# Some refinements of operator reverse AM-GM mean inequalities

## Abstract

In this paper, we prove the operator inequalities as follows: Let $$A,B$$ be positive operators on a Hilbert space with $$0 < m \le A,B \le M$$ and $$\sqrt{\frac{M}{m}} \le2.314$$. Then for every positive unital linear map Φ,

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le\frac{{ ( {M + m} )^{2} }}{{4Mm}} \Phi^{2} ( {A\mathrel{\sharp} B} )$$

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le\frac{{ ( {M + m} )^{2} }}{{4Mm}} \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr)^{2}.$$

Moreover, we prove Lin’s conjecture when $$\sqrt{\frac{M}{m}} \le 2.314$$.

## 1 Introduction

Let $$\mathcal{B(H)}$$ be the $$C^{*}$$-algebra of all bounded linear operators on a Hilbert space $$\mathcal{H}$$. Throughout this paper, a capital letter denotes an operator in $$\mathcal{B(H)}$$, we identify a scalar with the identity operator I multiplied by this scalar. We write $$A\ge0$$ to mean that the operator A is positive. A is said to be strictly positive (denoted by $$A>0$$ ) if it is a positive invertible operator. A linear map $$\Phi:\mathcal{B(H)} \to\mathcal {B(K)}$$ is called positive if $$A\ge0$$ implies $$\Phi(A)\ge0$$. It is said to be unital if $$\Phi(I)=I$$. For $$A,B>0$$, the geometric mean $$A\mathrel{\sharp} B$$ is defined by

$$A\mathrel{\sharp} B = A^{\frac{1}{2}} \bigl(A^{ - \frac{1}{2}} BA^{ - \frac {1}{2}} \bigr)^{{\frac{1}{2}} } A^{\frac{1}{2}}.$$

Let $$0 < m \le A,B \le M$$. Tominaga [1] showed that the following operator reverse AM-GM inequality holds:

$$\frac{{A + B}}{2} \le S ( h )A\mathrel{\sharp} B,$$
(1.1)

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

$$S ( h ) \le\frac{{ ( {M + m} )^{2} }}{{4Mm}} \le S^{2} ( h ) \quad ( {h \ge1} )$$
(1.2)

was observed by Lin [2, (3.3)].

Let Φ be a positive linear map and $$A,B > 0$$. Ando [3] gave the following inequality:

$$\Phi ( {A\mathrel{\sharp} B} ) \le\Phi ( A )\mathrel{\sharp} \Phi ( B ).$$
(1.3)

By (1.1), (1.2) and (1.3), it is easy to obtain the following inequalities:

$$\Phi \biggl( {\frac{{A + B}}{2}} \biggr) \le\frac{{ ( {M + m} )^{2} }}{{4Mm}}\Phi ( {A\mathrel{\sharp} B} )$$
(1.4)

and

$$\Phi \biggl( {\frac{{A + B}}{2}} \biggr) \le\frac{{ ( {M + m} )^{2} }}{{4Mm}} \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr).$$
(1.5)

Lin [2] proved that (1.4) and (1.5) can be squared:

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le \biggl({\frac{{ ( {M + m} )^{2} }}{{4Mm}}} \biggr)^{2} \Phi^{2} ( {A\mathrel{\sharp} B} )$$
(1.6)

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le \biggl({\frac{{ ( {M + m} )^{2} }}{{4Mm}}} \biggr)^{2} \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr)^{2}.$$
(1.7)

Meanwhile, Lin [2] conjectured that the following inequalities hold:

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le S^{2} ( h )\Phi ^{2} ( {A\mathrel{\sharp} B} )$$
(1.8)

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le S^{2} ( h ) \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr)^{2}.$$
(1.9)

In this paper, we will present some operator reverse AM-GM inequalities which are refinements of (1.1), (1.6) and (1.7). Furthermore, we will prove (1.8) and (1.9) if the condition number $$\sqrt{\frac{M}{m}}$$ is not too big.

## 2 Main results

We begin this section with the following lemmas.

### Lemma 1

([8])

Let $$A,B>0$$. Then the following norm inequality holds:

$$\Vert {AB} \Vert \le\frac{1}{4} \Vert {A+B} \Vert ^{2}.$$
(2.1)

### Lemma 2

([9])

Let $$A>0$$. Then for every positive unital linear map Φ,

$$\Phi\bigl(A^{-1}\bigr)\ge\Phi^{-1}(A).$$
(2.2)

### Theorem 1

If $$0 < m \le A,B \le M$$ for some scalars $$m\le M$$, then

$$\frac{{A + B}}{2} \le\frac{{M + m}}{{2\sqrt{Mm} }}A\mathrel{\sharp} B.$$
(2.3)

### Proof

Put $$C = A^{ - \frac{1}{2}} BA^{-\frac{1}{2}}$$. Since $$\frac{m}{M} \le C \le\frac{M}{m}$$, it follows that

$$\biggl[ {C^{\frac{1}{2}} - \frac{1}{2} \biggl( {\sqrt{\frac{m}{M}} + \sqrt{\frac{M}{m}} } \biggr)} \biggr]^{2} \le\frac{1}{4} \biggl( {\sqrt {\frac{M}{m}} - \sqrt{\frac{m}{M}} } \biggr)^{2},$$

and hence

$$C + 1 \le \biggl( {\sqrt{\frac{M}{m}} + \sqrt{\frac{m}{M}} } \biggr)C^{\frac{1}{2}}.$$

This implies

$$B + A \le \biggl( {\sqrt{\frac{M}{m}} + \sqrt{\frac{m}{M}} } \biggr)A\mathrel{\sharp} B.$$

Thus

$$\frac{{A + B}}{2} \le\frac{{M + m}}{{2\sqrt{Mm} }}A\mathrel{\sharp} B.$$

This completes the proof. □

### Remark 1

By (1.2), it is easy to know that (2.3) is tighter than (1.1).

### Theorem 2

If $$0 < m \le A,B \le M$$ and $$\sqrt{\frac {M}{m}} \le2.314$$ for some scalars $$m\le M$$, then

$$\biggl( {\frac{{A + B}}{2}} \biggr)^{2} \le \biggl( {\frac{{M + m}}{{2\sqrt {Mm} }}} \biggr)^{2} ( {A\mathrel{\sharp} B} )^{2}.$$
(2.4)

### Proof

Inequality (2.4) is equivalent to

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{M + m}}{{2\sqrt{Mm} }}.$$
(2.5)

If $$0 < m \le A,B \le\frac{{M + m}}{2}$$, we have

$$A + \frac{{M + m}}{2}mA^{ - 1} \le\frac{{M + m}}{2} + m$$
(2.6)

and

$$B + \frac{{M + m}}{2}mB^{ - 1} \le\frac{{M + m}}{2} + m.$$
(2.7)

Compute

\begin{aligned} \biggl\Vert {\frac{{A + B}}{2} \frac{{M + m}}{2}m ( {A\mathrel {\sharp} B} )^{ - 1} } \biggr\Vert &\le \frac{1}{4} \biggl\Vert {\frac{{A + B}}{2} + \frac{{M + m}}{2}m ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.1)}\bigr) \\ &\le \frac{1}{4} \biggl\Vert {\frac{{A + B}}{2} + \frac{{M + m}}{2}m \frac{{A^{ - 1} + B^{ - 1} }}{2}} \biggr\Vert ^{2} \\ &\le \frac{1}{4} \biggl( {\frac{{M + m}}{2} + m} \biggr)^{2} \quad\bigl(\mbox{by (2.6), (2.7)}\bigr). \end{aligned}

That is,

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{ ( {\frac{{M + m}}{2} + m} )^{2} }}{{4 \frac{{M + m}}{2} m}}.$$

Since $$1 \le\sqrt{\frac{M}{m}} \le2.314$$, it follows that

$$\biggl( {\sqrt{\frac{M}{m}} - 1} \biggr)^{2} \biggl[ { \biggl( {\sqrt{\frac {M}{m}} } \biggr)^{3} - \frac{{2M}}{m} + \sqrt{\frac{M}{m}} - 4} \biggr] \le0.$$
(2.8)

It is easy to know that $$\frac{{ ( {\frac{{M + m}}{2} + m} )^{2} }}{{4 \frac{{M + m}}{2} m}} \le\frac{{M + m}}{{2\sqrt{Mm} }}$$ is equivalent to (2.8).

Thus,

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{M + m}}{{2\sqrt{Mm} }}.$$

If $$\frac{{M + m}}{2} \le A,B \le M$$, we have

$$A + \frac{{M + m}}{2}MA^{ - 1} \le\frac{{M + m}}{2} + M$$
(2.9)

and

$$B + \frac{{M + m}}{2}MB^{ - 1} \le\frac{{M + m}}{2} + M.$$
(2.10)

Similarly, we get

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{ ( {\frac{{M + m}}{2} + M} )^{2} }}{{4 \frac{{M + m}}{2} M}} \le\frac{{ ( {\frac{{M + m}}{2} + m} )^{2} }}{{4 \frac{{M + m}}{2} m}} \le\frac{{M + m}}{{2\sqrt{Mm} }}.$$

If $$m \le A \le\frac{{M + m}}{2} \le B \le M$$, we have

\begin{aligned} \biggl\Vert {\frac{{A + B}}{2} \frac{{M + m}}{2} \sqrt{Mm} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert &\le \frac {1}{4} \biggl\Vert {\frac{{A + B}}{2} + \frac{{M + m}}{2}\sqrt{Mm} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.1)}\bigr) \\ &= \frac{1}{4} \biggl\Vert {\frac{{A + B}}{2} + \frac{{M + m}}{2} \bigl[ { \bigl( {mA^{ - 1} } \bigr)\mathrel{\sharp} \bigl({MB^{ - 1} } \bigr)} \bigr]} \biggr\Vert ^{2} \\ &\le \frac{1}{4} \biggl\Vert {\frac{{A + B}}{2} + \frac{{M + m}}{2} \frac{{mA^{ - 1} + MB^{ - 1} }}{2}} \biggr\Vert ^{2} \\ &\le \frac{1}{4} ( {M + m} )^{2} \quad\bigl(\mbox{by (2.6), (2.10)}\bigr). \end{aligned}

That is,

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{ ( {M + m} )^{2} }}{{4 \frac{{M + m}}{2} \sqrt{Mm} }} = \frac{{M + m}}{{2\sqrt{Mm} }}.$$

If $$m \le B \le\frac{{M + m}}{2} \le A \le M$$, similarly, by (2.1), (2.7) and (2.9), we have

$$\biggl\Vert {\frac{{A + B}}{2} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr\Vert \le\frac{{M + m}}{{2\sqrt{Mm} }}.$$

This completes the proof. □

### Theorem 3

Let Φ be a positive unital linear map. If $$0 < m \le A,B \le M$$ and $$\sqrt{\frac{M}{m}} \le2.314$$ for some scalars $$m\le M$$, then

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le \frac{{ ( {M + m} )^{2} }}{{4Mm}}\Phi^{2} ( {A\mathrel{\sharp} B} )$$
(2.11)

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le \frac{{ ( {M + m} )^{2} }}{{4Mm}} \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr)^{2}.$$
(2.12)

### Proof

Inequality (2.11) is equivalent to

$$\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert \le\frac{{M + m}}{{2\sqrt {Mm} }}.$$

If $$0 < m \le A,B \le\frac{{M + m}}{2}$$, compute

\begin{aligned} &\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\frac{{M + m}}{2}m\Phi^{ - 1} ( {A\mathrel{\sharp} B} )} \biggr\Vert \\ &\quad \le\frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr) + \frac{{M + m}}{2}m\Phi^{ - 1} ( {A\mathrel{\sharp} B} )} \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.1)}\bigr) \\ &\quad \le \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr) + \frac{{M + m}}{2}m\Phi \bigl( { ( {A\mathrel{\sharp} B} )^{- 1} } \bigr)} \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.2)}\bigr) \\ &\quad = \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2} + \frac{{M + m}}{2}m ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr)} \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2} + \frac {{M + m}}{2}m\frac{{A^{ - 1} + B^{ - 1} }}{2}} \biggr)} \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl( {\frac{{M + m}}{2} + m} \biggr)^{2} \quad\bigl(\mbox{by (2.6), (2.7)}\bigr). \end{aligned}

By $$1 \le\sqrt{\frac{M}{m}} \le2.314$$ and (2.8), we have

$$\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert \le\frac{{M + m}}{{2\sqrt {Mm} }}.$$

If $$0 < \frac{{M + m}}{2} \le A,B \le M$$, similarly, by (2.1), (2.2), (2.8), (2.9), (2.10) and $$\frac {{ ( {\frac{{M + m}}{2} + M} )^{2} }}{M} \le\frac{{ ({\frac{{M + m}}{2} + m} )^{2} }}{m}$$, we have

$$\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert \le\frac{{M + m}}{{2\sqrt {Mm} }}.$$

If $$m \le A \le\frac{{M + m}}{2} \le B \le M$$, we have

\begin{aligned} &\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\frac{{M + m}}{2}\sqrt{Mm} \Phi^{ - 1} ( {A\mathrel{\sharp} B} )} \biggr\Vert \\ &\quad \le\frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr) + \frac{{M + m}}{2}\sqrt{Mm} \Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.1)}\bigr) \\ &\quad \le \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr) + \frac{{M + m}}{2}\sqrt{Mm} \Phi \bigl( { ( {A\mathrel{\sharp} B} )^{ - 1} } \bigr)} \biggr\Vert ^{2} \quad\bigl(\mbox{by (2.2)}\bigr) \\ &\quad = \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2} + \frac{{M + m}}{2}\sqrt{Mm} ( {A\mathrel{\sharp} B} )^{ - 1} } \biggr)} \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2} + \frac {{M + m}}{2} \bigl( {mA^{ - 1} \mathrel{\sharp} MB^{ - 1} } \bigr)} \biggr)} \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} \biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2} + \frac {{M + m}}{2}\frac{{mA^{ - 1} + MB^{ - 1} }}{2}} \biggr)} \biggr\Vert ^{2} \\ &\quad \le \frac{1}{4} ( {M + m} )^{2} \quad\bigl(\mbox{by (2.6), (2.10)}\bigr). \end{aligned}

That is,

$$\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert \le\frac{{M + m}}{{2\sqrt {Mm} }}.$$

If $$m \le B \le\frac{{M + m}}{2} \le A \le M$$, similarly, by (2.1), (2.2), (2.7), (2.9), we have

$$\biggl\Vert {\Phi \biggl( {\frac{{A + B}}{2}} \biggr)\Phi^{ - 1} ( {A \mathrel{\sharp} B} )} \biggr\Vert \le\frac{{M + m}}{{2\sqrt {Mm} }}.$$

Thus (2.11) holds.

A and B are replaced by $$\Phi ( A )$$ and $$\Phi (B )$$ in (2.4), respectively, we get (2.12).

This completes the proof. □

### Remark 2

Since $$0 < m \le M$$, then $$\frac{{ ( {M + m} )^{2} }}{{4Mm}} \le [ {\frac{{ ( {M + m} )^{2} }}{{4Mm}}} ]^{2}$$. Thus (2.11) and (2.12) are refinements of (1.6) and (1.7), respectively, when $$\sqrt{\frac{M}{m}} \le2.314$$.

By (1.2) and Theorem 3, we know that Lin’s conjecture (1.8) and (1.9) hold when $$\sqrt{\frac{M}{m}} \le2.314$$.

### Corollary 1

Let Φ be a positive unital linear map. If $$0 < m \le A,B \le M$$ and $$\sqrt{\frac{M}{m}} \le2.314$$ for some scalars $$m\le M$$, then

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le S^{2} ( h ) \Phi ^{2} ( {A\mathrel{\sharp} B} )$$

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le S^{2} ( h ) \bigl( {\Phi ( A )\mathrel{\sharp} \Phi ( B )} \bigr)^{2},$$

where $$S ( h ) = \frac{{h^{\frac{1}{{h - 1}}} }}{{e\log h^{\frac{1}{{h - 1}}} }}$$, $$h = \frac{M}{m}$$.

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## Acknowledgements

This research was supported by the Scientific Research Fund of Yunnan Provincial Education Department (No. 2014Y645).

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Correspondence to Jianming Xue.

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Xue, J. Some refinements of operator reverse AM-GM mean inequalities. J Inequal Appl 2017, 283 (2017). https://doi.org/10.1186/s13660-017-1557-y