1 Introduction

Let \(M_{n} \) be the space of \(n\times n\) complex matrices. Let \(A,B\in M_{n} \) be positive definite, the weighted geometric mean of A and B, denoted by \(A\# B\), is defined as

$$A\#_{t}B = A^{1/2} \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{t} A^{1/2}. $$

When \(t=\frac{1}{2}\), this is the geometric mean, denoted by \(A\# B\). For \(A\in M_{n}\), we denote the vector of eigenvalues by \(\lambda ( A ) = ( {\lambda_{1} ( A ),\lambda_{2} ( A ), \ldots,\lambda_{n} ( A ) } ) \), and we assume that the components of \(\lambda ( A ) \) are in descending order. Let \(\Vert \cdot \Vert \) denote any unitarily invariant norm on \(M_{n} \).

Recently, Bhatia, Lim, and Yamazaki proved in [1] that if \(A,B\in M _{n} \) are positive definite, then

$$ \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) $$
(1.1)

and

$$ \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4} } \bigr). $$
(1.2)

These authors also have shown in [1] that if \(A,B\in M_{n} \) are positive definite and \(0 < t < 1\), then

$$ \operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) \le \operatorname{tr} \bigl( {A^{1 - t} B ^{t} + A^{t} B^{1 - t} } \bigr) $$
(1.3)

and

$$ \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{2} } \bigr) \le \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$
(1.4)

In this short note, we first obtain a trace inequality, which is similar to inequality (1.1). Meanwhile, we also obtain generalizations of inequalities (1.1), (1.2), (1.3), and (1.4).

2 Main results

In this section, we first give a trace inequality, which is similar to inequality (1.1). To do this, we need the following lemmas.

Lemma 2.1

[2]

Let \(A,B \in M_{n}\) be positive definite. Then

$$\prod_{j = 1}^{k} {\lambda_{j} ( {AB} ) } \le \prod_{j = 1}^{k} { \lambda_{j}^{1/2} \bigl( {A^{2} B^{2} } \bigr) },\quad 1 \le k \le n. $$

Lemma 2.2

[3]

Let \(A,B \in M_{n}\). If \(\lambda ( A ), \lambda ( B ) > 0\) such that

$$\prod_{j = 1}^{k} {\lambda_{j} ( A ) } \le \prod_{j = 1}^{k} { \lambda_{j} ( B ) },\quad 1 \le k \le n, $$

then

$$\det ( {I + A} ) \le \det ( {I + B} ). $$

Theorem 2.1

Let A and B be positive definite. Then

$$\operatorname{tr} \bigl( {\log \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( {\log \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr). $$

Proof

By Lemma 2.1, we have

$$\begin{aligned} \prod_{j = 1}^{k} {\lambda_{j} \bigl( {A^{ - 1/2} B^{1/2} } \bigr) } &= \prod _{j = 1}^{k} {\lambda_{j} \bigl( {B^{1/2} A ^{ - 1/2} } \bigr) } \\ &\le \prod_{j = 1}^{k} {\lambda_{j} \bigl( { \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr) },\quad 1 \le k \le n. \end{aligned}$$

Using Lemma 2.2, we get

$$ \det \bigl( {I + A^{ - 1/2} B^{1/2} } \bigr) \le \det \bigl( {I + \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr) $$
(2.1)

and

$$ \det \bigl( {I + B^{1/2} A^{ - 1/2} } \bigr) \le \det \bigl( {I + \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr). $$
(2.2)

It follows from (2.1) and (2.2) that

$$\begin{aligned} \det \bigl( {I + A^{ - 1/2} B^{1/2} } \bigr) \det \bigl( {I + B^{1/2} A^{ - 1/2} } \bigr) \le \det \bigl( {I + 2 \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr), \end{aligned}$$

which is equivalent to

$$\begin{aligned} &\det \bigl( {I + A^{ - 1/2} B^{1/2} + B^{1/2} A^{ - 1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr) \\ &\quad \le \det \bigl( {I + 2 \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr). \end{aligned}$$
(2.3)

Multiplying \(\det A^{1/2} \) both sides in inequality (2.3), we have

$$ \det \bigl( {A + B + A^{1/2} B^{1/2} + B^{1/2} A^{1/2} } \bigr) \le \det \bigl( {A + B + 2A^{1/2} \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} A^{1/2} } \bigr). $$
(2.4)

Note that \(\log \det X = \operatorname{tr}\log X\), inequality (2.4) implies

$$\operatorname{tr} \bigl( {\log \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( {\log \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr). $$

This completes the proof. □

Next, we show generalizations of inequalities (1.1), (1.2), (1.3), and (1.4). To do this, we need the following lemma.

Lemma 2.3

[2]

Let \(A,B \in M_{n}\) and \(\frac{1}{p} + \frac{1}{q} = 1\), \(p,q > 0\). Then

$$\Vert {AB} \Vert \le \bigl\Vert {\vert A \vert ^{p} } \bigr\Vert ^{1/p} \bigl\Vert {\vert B \vert ^{q} } \bigr\Vert ^{1/q}. $$

This is the Hölder inequality of unitary invariant norms for matrices. For more information on this inequality and its applications the reader is referred to [4] and the references therein.

Theorem 2.2

Let A and B be positive definite and \(1 \le r \le 2\). Then

$$ \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} + ( {r - 1} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4}. $$
(2.5)

Proof

Let

$$p = \frac{1}{{2 - r}},\qquad q = \frac{1}{{r - 1}}, $$

then

$$\frac{1}{p} + \frac{1}{q} = 1,\quad p,q > 0. $$

By Lemma 2.3, we obtain

$$ \begin{aligned}[b] &\operatorname{tr} \bigl( \bigl( A + B + 2 ( A \# B ) \bigr)^{r} \bigr) \\ &\quad = \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2 - r} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2r - 2} } \bigr) \\ &\quad \le \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{p ( {2 - r} ) } } \bigr) ^{1/p} \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{q ( {2r - 2} ) } } \bigr) ^{1/q} \\ &\quad = \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2} } \bigr) ^{r - 1}. \end{aligned} $$
(2.6)

It follows from (1.1), (1.2), and (2.6) that

$$\operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le \bigl( {\operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) ^{2 - r} \bigl( { \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4} } \bigr) ^{r - 1}. $$

By Young’s inequality, we have

$$\operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} + ( {r - 1} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4}. $$

This completes the proof. □

Remark 2.1

Putting \(r=1\) in (2.5), we get (1.1). Putting \(r=2\) in (2.5), we get (1.2). Therefore, inequality (2.5) is a generalization of inequalities (1.1) and (1.2).

Remark 2.2

Let A and B be positive definite. By the concavity of \(f ( x ) = x^{r}\), \(x \ge 0\), \(0 < r<1\), then we have

$$n^{r - 1} \operatorname{tr}f ( X ) \le f ( {\operatorname{tr}X} ), $$

where X is positive definite. It follows from this last inequality and inequality (1.1) that

$$\begin{aligned} n^{r - 1} \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} & \le \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr) ^{r} \\ &\le \bigl( {\operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) } \bigr) ^{r}. \end{aligned}$$

Meanwhile, we also have

$$f ( {\operatorname{tr}X} ) \le \operatorname{tr}f ( X ), $$

which implies

$$n^{r - 1} \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2r} } \bigr). $$

This is a complement of (1.1) for \(0 < r<1\).

Theorem 2.3

Let A and B be positive definite and \(1 \le r \le 2\). Then

$$ \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) + ( {r - 1} ) \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$
(2.7)

Proof

Let

$$p = \frac{1}{{2 - r}},\qquad q = \frac{1}{{r - 1}}, $$

then

$$\frac{1}{p} + \frac{1}{q} = 1,\quad p,q > 0. $$

By Lemma 2.3, we obtain

$$ \begin{aligned}[b] \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) &= \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{2 - r} ( {A\# _{t} B + B\# _{t} A} ) ^{2r - 2} } \bigr) \\ &\le \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{p ( {2 - r} ) } } \bigr) ^{1/p} \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{q ( {2r - 2} ) } } \bigr) ^{1/q} \\ &= \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{2} } \bigr) ^{r - 1}. \end{aligned} $$
(2.8)

It follows from (1.3), (1.4), and (2.8) that

$$\operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le \bigl( {\operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B ^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr) } \bigr) ^{r - 1}. $$

By Young’s inequality, we have

$$\operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) + ( {r - 1} ) \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$

This completes the proof. □

Remark 2.3

Putting \(r=1\) in (2.7), we get (1.3). Putting \(r=2\) in (2.7), we get (1.4). Therefore, inequality (2.7) is a generalization of inequalities (1.3) and (1.4).