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 [4–7].

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