Open Access

Some refinements of operator reverse AM-GM mean inequalities

Journal of Inequalities and Applications20172017:283

https://doi.org/10.1186/s13660-017-1557-y

Received: 19 September 2017

Accepted: 31 October 2017

Published: 14 November 2017

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 \(\sqrt{\frac{M}{m}} \le 2.314\).

Keywords

operator inequalitiesreverse AM-GM means inequalitiespositive linear maps

MSC

47A6347A30

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)

For more information on operator inequalities, the reader is referred to [47].

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

Declarations

Acknowledgements

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

Authors’ contributions

The author read and approved the final manuscript.

Competing interests

The author declares that she has no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Oxbridge College, Kunming University of Science and Technology

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