# Weighted arithmetic–geometric operator mean inequalities

## Abstract

In this paper, we refine and generalize some weighted arithmetic–geometric operator mean inequalities due to Lin (Stud. Math. 215:187–194, 2013) and Zhang (Banach J. Math. Anal. 9:166–172, 2015) as follows: Let A and B be positive operators. If $$0< m\le A\le{m}'<{M}'\le B\le M$$ or $$0< m\le B\le{m}'<{M}'\le A\le M$$, then for a positive unital linear map Φ,

$$\begin{gathered} \Phi^{2} ( {A\nabla_{\alpha}B} ) \le \biggl[ { \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}} \biggr]^{2} \Phi^{2} ( {A \sharp_{\alpha}B} ), \\ \Phi^{2} ( {A\nabla_{\alpha}B} ) \le \biggl[ { \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}} \biggr]^{2} \bigl[ {\Phi ( A )\sharp_{\alpha}\Phi ( B )} \bigr]^{2}, \\ \Phi^{2p} ( {A\nabla_{\alpha}B} ) \le\frac{1}{{16}} \biggl[ { \frac{{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{p} \Phi^{2p} ( {A\sharp _{\alpha}B} ), \\ \Phi^{2p} ( {A\nabla_{\alpha}B} ) \le\frac{1}{{16}} \biggl[ { \frac{{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{p} \bigl[ {\Phi ( A )\sharp_{\alpha}\Phi ( B )} \bigr]^{2p},\end{gathered}$$

where $$\alpha \in [ {0,1} ]$$, $$K ( h ) = \frac {{ ( {h + 1} )^{2} }}{{4h}}$$, $$S(h') = \frac{{h^{\prime \frac{1}{{h' - 1}}} }}{{e\log h^{\prime \frac{1}{{h' - 1}}} }}$$, $$h = \frac{M}{m}$$, $$h' = \frac{{M'}}{{m'}}$$, $$r = \min \{ {\alpha,1 - \alpha} \}$$ and $$p \ge2$$.

## 1 Introduction

Let $$\mathcal{B(H)}$$ be the $$C^{*}$$-algebra of all bounded linear operators on a Hilbert space $$(\mathcal{H},\langle\cdot,\cdot\rangle)$$ and I be the identity operator. $$\Vert \cdot \Vert$$ is the operator norm. $$A\ge0$$ ($$A>0$$) implies that A is a positive (strictly positive) 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 α-weighted arithmetic mean and α-weighted geometric mean of A and B are defined, respectively, by

$$A\nabla_{\alpha}B = ( {1 - \alpha} )A + \alpha B,\qquad A\sharp _{\alpha}B = A^{\frac{1}{2}} \bigl( {A^{ - \frac{1}{2}} BA^{ - \frac {1}{2}} } \bigr)^{\alpha}A^{\frac{1}{2}},$$

where $$\alpha \in [ {0,1} ]$$. When $$\alpha = \frac{1}{2}$$, we write $$A\nabla B$$ and $$A\sharp B$$ for brevity for $$A\nabla_{\frac {1}{2}} B$$ and $$A\sharp_{\frac{1}{2}} B$$, respectively.

Let $$0 < m\le A,B \le M$$, and Φ be a positive unital linear map. Tominaga  showed that the following operator inequality holds:

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

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

$$S(h) \le K(h) = \frac{{ ( {h + 1} )^{2} }}{{4h}} \le S^{2} (h)\quad (h \ge1)$$
(1.2)

was observed by Lin [1, (3.3)].

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

$$\Phi \biggl( {\frac{{A + B}}{2}} \biggr) \le K ( h )\Phi ( {A \sharp B} ).$$
(1.3)

Lin [1, Theorem 2.1] proved that (1.3) can be squared as follows:

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le K^{2} ( h )\Phi ^{2} ( {A\sharp B} )$$
(1.4)

and

$$\Phi^{2} \biggl( {\frac{{A + B}}{2}} \biggr) \le K^{2} ( h ) \bigl[ {\Phi ( A )\sharp\Phi ( B )} \bigr]^{2}.$$
(1.5)

Zhang  generalized (1.4) and (1.5) when $$p \ge2$$

$$\Phi^{2p} \biggl( {\frac{{A + B}}{2}} \biggr) \le \frac{{ [ {K ( h ) ( {M^{2} + m^{2} } )} ]^{2p} }}{{16M^{2p} m^{2p} }}\Phi^{2p} ( {A\sharp B} )$$
(1.6)

and

$$\Phi^{2p} \biggl( {\frac{{A + B}}{2}} \biggr) \le \frac{{ [ {K ( h ) ( {M^{2} + m^{2} } )} ]^{2p} }}{{16M^{2p} m^{2p} }} \bigl[ {\Phi ( A )\sharp\Phi ( B )} \bigr]^{2p}.$$
(1.7)

A great number of results on operator inequalities have been given in the literature, for example, see  and the references therein.

In this paper, motivated by the aforementioned discussion, we extend (1.4)–(1.7) to the weighted arithmetic–geometric mean. In order to prove our results, we show a new operator weighted arithmetic–geometric mean inequality. Manipulating this operator inequality enables us to refine and generalize (1.4)–(1.7). Furthermore, a numerical example is given to demonstrate the effectiveness of the theoretical results.

## 2 Main results

In this section, the main results of this paper will be given. To do this, the following lemmas are necessary.

### Lemma 1

()

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

()

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

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

### Lemma 3

()

Let $$A,B>0$$. Then for $$1\le r<\infty$$,

$$\bigl\Vert {A^{r}+B^{r}} \bigr\Vert \le \bigl\Vert {(A+B)^{r}} \bigr\Vert .$$
(2.3)

### Lemma 4

()

Let $$0< m\le A\le{m}'<{M}'\le B\le M$$ or $$0< m\le B\le{m}'<{M}'\le A\le M$$. Then for each $$\alpha \in [ {0,1} ]$$,

$$A\nabla_{\alpha}B \ge S \bigl( {h^{\prime r} } \bigr)A\sharp_{\alpha}B,$$
(2.4)

where $$S(h') = \frac{{h^{\prime \frac{1}{{h' - 1}}} }}{{e\log h^{\prime \frac {1}{{h' - 1}}} }}$$, $$h' = \frac{{M'}}{{m'}}$$ and $$r = \min \{ {\alpha,1 - \alpha} \}$$.

### Theorem 1

Let $$0< m\le A\le{m}'<{M}'\le B\le M$$ or $$0< m\le B\le{m}'<{M}'\le A\le M$$. Then for each $$\alpha \in [ {0,1} ]$$,

$$A\nabla_{\alpha}B + MmS \bigl( {h^{\prime r} } \bigr) ( {A\sharp_{\alpha}B} )^{ - 1} \le M + m,$$
(2.5)

where $$S(h') = \frac{{h^{\prime \frac{1}{{h' - 1}}} }}{{e\log h^{\prime \frac {1}{{h' - 1}}} }}$$, $$h' = \frac{{M'}}{{m'}}$$ and $$r = \min \{ {\alpha,1 - \alpha} \}$$.

### Proof

Since

$$0 < m \le A \le M,$$

then

$$( {1 - \alpha} ) ( {M - A} ) ( {m - A} )A^{ - 1} \le0.$$

That is,

$$( {1 - \alpha} ) \bigl( {A + MmA^{ - 1} } \bigr) \le ( {1 - \alpha} ) ( {M + m} ).$$
(2.6)

Similarly, we get

$$\alpha \bigl( {B + MmB^{ - 1} } \bigr) \le\alpha ( {M + m} ).$$
(2.7)

Summing up inequalities (2.6) and (2.7), we get

$$A\nabla_{\alpha}B + MmA^{ - 1} \nabla_{\alpha}B^{ - 1} \le M + m.$$

By $$( {A\sharp_{\alpha}B} )^{ - 1} = A^{ - 1} \sharp _{\alpha}B^{ - 1}$$ and (2.4), we have

\begin{aligned} A\nabla_{\alpha}B + MmS \bigl( {h^{\prime r} } \bigr) ( {A\sharp _{\alpha}B} )^{ - 1} &\le A\nabla_{\alpha}B + MmA^{ - 1} \nabla_{\alpha}B^{ - 1} \\ &\le M + m. \end{aligned}

This completes the proof. □

### Theorem 2

Let Φ be a positive unital linear map and let A and B be positive operators. If $$0< m\le A\le{m}'<{M}'\le B\le M$$ or $$0< m\le B\le{m}'<{M}'\le A\le M$$, then for each $$\alpha \in [ {0,1} ]$$,

\begin{aligned}& \Phi^{2} ( {A\nabla_{\alpha}B} ) \le \biggl[ { \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}} \biggr]^{2} \Phi^{2} ( {A \sharp_{\alpha}B} ), \end{aligned}
(2.8)
\begin{aligned}& \Phi^{2} ( {A\nabla_{\alpha}B} ) \le \biggl[ { \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}} \biggr]^{2} \bigl[ {\Phi ( A )\sharp_{\alpha}\Phi ( B )} \bigr]^{2}, \end{aligned}
(2.9)

where $$K ( h ) = \frac{{ ( {h + 1} )^{2} }}{{4h}}$$, $$S(h') = \frac{{h^{\prime \frac{1}{{h' - 1}}} }}{{e\log h^{\prime \frac{1}{{h' - 1}}} }}$$, $$h = \frac{M}{m}$$, $$h' = \frac{{M'}}{{m'}}$$ and $$r = \min \{ {\alpha,1 - \alpha} \}$$.

### Proof

Inequality (2.8) is equivalent to

$$\bigl\Vert {\Phi ( {A\nabla_{\alpha}B} )\Phi^{ - 1} ( {A \sharp_{\alpha}B} )} \bigr\Vert \le\frac{{K ( h )}}{{S ( {h^{\prime r} } )}}.$$

By (2.1), (2.2) and (2.5), we have

\begin{aligned} \bigl\Vert {\Phi ( {A\nabla_{\alpha}B} )MmS \bigl( {h^{\prime r} } \bigr)\Phi^{ - 1} ( {A \sharp_{\alpha}B} )} \bigr\Vert &\le \frac{1}{4} \bigl\Vert { \Phi ( {A\nabla_{\alpha}B} ) + MmS \bigl( {h^{\prime r} } \bigr)\Phi^{ - 1} ( {A\sharp_{\alpha}B} )} \bigr\Vert ^{2} \\ &\le \frac{1}{4} \bigl\Vert {\Phi ( {A\nabla_{\alpha}B} ) + MmS \bigl( {h^{\prime r} } \bigr)\Phi \bigl[ { ( {A \sharp_{\alpha}B} )^{ - 1} } \bigr]} \bigr\Vert ^{2} \\ &= \frac{1}{4} \bigl\Vert {\Phi \bigl[ { ( {A\nabla_{\alpha}B} ) + MmS \bigl( {h^{\prime r} } \bigr) ( {A\sharp_{\alpha}B} )^{ - 1} } \bigr]} \bigr\Vert ^{2} \\ &\le \frac{1}{4} \bigl\Vert {\Phi ( {M + m} )} \bigr\Vert ^{2} \\ &= \frac{1}{4} ( {M + m} )^{2}. \end{aligned}

That is,

$$\bigl\Vert {\Phi ( {A\nabla_{\alpha}B} )\Phi^{ - 1} ( {A \sharp_{\alpha}B} )} \bigr\Vert \le\frac{{ ( {M + m} )^{2} }}{{4MmS ( {h^{\prime r} } )}} = \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}.$$

Thus, (2.8) holds.

Inequality (2.9) is equivalent to

$$\bigl\Vert {\Phi ( {A\nabla_{\alpha}B} ) \bigl[ {\Phi ( A ) \sharp_{\alpha}\Phi ( B )} \bigr]^{ - 1} } \bigr\Vert \le \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}.$$

By (2.1) and (2.5), we have

$$\begin{gathered} \bigl\Vert {\Phi ( {A\nabla_{\alpha}B} )MmS \bigl( {h^{\prime r} } \bigr) \bigl[ {\Phi ( A ) \sharp_{\alpha}\Phi ( B )} \bigr]^{ - 1} } \bigr\Vert \\ \quad\le\frac{1}{4} \bigl\Vert {\Phi ( {A\nabla_{\alpha}B} ) + MmS \bigl( {h^{\prime r} } \bigr) \bigl[ {\Phi ( A ) \sharp_{\alpha}\Phi ( B )} \bigr]^{ - 1} } \bigr\Vert ^{2} \\ \quad= \frac{1}{4} \bigl\Vert {\Phi ( A )\nabla_{\alpha}\Phi ( B ) + MmS \bigl( {h^{\prime r} } \bigr) \bigl[ {\Phi ( A ) \sharp_{\alpha}\Phi ( B )} \bigr]^{ - 1} } \bigr\Vert ^{2} \\ \quad\le\frac{1}{4} ( {M + m} )^{2}. \end{gathered}$$

That is,

$$\bigl\Vert {\Phi ( {A\nabla_{\alpha}B} ) \bigl[ {\Phi ( A ) \sharp_{\alpha}\Phi ( B )} \bigr]^{ - 1} } \bigr\Vert \le \frac{{K ( h )}}{{S ( {h^{\prime r} } )}}.$$

Thus, (2.9) holds.

This completes the proof. □

### Theorem 3

Let Φ be a positive unital linear map and let A and B be positive operators. If $$0< m\le A\le{m}'<{M}'\le B\le M$$ or $$0< m\le B\le{m}'<{M}'\le A\le M$$ and $$2 \le p < \infty$$, then for each $$\alpha \in [ {0,1} ]$$,

\begin{aligned}& \Phi^{2p} ( {A\nabla_{\alpha}B} ) \le \frac{1}{{16}} \biggl[ {\frac{{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{p} \Phi^{2p} ( {A\sharp _{\alpha}B} ), \end{aligned}
(2.10)
\begin{aligned}& \Phi^{2p} ( {A\nabla_{\alpha}B} ) \le \frac{1}{{16}} \biggl[ {\frac{{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{p} \bigl[ {\Phi ( A )\sharp_{\alpha}\Phi ( B )} \bigr]^{2p}, \end{aligned}
(2.11)

where $$K ( h ) = \frac{{ ( {h + 1} )^{2} }}{{4h}}$$, $$S(h') = \frac{{h^{\prime \frac{1}{{h' - 1}}} }}{{e\log h^{\prime \frac{1}{{h' - 1}}} }}$$, $$h = \frac{M}{m}$$, $$h' = \frac{{M'}}{{m'}}$$ and $$r = \min \{ {\alpha,1 - \alpha} \}$$.

### Proof

By (2.8), we have

$$\Phi^{ - 2} ( {A\sharp_{\alpha}B} ) \le L^{2} \Phi^{ - 2} ( {A\nabla_{\alpha}B} ),$$
(2.12)

where $$L = \frac{{K(h)}}{{S(h^{\prime r} )}}$$.

Inequality (2.10) is equivalent to

$$\bigl\Vert {\Phi^{p} ( {A\nabla_{\alpha}B} ) \Phi^{ - p} ( {A\sharp_{\alpha}B} )} \bigr\Vert \le \frac{1}{4} \biggl[ {\frac {{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{\frac{p}{2}}.$$

By (2.1), (2.3) and (2.12), we have

\begin{aligned}& \bigl\Vert {\Phi^{p} ( {A \nabla_{\alpha}B} )M^{p} m^{p} \Phi^{ - p} ( {A \sharp_{\alpha}B} )} \bigr\Vert \\& \quad\le\frac{1}{4} \biggl\Vert {L^{\frac{p}{2}} \Phi^{p} ( {A\nabla _{\alpha}B} ) + \biggl( {\frac{{M^{2} m^{2} }}{L}} \biggr)^{\frac {p}{2}} \Phi^{ - p} ( {A\sharp_{\alpha}B} )} \biggr\Vert ^{2} \\& \quad\le\frac{1}{4} \biggl\Vert {L\Phi^{2} ( {A \nabla_{\alpha}B} ) + \frac{{M^{2} m^{2} }}{L}\Phi^{ - 2} ( {A \sharp_{\alpha}B} )} \biggr\Vert ^{p} \\& \quad\le \frac{1}{4} \bigl\Vert {L\Phi^{2} ( {A \nabla_{\alpha}B} ) + LM^{2} m^{2} \Phi^{ - 2} ( {A\nabla_{\alpha}B} )} \bigr\Vert ^{p} \\& \quad\le\frac{1}{4} \bigl[ {L \bigl( {M^{2} + m^{2} } \bigr)} \bigr]^{p}. \end{aligned}

That is,

$$\bigl\Vert {\Phi^{p} ( {A\nabla_{\alpha}B} ) \Phi^{ - p} ( {A\sharp_{\alpha}B} )} \bigr\Vert \le \frac{1}{4} \biggl[ {\frac {{L ( {M^{2} + m^{2} } )}}{{Mm}}} \biggr]^{p} = \frac{1}{4} \biggl[ {\frac{{K^{2} ( h ) ( {M^{2} + m^{2} } )^{2} }}{{S^{2} ( {h^{\prime r} } )M^{2} m^{2} }}} \biggr]^{\frac{p}{2}}.$$

Thus, (2.10) holds.

Similarly, (2.11) holds by inequality (2.9).

This completes the proof. □

### Remark 1

When $$\alpha = \frac{1}{2}$$, because of $$\frac {{K ( h )}}{{S ( {\sqrt{h'} } )}} < K ( h )$$, inequalities (2.8), (2.9), (2.10) and (2.11) are sharper than (1.4), (1.5), (1.6) and (1.7), respectively.

In what follows, when $$\alpha=\frac{1}{2}$$, we present an example showing that inequalities (2.8)–(2.11) are sharper than (1.4)–(1.7), respectively.

### Example 1

Take $$A = \bigl[ { {\scriptsize\begin{matrix}{} {\frac{2}{3}} & 0 \cr 0 & {\frac{5}{7}} \end{matrix}} } \bigr]$$ and $$B = \bigl[ { {\scriptsize\begin{matrix}{} {\frac{10}{3}} & 0 \cr 0 & {\frac{23}{7}} \end{matrix}} } \bigr]$$. We find $$\frac{1}{2} < A < \frac{3}{4} < 3 < B < 4$$. A calculation shows $$\frac{{K(8)}}{{S(2)}} \approx2.3847 < K(8) \approx2.5313$$.

## 3 Conclusions

In this paper, we have presented some new weighted arithmetic–geometric operator mean inequalities. These inequalities are refinements and generalizations of some corresponding results of [1, 2].

## References

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## Acknowledgements

The author would like to express her sincere thanks to referees and editor for their enthusiastic guidance and help.

## Funding

This research was supported by the Scientific Research Fund of Yunnan Provincial Education Department (Grant Nos. 2014Y645, 2018JS747).

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The author read and approved the final manuscript.

### Corresponding author

Correspondence to Jianming Xue.

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