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 [4–6] 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].