1 Introduction

For any \(n\times n\) complex matrix X, the eigenvalues and the singular values of X are denoted by \(\lambda _{i}(X)\) and \(\sigma _{i}(X)\) for \( i=1,2,\ldots ,n,\) and arranged in such a way that \(\left| \lambda _{1}(X)\right| \ge \left| \lambda _{2}(X)\right| \ge \cdots \ge \left| \lambda _{n}(X)\right| \) and \(\sigma _{1}(X)\ge \sigma _{2}(X)\ge \cdots \ge \sigma _{n}(X).\) Thus, \(\sigma _{i}(X)=\lambda _{i}(\left| X\right| )\) for \(i=1,2,...,n\), where \(\left| X\right| =(X^{*}X)^{1/2}\) is the absolute value of X.

Recall that for any two \(n\times n\) complex matrices X and Y, we have \( \lambda _{i}(XY)=\lambda _{i}(YX)\) for \(i=1,2,\ldots ,n,\) and \(\left\| X\right\| _{2}=\left( \underset{i=1}{\overset{n}{\sum }}\sigma _{i}^{2}(X)\right) ^{1/2}=\left( \textrm{tr}\left| X\right| ^{2}\right) ^{1/2}\) and \(\left\| X\right\| _{1}=\underset{i=1}{\overset{n}{\sum }}\sigma _{i}(X)=\textrm{tr}\left| X\right| \) are the Hilbert–Schmidt norm and the trace norm of X, respectively. Moreover, any unitarily invariant norm is an increasing function of singular values.

Let A and B be positive semidefinite matrices, and let \(|||\cdot |||\) be any unitarily invariant norm. Bourin [5], in his paper on the subadditivity of concave functions of positive semidefinite matrices, asked whether the inequality

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le |||A+B|||,~v\in [0,1], \end{aligned}$$
(1.1)

is true.

In their works on the aforesaid conjecture, Bhatia [4] and Hayajneh and Kittaneh [9] proved that

$$\begin{aligned} ||A^{v}B^{1-v}+B^{v}A^{1-v}||_{2}\le ||A+B||_{2} \end{aligned}$$

is true whenever \(v\in \left[ \frac{1}{4},\frac{3}{4}\right] \).

A complete answer to Bourin\(^{,}\)s question for the trace norm \(\left\| \cdot \right\| _{1}\) has been given by Hayajneh and Kittaneh [7], that is,

$$\begin{aligned} \left\| A^{v}B^{1-v}+B^{v}A^{1-v}\right\| _{1}\le ||A+B||_{1} \end{aligned}$$

is true for \(v\in \left[ 0,1\right] .\) Several partial solutions to Bourin\( ^{,}\)s problem have been given in [10] and references therein.

In this paper, we prove that if f is a nonnegative concave function on \( \left[ 0,\infty \right) ,\) then

$$\begin{aligned} \left| \left| \left| f\left( \left| A^{^{\frac{v}{2}}}B^{1- \frac{v}{2}}+B^{^{\frac{v}{2}}}A^{1-\frac{v}{2}}\right| \right) \right| \right| \right|\le & {} \left| \left| \left| f\left( \frac{A+B}{2}\right) \right| \right| \right| \\{} & {} +\left| \left| \left| f\left( \frac{ B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}+A^{\frac{ v}{2}}B^{1-v}A^{\frac{v}{2}}}{2}\right) \right| \right| \right| \end{aligned}$$

for \(v\in \left[ 0,1\right] \). We also prove the following inequality related to the inequality (1.1)

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le |||(A^{1/r}+B^{1/r})^{r}||| \end{aligned}$$

for \(\frac{1}{2r}\le v\le \frac{2r-1}{2r},\) \(r\ge 1\).

2 Main results

The following lemmas are required in order to support the main results.

Lemma 2.1

[12] Given any positive semidefinite block matrix \( \begin{bmatrix} M &{} K \\ K^{*} &{} N \end{bmatrix} \), where M and N are \(m\times m\) and \(n\times n\) complex matrices, respectively, we have

$$\begin{aligned} 2\sigma _{i}(K)\le \sigma _{i}\left( \begin{bmatrix} M &{} K \\ K^{*} &{} N \end{bmatrix} \right) \end{aligned}$$

for \(i=1,\ldots ,r\) and \(r=\min \{m,n\}\).

Lemma 2.2

[3, p. 291] Let X be an \(n\times n\) complex matrix, and let f be a nonnegative increasing function on \(\left[ 0,\infty \right) .\) Then

$$\begin{aligned} f(\sigma _{i}(X))=\sigma _{i}(f(\left| X\right| )) \end{aligned}$$

for \(i=1,2,...,n.\)

Lemma 2.3

[6] Let X, Y,  and Z be \(n\times n\) complex matrices such that the block matrix \( \begin{bmatrix} X &{} Y \\ Y^{*} &{} Z \end{bmatrix} \) is positive semidefinite, and let f be a nonnegative concave function on \(\left[ 0,\infty \right) .\) Then

$$\begin{aligned} \left| \left| \left| f\left( \begin{bmatrix} X &{} Y \\ Y^{*} &{} Z \end{bmatrix} \right) \right| \right| \right| \le \left| \left| \left| f(X)\right| \right| \right| +\left| \left| \left| f(Z)\right| \right| \right| \text {.} \end{aligned}$$

In particular,

$$\begin{aligned} \left| \left| \left| \begin{bmatrix} X &{} Y \\ Y^{*} &{} Z \end{bmatrix} \right| \right| \right| \le \left| \left| \left| X\right| \right| \right| +\left| \left| \left| Z\right| \right| \right| . \end{aligned}$$

Using Lemma 2.1, Lemma 2.2, and Lemma 2.3, we prove our first main result.

Theorem 2.4

Let A and B be positive semidefinite matrices, and let f be a nonnegative concave function on \(\left[ 0,\infty \right) .\) Then

$$\begin{aligned} \left| \left| \left| f\left( \left| A^{^{\frac{v}{2}}}B^{1- \frac{v}{2}}+B^{^{\frac{v}{2}}}A^{1-\frac{v}{2}}\right| \right) \right| \right| \right|\le & {} \left| \left| \left| f\left( \frac{A+B}{2}\right) \right| \right| \right| \\{} & {} +\left| \left| \left| f\left( \frac{ B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}+A^{\frac{ v}{2}}B^{1-v}A^{\frac{v}{2}}}{2}\right) \right| \right| \right| \end{aligned}$$

for \(v\in \left[ 0,1\right] .\)

Proof

Let \(r\ge 0,\) and let \(X= \begin{bmatrix} A^{\frac{r}{2}} &{} B^{\frac{r}{2}} \\ B^{\frac{r}{2}} &{} A^{\frac{r}{2}} \end{bmatrix} \) and \(Y= \begin{bmatrix} A &{} 0 \\ 0 &{} B \end{bmatrix} \). Then

$$\begin{aligned} X^{*}YX= \begin{bmatrix} A^{r+1}+B^{r+1} &{} A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}} \\ \left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2} }\right) ^{*} &{} B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{\frac{ r}{2}} \end{bmatrix} \end{aligned}$$

is positive semidefinite, and hence by using Lemma 2.1, we have

$$\begin{aligned}{} & {} \sigma _{i}\left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}}\right) \nonumber \\{} & {} \quad \le \frac{1}{2}\sigma _{i}\left( \begin{bmatrix} A^{r+1}+B^{r+1} &{} A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}} \\ \left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2} }\right) ^{*} &{} B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{\frac{ r}{2}} \end{bmatrix} \right) \end{aligned}$$
(2.1)

for \(i=1,2,...,n.\)

Now, for \(i=1,2,...,n,\) we have

$$\begin{aligned}{} & {} \sigma _{i}\left( f\left( \left| A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{ \frac{r}{2}+1}A^{\frac{r}{2}}\right| \right) \right) \\{} & {} \quad =f\left( \sigma _{i}\left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2} +1}A^{\frac{r}{2}}\right) \right) \text { (by Lemma~2.2)} \\{} & {} \quad \le f\left( \frac{1}{2}\sigma _{i}\left( \begin{bmatrix} A^{r+1}+B^{r+1} &{} A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}} \\ \left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2} }\right) ^{*} &{} B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{\frac{ r}{2}} \end{bmatrix} \right) \right) \\{} & {} \quad =\sigma _{i}\left( f\left( \frac{1}{2} \begin{bmatrix} A^{r+1}+B^{r+1} &{} A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}} \\ \left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2} }\right) ^{*} &{} B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{\frac{ r}{2}} \end{bmatrix} \right) \right) \\{} & {} \qquad \qquad \qquad \qquad { (by~Lemma~2.2).} \end{aligned}$$

So,

$$\begin{aligned}{} & {} \left| \left| \left| f\left( \left| A^{\frac{r}{2}+1}B^{ \frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2}}\right| \right) \right| \right| \right| \\{} & {} \quad \le \left| \left| \left| f\left( \frac{1}{2} \begin{bmatrix} A^{r+1}+B^{r+1} &{} A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{ r}{2}} \\ \left( A^{\frac{r}{2}+1}B^{\frac{r}{2}}+B^{\frac{r}{2}+1}A^{\frac{r}{2} }\right) ^{*} &{} B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{\frac{ r}{2}} \end{bmatrix} \right) \right| \right| \right| \\{} & {} \quad \le \left| \left| \left| f\left( \frac{A^{r+1}+B^{r+1}}{2} \right) \right| \right| \right| +\left| \left| \left| f\left( \frac{B^{\frac{r}{2}}AB^{\frac{r}{2}}+A^{\frac{r}{2}}BA^{ \frac{r}{2}}}{2}\right) \right| \right| \right| \\{} & {} \qquad \qquad \qquad \qquad { (by~Lemma~2.3).} \end{aligned}$$

Replacing A, B by \(A^{\frac{1}{r+1}},B^{\frac{1}{r+1}},\) respectively, and taking \(v=\frac{r}{r+1}\), we obtain

$$\begin{aligned} \left| \left| \left| f\left( \left| A^{^{\frac{v}{2}}}B^{1- \frac{v}{2}}+B^{^{\frac{v}{2}}}A^{1-\frac{v}{2}}\right| \right) \right| \right| \right|\le & {} \left| \left| \left| f\left( \frac{A+B}{2}\right) \right| \right| \right| \\{} & {} +\left| \left| \left| f\left( \frac{B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}+A^{\frac{ v}{2}}B^{1-v}A^{\frac{v}{2}}}{2}\right) \right| \right| \right| \end{aligned}$$

for \(v\in [0,1]\). Here, we have used the fact that \(\left| \left| \left| f\left( \left| X\right| \right) \right| \right| \right| =\left| \left| \left| f\left( \left| X^{*}\right| \right) \right| \right| \right| \) for any complex matrix X. This completes the proof of the theorem.

Taking \(v=1\) in the aforementioned Theorem 2.4, we have the following corollary.

Corollary 2.5

Let A and B be positive semidefinite matrices, and let f be a nonnegative concave function on \(\left[ 0,\infty \right) .\) Then

$$\begin{aligned} \left| \left| \left| f\left( \left| A^{\frac{1}{2}}B^{\frac{1}{2}} +B^{\frac{1}{2}}A^{\frac{1}{2}}\right| \right) \right| \right| \right| \le 2\left| \left| \left| f\left( \frac{A+B}{2}\right) \right| \right| \right| . \end{aligned}$$

As an application of Theorem 2.4, we give a different solution to Bourin\(^{,}\)s question for the Hilbert–Schmidt norm. To achieve this, we need the following lemmas.

Lemma 2.6

[9] Let A and B be positive semidefinite matrices, and let \(v\in \left[ 0,1\right] \). Then

$$\begin{aligned} \textrm{tr}\left( \left( B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}\right) ^{2}+\left( A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right) ^{2}\right) \le \textrm{tr}\left( A^{2}+B^{2}\right) . \end{aligned}$$

Lemma 2.7

[9] Let A and B be positive semidefinite matrices, and let \(v\in \left[ \frac{1}{2},1\right] \). Then

$$\begin{aligned} \textrm{tr}~B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}A^{\frac{v}{2}}B^{1-v}A^{ \frac{v}{2}}\le \textrm{tr}~AB. \end{aligned}$$

An equivalent form of the following lemma has been given in [9]. For the reader\(^{,}\)s convenience, we give a short proof of this lemma based on Lemma 2.6 and Lemma 2.7.

Lemma 2.8

Let A and B be positive semidefinite matrices, and let \( v\in \left[ \frac{1}{2},1\right] \). Then

$$\begin{aligned} \left| \left| B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}+A^{\frac{v}{2} }B^{1-v}A^{\frac{v}{2}}\right| \right| _{2}\le ||A+B||_{2}. \end{aligned}$$
(2.2)

Proof

We can easily check that the square of the left hand side of the inequality ( 2.2) is equal to

$$\begin{aligned} \mathrm {tr~}\left( \left( B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}\right) ^{2}+\left( A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right) ^{2}+2B^{\frac{v}{2} }A^{1-v}B^{\frac{v}{2}}A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right) . \end{aligned}$$

Hence, the inequality (2.2) is equivalent to the inequality

$$\begin{aligned} \mathrm {tr~}&\left( \left( B^{\frac{v}{2}}A^{1-v}B^{\frac{v}{2}}\right) ^{2}+\left( A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right) ^{2}+2B^{\frac{v}{2} }A^{1-v}B^{\frac{v}{2}}A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right) \\&\le \mathrm {tr~}\left( A^{2}+B^{2}+2AB\right) . \end{aligned}$$

In view of Lemma 2.8 and Theorem 2.4, applied to the Hilbert–Schmidt norm and the case \(f(t)=t,\) we have the following corollary.

Corollary 2.9

Let A and B be positive semidefinite matrices, and let \( v\in \left[ \frac{1}{2},1\right] \). Then

$$\begin{aligned} \left| \left| A^{\frac{v}{2}}B^{1-\frac{v}{2}}+B^{\frac{v}{2}}A^{1- \frac{v}{2}}\right| \right| _{2}&\le \frac{1}{2}\left( \left| \left| A+B\right| \right| _{2}+\left| \left| B^{\frac{v}{ 2}}A^{1-v}B^{\frac{v}{2}}+A^{\frac{v}{2}}B^{1-v}A^{\frac{v}{2}}\right| \right| _{2}\right) \\&\le ||A+B||_{2}. \end{aligned}$$

It should be mentioned here that Corollary 2.9 can be concluded from Theorem 2.7 in [8], using a completely different analysis.

Corollary 2.10

Let A and B be positive semidefinite matrices, and let \(v\in \left[ \frac{1}{4},\frac{3}{4}\right] \). Then

$$\begin{aligned} \left| \left| A^{v}B^{1-v}+B^{v}A^{1-v}\right| \right| _{2}\le ||A+B||_{2}. \end{aligned}$$
(2.3)

Proof

Using Corollary 2.9, we have

$$\begin{aligned} \left| \left| A^{\frac{v}{2}}B^{1-\frac{v}{2}}+B^{\frac{v}{2}}A^{1- \frac{v}{2}}\right| \right| _{2}\le \left| \left| A+B\right| \right| _{2} \end{aligned}$$

for \(v\in \left[ \frac{1}{2},1\right] \). Hence, the inequality (2.3) is valid for \(v\in \left[ \frac{1}{4},\frac{1}{2}\right] \). Therefore,

$$\begin{aligned} ||A^{v}B^{1-v}+B^{v}A^{1-v}||_{2}&=||(A^{v}B^{1-v}+B^{v}A^{1-v})^{*}||_{2} \\&=||A^{1-v}B^{v}+B^{1-v}A^{v}||_{2} \\&\le ||A+B||_{2} \end{aligned}$$

is also valid for \(1-v\in \left[ \frac{1}{4},\frac{1}{2}\right] \), i.e., \( v\in \left[ \frac{1}{2},\frac{3}{4}\right] \). Hence, the inequality (2.3) is valid for \(v\in \left[ \frac{1}{4},\frac{3}{4}\right] \).

To prove our second main result in this paper, we need the following lemmas.

Lemma 2.11

(Matrix Young Inequality) [1] Let A and B be \(n\times n\) complex matrices. Then

$$\begin{aligned} \sigma _{i}\left( AB^{*}\right) \le \sigma _{i}\left( {\frac{1}{p} \left| A\right| ^{p}}+{\frac{1}{q}\left| B\right| ^{q}} \right) \end{aligned}$$

for \(i=1,2,...,n,\) and \(p,q>1\) with \(\frac{1}{p}+\frac{1}{q}=1\).

Lemma 2.12

[2] Let A and B be positive semidefinite matrices. Then

$$\begin{aligned} |||A^{r}+B^{r}|||\le |||(A+B)^{r}|||~\mathrm {for~}r\ge 1 \end{aligned}$$

and

$$\begin{aligned} |||(A+B)^{r}|||\le |||A^{r}+B^{r}|||~\mathrm {for~}0<r\le 1. \end{aligned}$$

Theorem 2.13

Let A and B be positive semidefinite matrices, and let \(r\ge 1\). Then

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le |||(A^{1/r}+B^{1/r})^{r}||| \end{aligned}$$

for \(\frac{1}{2r}\le v\le \frac{2r-1}{2r}.\)

Proof

Let \(X= \begin{bmatrix} A^{v} &{} B^{v} \\ 0 &{} 0 \end{bmatrix} \) and \(Y= \begin{bmatrix} B^{1-v} &{} A^{1-v} \\ 0 &{} 0 \end{bmatrix} \). Then

$$\begin{aligned} XY^{*}= \begin{bmatrix} A^{v}B^{1-v}+B^{v}A^{1-v} &{} 0 \\ 0 &{} 0 \end{bmatrix} \end{aligned}$$

and

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||=|||XY^{*}|||&=|||~|X||Y|~||| \\&\le |||v|X|^{\frac{1}{v}}+(1-v)|Y|^{\frac{1}{(1-v)}}||| \\&\le v|||~|X|^{\frac{1}{v}}|||+(1-v)|||~|Y|^{\frac{1}{(1-v)}}||| \\&=v|||~|X^{*}|^{\frac{1}{v}}|||+(1-v)|||~|Y^{*}|^{\frac{1}{(1-v)}}|||, \end{aligned}$$

where the first inequality follows from Lemma 2.11, the second inequality follows from the triangle inequality, and the last equality follows using the fact that \(|||~|X|^{r}|||=|||~|X^{*}|^{r}|||\) for any complex matrix X and for \(r>0\). Hence,

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le & {} v|||(A^{2v}+B^{2v})^{\frac{1}{2v}}|||\\{} & {} +(1-v)|||(A^{2(1-v)}+B^{2(1-v)})^{\frac{1}{2(1-v)}}|||. \end{aligned}$$

Assume that \(\frac{1}{2}\le v\le \frac{2r-1}{2r}\). Since \(\frac{1}{2v}\le 1\), it follows, by Lemma 2.12, that

$$\begin{aligned} |||(A^{2v}+B^{2v})^{1/2v}|||\le |||A+B|||. \end{aligned}$$

It is known [3, p. 95] that \(|||X|||_{(r)}:=|||~|X|^{r}|||^{1/r}\) is a unitarily invariant norm for \(r\ge 1\). Hence, again using Lemma 2.12, we have

$$\begin{aligned} |||(A^{2(1-v)}+B^{2(1-v)})^{\frac{1}{2(1-v)}}|||&=|||~|(A^{2(1-v)}+B^{2(1-v)})^{\frac{1}{2r(1-v)}}|~|||_{(r)}^{r} \\&\le |||A^{1/r}+B^{1/r}|||_{(r)}^{r}=|||(A^{1/r}+B^{1/r})^{r}|||. \end{aligned}$$

Hence, when \(\frac{1}{2}\le v\le \frac{2r-1}{2r},\) we obtain

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le v|||A+B|||+(1-v)|||(A^{1/r}+B^{1/r})^{r}|||. \end{aligned}$$

Moreover, by Lemma 2.12, we have \(|||A+B|||\le |||(A^{1/r}+B^{1/r})^{r}|||\) and this gives the following inequality

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le |||(A^{1/r}+B^{1/r})^{r}|||. \end{aligned}$$
(2.4)

Similarly, when \(\frac{1}{2r}\le v\le \frac{1}{2}\), i.e., \(\frac{1}{2}\le 1-v\le \frac{2r-1}{2r}\), we again have the inequality (2.4). Therefore,

$$\begin{aligned} |||A^{v}B^{1-v}+B^{v}A^{1-v}|||\le |||(A^{1/r}+B^{1/r})^{r}||| \end{aligned}$$

for \(\frac{1}{2r}\le v\le \frac{2r-1}{2r}\).

We conclude the paper with the following remark.

Remark 2.14

The case \(v=\frac{1}{2}\) and \(r=1\) in Theorem 2.13 is the inequality

$$\begin{aligned} |||A^{1/2}B^{1/2}+B^{1/2}A^{1/2}|||\le \left| \left| \left| A+B\right| \right| \right| , \end{aligned}$$

which can also be concluded from the triangle inequality and the arithmetic–geometric mean inequality for unitarily invariant norms (see, e.g., [11]).