (A\sharp B)\quad \text{(by Lemma 2.1).} \end{aligned}

This completes the proof. □

### Remark 2.17

Maybe it is just a clerical error in Theorem 1.2 of their work . However, the authors present the following inequalities in their proof:

\begin{aligned} \Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) &\le \sec ^{2}(\alpha ) \bigl( \Re \bigl({A^{-1}} \bigr) \bigr)^{-1}\sharp \bigl( \Re \bigl({B^{-1}} \bigr) \bigr)^{-1} \\ &\le \sec ^{2}(\alpha ) \bigl(\Re (A)\sharp \Re (B) \bigr). \end{aligned}

Obviously, such a deduction in their proof collapses given the property of geometric mean for positive definite matrices. Thus we give Theorem 2.16 and the proof.

Let $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B)\subset S_{\alpha }$$, $$0< {mI}_{n}\le \Re (A^{-1}),\Re (B^{-1})\le {MI}_{n}$$ and Φ be a unital positive linear map. As a complement of inequalities (13) and (24), we have the following reverse harmonic–geometric mean inequality:

\begin{aligned} \varPhi \bigl(\Re (A\sharp B) \bigr)\le \sec ^{2}( \alpha )K(h)\varPhi \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr), \end{aligned}
(26)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

Compute

\begin{aligned} \Re (A\sharp B) &=\Re \bigl( \bigl(A^{-1}\sharp B^{-1} \bigr)^{-1} \bigr) \\ &\le \Re \bigl(A^{-1}\sharp B^{-1} \bigr)^{-1} \\ &\le K(h)\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \\ &\le \sec ^{2}(\alpha )K(h)\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr), \end{aligned}

in which the first inequality is by Lemma 2.3, the second one is by inequality (13) and the last one is by Lemma 2.4.

Imposing Φ on both sides of the inequalities above, we thus obtain inequality (26). □

As an analog of Theorem 2.12, we shall present a squaring version of inequality (26).

### Theorem 2.18

If $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B) \subset S_{\alpha }$$ and $$0< {mI}_{n}\le \Re (A^{-1}), \Re (B^{-1}) \le {MI}_{n}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \bigl(\Re (A\sharp B) \bigr)\le \bigl(\sec ^{4}(\alpha )K(h) \bigr)^{2}\varPhi ^{2} \biggl( \Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr), \end{aligned}
(27)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

From Theorem 2.10 we have

\begin{aligned} \frac{1}{2}\Re \bigl(A^{-1} \bigr)+ \frac{1}{2}Mm\Re \bigl(A^{-1} \bigr)^{-1}\le \frac{1}{2}(M+m)I _{n} \end{aligned}
(28)

and

\begin{aligned} \frac{1}{2}\Re \bigl(B^{-1} \bigr)+ \frac{1}{2}Mm\Re \bigl(B^{-1} \bigr)^{-1}\le \frac{1}{2}(M+m)I _{n}. \end{aligned}
(29)

Summing up inequalities (28) and (29), we get

\begin{aligned} &\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)+Mm\Re \biggl(\frac{A+B}{2} \biggr) \\ &\quad \le \Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)+Mm \biggl( \frac{ \Re (A^{-1})^{-1}+\Re (B^{-1})^{-1}}{2} \biggr) \\ &\quad \le (M+m)I_{n}. \end{aligned}

Inequality (27) is equivalent to

\begin{aligned} \biggl\Vert \varPhi \bigl(\Re (A\sharp B) \bigr)\varPhi ^{-1} \biggl( \Re \biggl( \biggl( \frac{A ^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \le \sec ^{4}(\alpha )K(h). \end{aligned}

By computation, we have

\begin{aligned} & \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)\varPhi ^{-1} \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\varPhi ^{-1} \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.6)} \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\varPhi \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr)^{-1} \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.7)} \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\sec ^{2}(\alpha )\varPhi \biggl(\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.4)} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{2}(\alpha )Mm \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+\sec ^{2}( \alpha )\varPhi \biggl(\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2}\quad \bigl(\text{by (6)}\bigr) \\ &\quad =\frac{1}{4} \biggl\Vert \sec ^{2}(\alpha )\varPhi \biggl(Mm\Re \biggl( \frac{A+B}{2} \biggr)+\Re \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) \biggr) \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4}\sec ^{4}(\alpha ) (M+m)^{2}. \end{aligned}

That is,

\begin{aligned} \biggl\Vert \varPhi \bigl(\Re (A\sharp B) \bigr)\varPhi ^{-1} \biggl( \Re \biggl( \biggl( \frac{A ^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \le \sec ^{4}(\alpha )K(h). \end{aligned}

This completes the proof. □

Obviously, the optimal factor in Theorem 2.18 should be $$(\sec ^{2}( \alpha )K(h))^{2}$$. We note that it is affirmative under the condition $${mI}_{n}\le \Re (A^{-1})\le \Re (A)^{-1}\le {MI}_{n}$$ and $${mI}_{n}\le \Re (B^{-1})\le \Re (B)^{-1}\le {MI}_{n}$$ by presenting the following theorem.

### Theorem 2.19

If $$A, B\in \mathbb{M}_{n}$$ with $$W(A), W(B) \subset S_{\alpha }$$ and $$0< {mI}_{n}\le \Re (A)^{-1}, \Re (B)^{-1} \le {MI}_{n}$$, then, for every positive unital linear map Φ,

\begin{aligned} \varPhi ^{2} \bigl(\Re (A\sharp B) \bigr)\le \bigl(\sec ^{2}(\alpha )K(h) \bigr)^{2}\varPhi ^{2} \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr), \end{aligned}
(30)

where $$K(h)=\frac{(h+1)^{2}}{4h}$$ and $$h=\frac{M}{m}$$.

### Proof

From Theorem 2.10 we obtain

\begin{aligned} \frac{{\Re (A)}^{-1}+{\Re (B)}^{-1}}{2}+Mm\Re \biggl( \frac{A+B}{2} \biggr) \le (M+m)I_{n}. \end{aligned}
(31)

Inequality (30) is equivalent to

\begin{aligned} \biggl\Vert \varPhi \bigl(\Re (A\sharp B) \bigr)\varPhi ^{-1} \biggl( \Re \biggl( \biggl( \frac{A ^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \le \sec ^{2}(\alpha )K(h). \end{aligned}

By computation, we have

\begin{aligned} & \biggl\Vert \sec ^{2}(\alpha )Mm\varPhi \bigl(\Re (A\sharp B) \bigr) \varPhi ^{-1} \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\sec ^{2}(\alpha )\varPhi ^{-1} \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr) ^{-1} \biggr) \biggr) \biggr\Vert ^{2} \quad \text{(by Lemma 2.6)} \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\sec ^{2}(\alpha )\varPhi \biggl(\Re \biggl( \biggl( \frac{A^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) ^{-1} \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.7)} \\ &\quad \le \frac{1}{4} \biggl\Vert Mm\varPhi \bigl(\Re (A\sharp B) \bigr)+\sec ^{2}(\alpha )\varPhi \biggl(\frac{{\Re (A)}^{-1}+{\Re (B)}^{-1}}{2} \biggr) \biggr\Vert ^{2}\quad \text{(by Lemma 2.2)} \\ &\quad \le \frac{1}{4} \biggl\Vert \sec ^{2}(\alpha )Mm \varPhi \biggl(\Re \biggl( \frac{A+B}{2} \biggr) \biggr)+\sec ^{2}( \alpha )\varPhi \biggl(\frac{ {\Re (A)}^{-1}+{\Re (B)}^{-1}}{2} \biggr) \biggr\Vert ^{2} \quad \bigl(\text{by (6)}\bigr) \\ &\quad =\frac{1}{4} \biggl\Vert \sec ^{2}(\alpha )\varPhi \biggl(Mm\Re \biggl( \frac{A+B}{2} \biggr)+\frac{{\Re (A)}^{-1}+{\Re (B)}^{-1}}{2} \biggr) \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4}\sec ^{4}(\alpha ) (M+m)^{2} \quad \bigl(\text{by (31)}\bigr). \end{aligned}

That is,

\begin{aligned} \biggl\Vert \varPhi \bigl(\Re (A\sharp B) \bigr)\varPhi ^{-1} \biggl( \Re \biggl( \biggl( \frac{A ^{-1}+B^{-1}}{2} \biggr)^{-1} \biggr) \biggr) \biggr\Vert \le \sec ^{2}(\alpha )K(h). \end{aligned}

This completes the proof. □

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## Funding

This research is supported by the National Natural Science Foundation of P.R. China (No. 11571247).

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All authors contributed almost the same amount of work to manuscript. All authors read and approved the final manuscript.

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Correspondence to Fangyan Lu.

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