1 Introduction

Throughout this paper, all matrices are assumed to be complex square matrices of the same size. Let A and B be positive semidefinite matrices. For \(t\in \left[ 0,1\right] \), let

$$\begin{aligned} b_{t}=A^{t}B^{1-t}+B^{t}A^{1-t}. \end{aligned}$$

In this paper, \({\left| \left| \left| . \right| \right| \right| }\) denotes any unitarily invariant norm on the space of matrices. Among the most important unitarily invariant norms are the usual operator (or the spectral) norm \(\left\| .\right\| _{\infty }\) and the Schatten \(p-\)norm \(\left\| .\right\| _{p}\) for \(1\le p<\infty \).

In [10], and in his work on the subadditivity of concave functions of positive semidefinite matrices, J.C. Bourin asked the following question.

Question 1.1

Let A and B be any two positive semidefinite matrices, and let \(t\in \left[ 0,1\right] \). Is it true that

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le {\left| \left| \left| A+B \right| \right| \right| }? \end{aligned}$$

Question 1.1 has been answered for the Hilbert-Schmidt norm under the condition \(t\in \left[ \frac{1}{4},\frac{3}{4}\right] \), see [5, 19], and [21].

In [13], Hayajneh and Kittaneh proved the following norm inequality for all unitarily invariant norms:

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le {\left| \left| \left| A \right| \right| \right| }+{\left| \left| \left| B \right| \right| \right| }, \end{aligned}$$

where A and B are positive semidefinite matrices and \(t\in \left[ 0,1\right] \). Consequently,

$$\begin{aligned} \left\| b_{t}\right\| _{1}\le \left\| A+B\right\| _{1} \end{aligned}$$

for \(t\in \left[ 0,1\right] \). This gives an affirmative answer to Question 1.1 for the trace norm.

In [18], the authors proved the following stronger norm inequality for all unitarily invariant norms:

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le \sqrt{{\left| \left| \left| A+B \right| \right| \right| }\left( {\left| \left| \left| A \right| \right| \right| }+{\left| \left| \left| B \right| \right| \right| }\right) }, \end{aligned}$$

where A and B are positive semidefinite matrices and \(t\in \left[ 0,1\right] \).

They also gave an affirmative answer to Question 1.1 in the case of \(t=\frac{1}{4}\) and \(t=\frac{3}{4}\) for all unitarily invariant norms. Clearly, Question 1.1 is true for \(t=0\) and \(t=1\).

Moreover, a partial answer to this question in the wider class of unitarily invariant norms has been given in [13] by proving that

$$\begin{aligned} {\left| \left| \left| \hbox {Re }b_{t} \right| \right| \right| }\le {\left| \left| \left| A+B \right| \right| \right| } \end{aligned}$$

and

$$\begin{aligned} {\left| \left| \left| \hbox {Im }b_{t} \right| \right| \right| }\le {\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

In [1], Alakhrass proved the following norm inequality for all unitarily invariant norms:

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{\left| \frac{1}{2}-t\right| }{\left| \left| \left| A+B \right| \right| \right| }, \end{aligned}$$
(1.1)

where A and B are positive semidefinite matrices and \(t\in \left[ 0,1\right] \).

In [20], and in order to study Question 1.1, the authors proved the following inequality:

$$\begin{aligned} \left\| b_{t}\right\| _{p}\le 2^{\frac{1}{2}-\frac{1}{2p}} \left\| A+B\right\| _{p}, \end{aligned}$$
(1.2)

where A and B are positive semidefinite matrices, \(t\in \left[ 0,1\right] \), and \(p\ge 1\). In particular,

$$\begin{aligned} \left\| b_{t}\right\| _{\infty }\le 2^{\frac{1}{2}} \left\| A+B\right\| _{\infty }. \end{aligned}$$

Note that for the trace norm, the inequality (1.2) is sharper than the inequality (1.1) for all \(t \in [0,1].\) For the Hilbert-Schmidt norm, the inequality (1.2) is sharper than the inequality (1.1) for \(t \in [0,\frac{1}{4}]\) and \(t \in [\frac{3}{4},1].\) However, the inequality (1.1) is sharper than the inequality (1.2) for \(t \in [\frac{1}{4},\frac{3}{4}].\) But we already know that Bourin’s question is true for \(t \in [\frac{1}{4},\frac{3}{4}]\) for the Hilbert-Schmidt norm.

In [12], and in order to try to answer Question 1.1, the authors proved the following inequality involving the spectral radius \(r\left( .\right) \):

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le {\left| \left| \left| A+B \right| \right| \right| } \sqrt{\max \left( r\left( B^{2t-1}A^{1-2t}\right) ,r\left( A^{2t-1}B^{1-2t}\right) \right) }, \end{aligned}$$

where A and B are positive definite matrices and \(t\in \left[ 0,1\right] \).

Consequently, the authors proved that if \(B\le A \le \left( 1+\epsilon \right) ^{2}B\), then

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le \left( 1+\epsilon \right) {\left| \left| \left| A+B \right| \right| \right| } \end{aligned}$$

for \(t\in \left[ 0,1\right] ,\epsilon >0\), and for every unitarily invariant norm. Equivalently, if \(\alpha \ge 1\) and if the spectrum of \(AB^{-1}\) lies in the interval \(\left[ 1,\alpha \right] \), then

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le \sqrt{\alpha }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

For a comprehensive account on related trace and norm inequalities, we refer to [2, 3, 7,8,9,10,11,12,13,14,15,16,17,18, 20,21,22, 25,26,28], and references therein.

In Section 2, we introduce a new way of proving the inequality (1.1) without using the notion of majorization.

In Section 2, we will prove the following norm inequalities for all unitarily invariant norms:

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{2\left( t-\frac{3}{4}\right) }{\left| \left| \left| A+B \right| \right| \right| }, \end{aligned}$$

where A and B are positive semidefinite matrices and \(t\in \left[ \frac{3}{4},1\right] \), and

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{2\left( \frac{1}{4}-t\right) }{\left| \left| \left| A+B \right| \right| \right| }, \end{aligned}$$

where A and B are positive semidefinite matrices and \(t\in \left[ 0,\frac{1}{4}\right] \). These norm inequalities are sharper than the norm inequality (1.1) and closely related to Question 1.1. In fact, they lead to solve Question 1.1 in the case of \(t=\frac{3}{4}\) and \(t=\frac{1}{4}\) for all unitarily invariant norms, which is a result due to Hayajneh and Kittaneh [18].

2 Main Result

We begin with the following lemmas that will be used in proving our main result.

The following lemma can be found in [4, p. 95]. It contains the celebrated Hölder inequality for all unitarily invariant norms.

Lemma 2.1

Let A and B be any two matrices, \(p>1\), and \(\frac{1}{p}+\frac{1}{q}=1\). Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| AB \right| \right| \right| }\le {\left| \left| \left| \left| A\right| ^{p} \right| \right| \right| }^{\frac{1}{p}}{\left| \left| \left| \left| B\right| ^{q} \right| \right| \right| }^{\frac{1}{q}}. \end{aligned}$$

The following lemma, concerning the convexity and concavity of the power functions of positive semidefinite matrices, can be found in [23].

Lemma 2.2

Let A and B be any two positive semidefinite matrices. Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| \left( A+B\right) ^{r} \right| \right| \right| }\le 2^{r-1}{\left| \left| \left| A^{r}+B^{r} \right| \right| \right| }\hbox { for }r\ge 1 \end{aligned}$$

and

$$\begin{aligned} {\left| \left| \left| \left( A+B\right) ^{r} \right| \right| \right| }\le {\left| \left| \left| A^{r}+B^{r} \right| \right| \right| }\hbox { for }0\le r\le 1. \end{aligned}$$

The following lemma can be found in [4, p. 258].

Lemma 2.3

Let A and B be any two matrices. Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| \left( ABA\right) ^{r} \right| \right| \right| }\le {\left| \left| \left| A^{r}B^{r}A^{r} \right| \right| \right| }\hbox { for }r\ge 1. \end{aligned}$$

The following lemma can be found in [4, p. 253].

Lemma 2.4

Let A and B be any two matrices such that the product AB is normal. Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| AB \right| \right| \right| }\le {\left| \left| \left| BA \right| \right| \right| }. \end{aligned}$$

The following lemma can be found in [6]. It contains the celebrated Heinz inequalities for all unitarily invariant norms.

Lemma 2.5

Let A and B be positive semidefinite matrices. Then for \(t\in \left[ 0,1\right] \) and for every unitarily invariant norm, we have

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

Convention 2.6

For any matrix T and for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| T\oplus 0 \right| \right| \right| }={\left| \left| \left| T \right| \right| \right| }. \end{aligned}$$

In the following theorem, we introduce a new way of proving the inequality (1.1) without using the notion of majorization.

Theorem 2.7

Let A and B be positive semidefinite matrices, and let \(t\in \left[ 0,1\right] \). Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{\left| \frac{1}{2}-t\right| }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

Proof

For the end points \(t=0\) and \(t=1\), the result is obvious.

Case 1: \(t\in \left( 0,\frac{1}{2}\right] \). In this case, let

$$\begin{aligned} X= \left[ \begin{array}{ccccc} A^{t} &{} \quad B^{t} \\ 0 &{} \quad 0 \end{array} \right] , Y= \left[ \begin{array}{ccccc} B^{1-t} &{} \quad 0 \\ A^{1-t} &{} \quad 0 \end{array} \right] . \end{aligned}$$

For the partition \(\frac{1}{p}+\frac{1}{q}=t+\left( 1-t\right) =1\) and noting that \(\frac{p}{2}=\frac{1}{2t}\ge 1\) and \(\frac{q}{2}=\frac{1}{2\left( 1-t\right) }\le 1\), we have

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }= & {} {\left| \left| \left| b_{t}\oplus 0 \right| \right| \right| }\\= & {} {\left| \left| \left| \left[ \begin{array}{ccccc} A^{t} &{} \quad B^{t} \\ 0 &{} \quad 0 \end{array} \right] \left[ \begin{array}{ccccc} B^{1-t} &{} \quad 0 \\ A^{1-t} &{} \quad 0 \end{array} \right] \right| \right| \right| }\\= & {} {\left| \left| \left| XY \right| \right| \right| }\\\le & {} {\left| \left| \left| \left| X\right| ^{p} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left| Y\right| ^{q} \right| \right| \right| }^{\frac{1}{q}}\left( \hbox {by Lemma 2.1}\right) \\= & {} {\left| \left| \left| \left( X^{*}X\right) ^{\frac{p}{2}} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }^{\frac{1}{q}}\\= & {} {\left| \left| \left| \left( XX^{*}\right) ^{\frac{p}{2}} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }^{\frac{1}{q}}\\= & {} {\left| \left| \left| \left( A^{2t}+B^{2t}\right) ^{\frac{p}{2}}\oplus 0 \right| \right| \right| }^{\frac{1}{p}}{\left| \left| \left| \left( A^{2\left( 1-t\right) }+B^{2\left( 1-t\right) }\right) ^{\frac{q}{2}}\oplus 0 \right| \right| \right| }^{\frac{1}{q}}\\= & {} {\left| \left| \left| \left( A^{2t}+B^{2t}\right) ^{\frac{p}{2}} \right| \right| \right| }^{\frac{1}{p}}{\left| \left| \left| \left( A^{2\left( 1-t\right) } +B^{2\left( 1-t\right) }\right) ^{\frac{q}{2}} \right| \right| \right| }^{\frac{1}{q}}\\\le & {} \left( 2^{\frac{p}{2}-1}\right) ^{\frac{1}{p}}{\left| \left| \left| A^{pt}+B^{pt} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| A^{q\left( 1-t\right) }+B^{q\left( 1-t\right) } \right| \right| \right| }^{\frac{1}{q}}\left( \hbox {by Lemma 2.2}\right) \\= & {} 2^{\frac{1}{2}-t}{\left| \left| \left| A+B \right| \right| \right| }^{\frac{1}{p}+\frac{1}{q}}\\= & {} 2^{\frac{1}{2}-t}{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

Case 2: \(t\in \left[ \frac{1}{2},1\right) \). In this case,

$$\begin{aligned}{\left| \left| \left| b_{t} \right| \right| \right| }= & {} {\left| \left| \left| b^{*}_{t} \right| \right| \right| }\\= & {} {\left| \left| \left| b_{1-t} \right| \right| \right| }\left( \hbox { because }\ b^{*}_{t}=b_{1-t}\right) \\\le & {} 2^{t-\frac{1}{2}}{\left| \left| \left| A+B \right| \right| \right| }\left( \hbox { since }1-t\in \left[ 0, \quad \frac{1}{2}\right] \hbox { and by using Case 1}\right) . \end{aligned}$$

This completes the proof. \(\square \)

To prove our main result, we state and prove the following lemma.

Lemma 2.8

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

$$\begin{aligned} X= \left[ \begin{array}{ccccc} A^{2t-1} &{} \quad B^{2t-1} \\ 0 &{} \quad 0 \end{array} \right] , Y= \left[ \begin{array}{ccccc} A^{1-t}B^{1-t} &{} \quad 0 \\ B^{1-t}A^{1-t} &{} \quad 0 \end{array} \right] . \end{aligned}$$

Then for every unitarily invariant norm and for \(p=\frac{1}{2t-1}\) and \(q=\frac{1}{2\left( 1-t\right) }\), we have

$$\begin{aligned} {\left| \left| \left| \left( XX^{*}\right) ^{\frac{p}{2}} \right| \right| \right| }\le {\left| \left| \left| A+B \right| \right| \right| } \hbox { and } {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }\le 2^{q\left( \frac{1}{2}-\frac{1}{q}\right) }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

Proof

Since \(t\in \left[ \frac{3}{4},1\right) \), it follows that \(\frac{p}{2} \le 1\) and \(\frac{q}{2}\ge 1\). In fact,

$$\begin{aligned} t\ge \frac{3}{4}\Longleftrightarrow 2\left( 2t-1\right) \ge 1 \Longleftrightarrow \frac{2}{p}\ge 1 \Longleftrightarrow \frac{p}{2}\le 1 \end{aligned}$$

and

$$\begin{aligned}t\ge \frac{3}{4}\Longleftrightarrow 4\left( 1-t\right) \le 1 \Longleftrightarrow \frac{2}{q}\le 1 \Longleftrightarrow \frac{q}{2}\ge 1. \end{aligned}$$

Now,

$$\begin{aligned} {\left| \left| \left| \left( XX^{*}\right) ^{\frac{p}{2}} \right| \right| \right| }= & {} {\left| \left| \left| \left( \left( A^{2\left( 2t-1\right) }+B^{2\left( 2t-1\right) }\right) \oplus 0\right) ^{\frac{p}{2}} \right| \right| \right| }\\= & {} {\left| \left| \left| \left( A^{2\left( 2t-1\right) }+B^{2\left( 2t-1\right) }\right) ^{\frac{p}{2}} \right| \right| \right| }\\= & {} {\left| \left| \left| \left( A^{\frac{2}{p}}+B^{\frac{2}{p}}\right) ^{\frac{p}{2}} \right| \right| \right| }\\\le & {} {\left| \left| \left| A^{\frac{2}{p}\frac{p}{2}}+B^{\frac{2}{p}\frac{p}{2}} \right| \right| \right| }\left( \hbox {by Lemma 2.2}\right) \\= & {} {\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

Also,

$$\begin{aligned} {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }= & {} {\left| \left| \left| \left( \left( B^{1-t}A^{2\left( 1-t\right) }B^{1-t} +A^{1-t}B^{2\left( 1-t\right) }A^{1-t}\right) \oplus 0\right) ^{\frac{q}{2}} \right| \right| \right| }\\= & {} {\left| \left| \left| \left( B^{1-t}A^{2\left( 1-t\right) }B^{1-t} +A^{1-t}B^{2\left( 1-t\right) }A^{1-t}\right) ^{\frac{q}{2}} \right| \right| \right| }\\= & {} {\left| \left| \left| \left( B^{\frac{1}{2q}}A^{\frac{1}{q}}B^{\frac{1}{2q}} +A^{\frac{1}{2q}}B^{\frac{1}{q}}A^{\frac{1}{2q}}\right) ^{\frac{q}{2}} \right| \right| \right| }\\\le & {} 2^{\frac{q}{2}-1}{\left| \left| \left| \left( B^{\frac{1}{2q}}A^ {\frac{1}{q}}B^{\frac{1}{2q}}\right) ^{\frac{q}{2}} +\left( A^{\frac{1}{2q}}B^{\frac{1}{q}}A^{\frac{1}{2q}}\right) ^{\frac{q}{2}} \right| \right| \right| } \left( \hbox {by Lemma 2.2}\right) \\\le & {} 2^{\frac{q}{2}-1}\left( {\left| \left| \left| \left( B^{\frac{1}{2q}}A^{\frac{1}{q}}B^ {\frac{1}{2q}}\right) ^{\frac{q}{2}} \right| \right| \right| }+{\left| \left| \left| \left( A^{\frac{1}{2q}}B^ {\frac{1}{q}}A^{\frac{1}{2q}}\right) ^{\frac{q}{2}} \right| \right| \right| }\right) \\\le & {} 2^{\frac{q}{2}-1}\left( {\left| \left| \left| B^{\frac{1}{4}}A^{\frac{1}{2}}B^ {\frac{1}{4}} \right| \right| \right| }+{\left| \left| \left| A^{\frac{1}{4}}B^{\frac{1}{2}}A^ {\frac{1}{4}} \right| \right| \right| }\right) \left( \hbox {by Lemma 2.3}\right) \\\le & {} 2^{\frac{q}{2}-1}\left( 2{\left| \left| \left| A^{\frac{1}{2}}B^ {\frac{1}{2}} \right| \right| \right| }\right) \left( \hbox {by Lemma 2.4}\right) \\\le & {} 2^{\frac{q}{2}-1}{\left| \left| \left| A+B \right| \right| \right| }\left( \hbox {by Lemma 2.5}\right) \\= & {} 2^{q\left( \frac{1}{2}-\frac{1}{q}\right) }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

This completes the proof. \(\square \)

Now, we are in a position to state and prove our main result.

Theorem 2.9

Let A and B be positive semidefinite matrices, and let \(t\in \left[ \frac{3}{4},1\right] \). Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{2\left( t-\frac{3}{4}\right) }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$
(2.1)

Proof

The result is obvious for \(t=1\). Now, for \(t\in \left[ \frac{3}{4},1\right) \), let

$$\begin{aligned} X= \left[ \begin{array}{ccccc} A^{2t-1} &{}\quad B^{2t-1} \\ 0&{}\quad 0 \end{array} \right] , Y= \left[ \begin{array}{ccccc} A^{1-t}B^{1-t} &{}\quad 0 \\ B^{1-t}A^{1-t}&{}\quad 0 \end{array} \right] . \end{aligned}$$

For the partition \(\frac{1}{p}+\frac{1}{q}=\left( 2t-1\right) +2\left( 1-t\right) =1\), noting that \(1\le p \le 2\) and \(2 \le q< \infty \), we have

$$\begin{aligned}{\left| \left| \left| b_{t} \right| \right| \right| }= & {} {\left| \left| \left| b_{t}\oplus 0 \right| \right| \right| }\\= & {} {\left| \left| \left| \left[ \begin{array}{ccccc} A^{2t-1} &{} B^{2t-1} \\ 0&{} 0 \end{array} \right] \left[ \begin{array}{ccccc} A^{1-t}B^{1-t} &{} 0 \\ B^{1-t}A^{1-t}&{} 0 \end{array} \right] \right| \right| \right| } \\= & {} {\left| \left| \left| XY \right| \right| \right| }\\\le & {} {\left| \left| \left| \left| X\right| ^{p} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left| Y\right| ^{q} \right| \right| \right| }^{\frac{1}{q}}\left( \hbox {by Lemma 2.1}\right) \\= & {} {\left| \left| \left| \left( X^{*}X\right) ^{\frac{p}{2}} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }^{\frac{1}{q}}\\= & {} {\left| \left| \left| \left( XX^{*}\right) ^{\frac{p}{2}} \right| \right| \right| }^{\frac{1}{p}} {\left| \left| \left| \left( Y^{*}Y\right) ^{\frac{q}{2}} \right| \right| \right| }^{\frac{1}{q}}\\\le & {} \left( {\left| \left| \left| A+B \right| \right| \right| }^{\frac{1}{p}}\right) \left( 2^{\frac{1}{2} -\frac{1}{q}}{\left| \left| \left| A+B \right| \right| \right| }^{\frac{1}{q}}\right) \left( \hbox {by Lemma 2.8}\right) \\= & {} 2^{\frac{1}{2}-\frac{1}{q}}{\left| \left| \left| A+B \right| \right| \right| }^{\frac{1}{p}+\frac{1}{q}}\\= & {} 2^{\frac{1}{2}-\frac{1}{q}}{\left| \left| \left| A+B \right| \right| \right| }\\= & {} 2^{\frac{1}{2}-2\left( 1-t\right) }{\left| \left| \left| A+B \right| \right| \right| }\\= & {} 2^{2\left( t-\frac{3}{4}\right) }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$

This completes the proof. \(\square \)

Based on Theorem 2.9, we have the following theorem.

Theorem 2.10

Let A and B be positive semidefinite matrices, and let \(t\in \left[ 0,\frac{1}{4}\right] \). Then for every unitarily invariant norm, we have

$$\begin{aligned} {\left| \left| \left| b_{t} \right| \right| \right| }\le 2^{2\left( \frac{1}{4}-t\right) }{\left| \left| \left| A+B \right| \right| \right| }. \end{aligned}$$
(2.2)

Proof

We have

$$\begin{aligned}{\left| \left| \left| b_{t} \right| \right| \right| }= & {} {\left| \left| \left| b^{*}_{t} \right| \right| \right| }\\= & {} {\left| \left| \left| b_{1-t} \right| \right| \right| }\left( \hbox {because}\ b^{*}_{t}=b_{1-t}\right) \\\le & {} 2^{2\left( \frac{1}{4}-t\right) }{\left| \left| \left| A+B \right| \right| \right| } \left( \hbox { since }1-t\in \left[ \frac{3}{4},1\right] \hbox { and by using Theorem 2.9}\right) . \end{aligned}$$

This completes the proof. \(\square \)

Remark 2.11

The inequality (2.1) is sharper than the inequality (1.1) for \(t\in \left[ \frac{3}{4},1\right] \), and the inequality (2.2) is sharper than the inequality (1.1) for \(t\in \left[ 0,\frac{1}{4}\right] \). In fact, for \(t\in \left[ \frac{3}{4},1\right] \), we have \(2\left( t-\frac{3}{4}\right) \le t-\frac{1}{2}\Longleftrightarrow t\le 1\), and for \(t\in \left[ 0,\frac{1}{4}\right] \), we have \(2\left( \frac{1}{4}-t\right) \le \frac{1}{2}-t\Longleftrightarrow 0\le t\).

Remark 2.12

Theorem 2.9 and Theorem 2.10 lead to an affirmative solution of Question 1.1 in the case of \(t=\frac{3}{4}\) and \(t=\frac{1}{4}\) for all unitarily invariant norms, which is a result due to Hayajneh and Kittaneh [18].