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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 [13], 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 [6] 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 [6]:

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

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 [6] 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 [15] 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 [3] 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 [3]. 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 [6], 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

([10])

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

([10])

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

([9])

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

([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

([9])

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

([2])

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

([1])

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

([1])

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 [12] 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 [7] 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 [6], 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 [1], 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. [11].

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 [11] 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 [11]. 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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Yang, C., Gao, Y. & Lu, F. Some reverse mean inequalities for operators and matrices. J Inequal Appl 2019, 115 (2019). https://doi.org/10.1186/s13660-019-2070-2

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MSC

  • 47A63
  • 15A15

Keywords

  • Positive linear maps
  • Arithmetic–geometric–harmonic mean
  • Sector matrix
  • Inequality