# 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 [3] 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 [2] 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 [46] 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

([7])

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

([8])

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

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

### Lemma 3

([9])

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

([10])

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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Correspondence to Jianming Xue.

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Xue, J. Weighted arithmetic–geometric operator mean inequalities. J Inequal Appl 2018, 154 (2018). https://doi.org/10.1186/s13660-018-1750-7