1 Introduction

On the space of \(n\times {n}\) complex matrices, denoted by \(M_{n}(\mathbb {C})\), \({\left| \left| \left| . \right| \right| \right| }\) stands for the unitarily invariant norm satisfying the property that \({\left| \left| \left| A \right| \right| \right| }={\left| \left| \left| UAV \right| \right| \right| }\) for all \(A,U,V \in {M_{n}(\mathbb {C})},\) where U and V are unitary. For \(A\in {M_{n}(\mathbb {C})}\), let \(s_{i}(A)\), for \(i = 1,2, \ldots ,n\), be the eigenvalues of the matrix \({\left| A \right| }=(A^{*}A)^{\frac{1}{2}}\), arranged so that \({s_i}(A) \ge {s_{i + 1}}(A)\), for \(i = 1,2, \ldots ,n - 1\). It is well known that for \(A\in {M_{n}(\mathbb {C})}\), \(s_{i}(A^{*})=s_{i}(A)\) and \(s_{i}({\left| A \right| })=s_{i}(A)\) for \(i=1,2,\ldots ,n\). For \(A\in {M_{n}(\mathbb {C})}\), the celebrated norm \({\left| \left| A \right| \right| }_{(k)}=\sum _{j=1}^{k} s_{j}(A)\) where \(k=1,2,\ldots ,n\) is called the Ky Fan norm of A, according to the Fan dominance theorem (see [4, p. 93]), \({\left| \left| \left| A \right| \right| \right| } \le {\left| \left| \left| B \right| \right| \right| }\) if and only if \({\left| \left| A \right| \right| }_{(k)}\le {\left| \left| B \right| \right| }_{(k)}\) for \(k=1,2,\ldots ,n\), where \(A,B\in {M_{n}(\mathbb {C})}\).

Throughout this paper, we confine our results to matrices regarded as operators acting on finite dimensional Hilbert space. Inequalities involving unitarily invariant norms have been of special interest; see e.g., [11].

The authors, in [7], obtained a matrix version of the arithmetic–geometric mean inequality; this version asserts that if \(A, B\in {M_{n}(\mathbb {C})}\), then

$$\begin{aligned} 2s_{i}(A^{*}B)\le s_{i}(AA^{*}+BB^{*}), \end{aligned}$$
(1)

for \(i=1,2,\ldots ,n\). Using the Fan dominance theorem and inequality (1), we get that

$$\begin{aligned} 2{\left| \left| \left| A^{*}B \right| \right| \right| } \le {\left| \left| \left| AA^{*}+BB^{*} \right| \right| \right| }. \end{aligned}$$
(2)

In [5, 9], a generalization of inequality (2) has been obtained, which asserts that if \(A, B, X\in {M_{n}(\mathbb {C})}\), then

$$\begin{aligned} 2{\left| \left| \left| A^{*}XB \right| \right| \right| } \le {\left| \left| \left| AA^{*}X+XBB^{*} \right| \right| \right| }. \end{aligned}$$
(3)

The Cauchy–Schwarz inequality for matrices says that if \(A, B\in {M_{n}(\mathbb {C})}\), then

$$\begin{aligned} {\left| \left| \left| A^{*}B \right| \right| \right| }^{2} \le {\left| \left| \left| AA^{*} \right| \right| \right| } {\left| \left| \left| BB^{*} \right| \right| \right| }. \end{aligned}$$
(4)

A generalization of inequality (4) asserts that if \(A, B, X\in {M_{n}(\mathbb {C})}\), then

$$\begin{aligned} {\left| \left| \left| A^{*}XB \right| \right| \right| }^{2}\le {\left| \left| \left| AA^{*}X \right| \right| \right| } {\left| \left| \left| XBB^{*} \right| \right| \right| }. \end{aligned}$$
(5)

We refer to [6, 8] and the references therein for more general forms of the Cauchy–Schwarz inequality.

Audenaert proved in [3] an inequality that interpolates between inequality (2) and inequality (4); this inequality says that if \(A, B\in {M_{n}(\mathbb {C})}\), then

$$\begin{aligned} {\left| \left| \left| A^{*}B \right| \right| \right| }^{2} \le {\left| \left| \left| pAA^{*}+(1-p)BB^{*} \right| \right| \right| } {\left| \left| \left| (1-p)AA^{*}+pBB^{*} \right| \right| \right| }, \end{aligned}$$
(6)

for all \(p\in [0,1]\).

In [10], the authors obtained an interpolation inequality between inequality (3) and inequality (5). This inequality, which is a generalization of inequality (6), asserts that if \(A, B, X\in M_{n}(\mathbb {C})\), then

$$\begin{aligned} {\left| \left| \left| A^{*}XB \right| \right| \right| }^{2} \le {\left| \left| \left| pAA^{*}X+(1-p)XBB^{*} \right| \right| \right| } {\left| \left| \left| (1-p)AA^{*}X+pXBB^{*} \right| \right| \right| }, \end{aligned}$$
(7)

for all \(p\in [0,1]\).

Very recently, it has been obtained in [2] that if \(A,B,S,T\in M_{n}(\mathbb {C})\), then

$$\begin{aligned} 2{s_i}(AS{T^*}{B^*}) \le {s_i}({S^*}{\left| A \right| ^2}S + {T^*}{\left| B \right| ^2}T), \end{aligned}$$
(8)

for \(i = 1,2, \ldots ,n\). The norm version of this inequality asserts that

$$\begin{aligned} 2 {\left| \left| \left| AS{T^*}{B^*} \right| \right| \right| } \le {\left| \left| \left| {S^*}{\left| A \right| ^2}S + {T^*}{\left| B \right| ^2}T \right| \right| \right| }. \end{aligned}$$
(9)

It has been concluded in the same paper that if \(A,B,S,T\in M_{n}(\mathbb {C})\) such that S and T are positive semidefinite, then

$$\begin{aligned} 2{s_i}(A{S^{\frac{1}{2}}}{T^{\frac{1}{2}}}{B^*}) \le {s_i}({S^{\frac{1}{2}}}{\left| A \right| ^2}{S^{\frac{1}{2}}} + {T^{\frac{1}{2}}}{\left| B \right| ^2}{T^{\frac{1}{2}}}), \end{aligned}$$
(10)

for \(i = 1,2, \ldots ,n\). The norm version of inequality (10) says that

$$\begin{aligned} 2{\left| \left| \left| A{S^{\frac{1}{2}}}{T^{\frac{1}{2}}}{B^*} \right| \right| \right| } \le {\left| \left| \left| {S^{\frac{1}{2}}}{\left| A \right| ^2}{S^{\frac{1}{2}}} + {T^{\frac{1}{2}}}{\left| B \right| ^2}{T^{\frac{1}{2}}} \right| \right| \right| }. \end{aligned}$$
(11)

Moreover, the author in this paper obtained that if \({A_i},{B_i},{S_i},{T_i}\in M_{n}(\mathbb {C})\) such that \({S_i}\) and \({T_i}\) are positive semidefinite for \(i = 1,2, \ldots ,n\), then

$$\begin{aligned} 2{s_i}\left( {\sum \limits _{i = 1}^n {{A_i}{S_i}^{\frac{1}{2}}{T_i}^{\frac{1}{2}}{B_i}^*} } \right) \le {s_i}^2\left( W \right) , \end{aligned}$$
(12)

for \(i = 1,2, \ldots ,n\), where \(W = \left[ {\begin{array}{*{20}{c}} {{A_1}{S_1}^{\frac{1}{2}}}&{}{{A_2}{S_2}^{\frac{1}{2}}}&{} \cdots &{}{{A_n}{S_n}^{\frac{1}{2}}}\\ {{B_1}{T_1}^{\frac{1}{2}}}&{}{{B_2}{T_2}^{\frac{1}{2}}}&{} \cdots &{}{{B_n}{T_n}^{\frac{1}{2}}} \end{array}} \right]\). Inequality (12) has the following norm version

$$\begin{aligned} 2{\left| \left| \left| {\sum \limits _{i = 1}^n {{A_i}{S_i}^{\frac{1}{2}}{T_i}^{\frac{1}{2}}{B_i}^*} } \right| \right| \right| } \le {\left| \left| \left| W \right| \right| \right| }^2. \end{aligned}$$
(13)

In [1], the authors obtained that if \(A_{i}, B_{i}, S_{i}\in {M_{n}(\mathbb {C})}\) for \(i=1, 2,\ldots , n\) such that each \(S_{i}\) is positive semidefinite and if \(p\in [0,1]\), then

$$\begin{aligned} {\left| \left| \left| \sum \limits _{i = 1}^n {{A_i}{S_i}{B_i}^*} \right| \right| \right| }^{2}\le & {} \mathop {\max }\limits _{1 \le i \le n} {\left| \left| {S_i} \right| \right| }^{2} \,\,{\left| \left| \left| \,{{\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p}{A_1}}&{}{\sqrt{p}{A_2}}&{} \cdots &{}{\sqrt{p}{A_n}}\\ {\sqrt{1-p}{B_1}}&{}{\sqrt{1-p}{B_2}}&{} \cdots &{}{\sqrt{1-p}{B_n}} \end{array}} \right] } \right| }^2} \right| \right| \right| } \nonumber \\&\quad \times \,\,{\left| \left| \left| \,{{\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1-p}{A_1}}&{}{\sqrt{1-p}{A_2}}&{} \cdots &{}{\sqrt{1-p}{A_n}}\\ {\sqrt{p}{B_1}}&{}{\sqrt{p}{B_2}}&{} \cdots &{}{\sqrt{p}{B_n}} \end{array}} \right] } \right| }^2} \right| \right| \right| }. \end{aligned}$$
(14)

In the second section, we give a generalizations of inequalities (9), (11), and a generalization of inequality (13) in the same manner in inequality (14).

2 Main results

Before we start our work, we need the following lemmas; the first can be found in [4], while the second is a direct consequence of Weyl’s monotonicity theorem (see [4, p. 63]) and Fan dominance theorem.

Lemma 1

If \(X \in {M_{n}(\mathbb {C})}\) is Hermitian, then

$$\begin{aligned} \pm X \le \left| X \right| . \end{aligned}$$

Lemma 2

Let \(A,B \in {M_{n}(\mathbb {C})}\) be positive semidefinite such that \(A \le B\). Then,

$$\begin{aligned} {\left| \left| \left| A \right| \right| \right| } \le {\left| \left| \left| B \right| \right| \right| }. \end{aligned}$$

We begin this section with our first result which can be considered as a generalization of inequality (9).

Lemma 3

Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\). Then,

$$\begin{aligned} {\left| \left| \left| AX{Y^*}{B^*} \right| \right| \right| }^2\le & {} {\left| \left| \left| p{X^*}{\left| A \right| ^2}X + \left( {1 - p} \right) {Y^*}{\left| B \right| ^2}Y \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left( {1 - p} \right) {X^*}{\left| A \right| ^2}X + p{Y^*}{\left| B \right| ^2}Y \right| \right| \right| }, \end{aligned}$$
(15)

for all \(p\in [0,1]\).

Proof

Replace A and B by \(X^{*}A^{*}\) and \(Y^{*}B^{*}\) respectively in inequality (6) to get the result.

Our second result is a generalization of inequality (11).

Lemma 4

Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be such that X and Y are positive semidefinite. Then,

$$\begin{aligned} {\left| \left| \left| A{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^*} \right| \right| \right| }^2\le & {} {\left| \left| \left| p{X^{\frac{1}{2}}}{\left| A \right| ^2}{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}{\left| B \right| ^2}{Y^{\frac{1}{2}}} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left( {1 - p} \right) {X^{\frac{1}{2}}}{\left| A \right| ^2}{X^{\frac{1}{2}}} + p{Y^{\frac{1}{2}}}{\left| B \right| ^2}{Y^{\frac{1}{2}}} \right| \right| \right| }, \end{aligned}$$
(16)

for all \(p\in [0,1]\).

Proof

Replace X and Y by \(X^\frac{1}{2}\) and \(Y^\frac{1}{2},\) respectively, in inequality (15) to get the result.

Note that substituting \(p=\frac{1}{2}\) and \(X=Y=I\) in inequalities (15) and (16) leads to the arithmetic–geometric mean inequality (2). Moreover, substituting \(p=0\) or 1 and \(X=Y=I\) in inequalities (15) and (16) leads to the Cauchy–Schwarz inequality (4).

Corollary 1

Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be positive semidefinite. Then,

$$\begin{aligned} {\left| \left| \left| A^{\frac{1}{2}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^{\frac{1}{2}}} \right| \right| \right| }^2\le & {} {\left| \left| \left| p{X^{\frac{1}{2}}}{A}{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}{B}{Y^{\frac{1}{2}}} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left( {1 - p} \right) {X^{\frac{1}{2}}}{A}{X^{\frac{1}{2}}} + p{Y^{\frac{1}{2}}}{B}{Y^{\frac{1}{2}}} \right| \right| \right| }, \end{aligned}$$
(17)

for all \(p\in [0,1]\).

Proof

Substitute \(A=A^\frac{1}{2}\) and \(B=B^\frac{1}{2}\) in inequality (16) to get the result.

Now, we will introduce the main result of our paper which can be considered as a generalization of inequality (13).

Theorem 1

Let \(A_{i}, B_{i}, X_{i}, Y_{i} \in {M_{n}(\mathbb {C})}\) be such that \(X_{i}\) and \(Y_{i}\) are positive semidefinite for \(i = 1,2, \ldots ,n\). Then,

$$\begin{aligned} {\left| \left| \left| \sum \limits _{i = 1}^n {{A_i}{X_i}^{\frac{1}{2}}{Y_i}^{\frac{1}{2}}{B_{i}^*}} \right| \right| \right| }^2 \le {\left| \left| \left| W(p) \right| \right| \right| } {\left| \left| \left| W(1-p) \right| \right| \right| }, \end{aligned}$$
(18)

for all \(p\in [0,1]\), where

$$\begin{aligned} W(t) = {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{t} {A_1}{X_1}^{\frac{1}{2}}}&{}{\sqrt{t} {A_2}{X_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{t} {A_n}{X_n}^{\frac{1}{2}}}\\ {\sqrt{1 - t} {B_1}{Y_1}^{\frac{1}{2}}}&{}{\sqrt{1 - t} {B_2}{Y_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{1 - t} {B_n}{Y_n}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2}. \end{aligned}$$

Proof

Consider the \(n^{2}\times n^{2}\) matrices

\(A = \left[ {\begin{array}{*{20}{c}} {{A_1}}&{}{{A_2}}&{} \cdots &{}{{A_n}}\\ 0&{}0&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}0 \end{array}} \right]\), \(B = \left[ {\begin{array}{*{20}{c}} {{B_1}}&{}{{B_2}}&{} \cdots &{}{{B_n}}\\ 0&{}0&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}0 \end{array}} \right]\), \(X = \left[ {\begin{array}{*{20}{c}} {{X_1}}&{}0&{} \cdots &{}0\\ 0&{}{{X_2}}&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}{{X_n}} \end{array}} \right]\), and \(Y = \left[ {\begin{array}{*{20}{c}} {{Y_1}}&{}0&{} \cdots &{}0\\ 0&{}{{Y_2}}&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}{{Y_n}} \end{array}} \right]\), and let \({M_{ij}} = p{X_i}^{\frac{1}{2}}{A_i}^*{A_j}{X_j}^{\frac{1}{2}} + (1 - p){Y_i}^{\frac{1}{2}}{B_i}^*{B_j}{Y_j}^{\frac{1}{2}}\). Then

$$\begin{aligned} A{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^*}=\left[ {\begin{array}{*{20}{c}} {\sum \limits _{i = 1}^n {{A_i}{X_i}^{\frac{1}{2}}{Y_i}^{\frac{1}{2}}{B^*_{i}}} }&{}0&{} \cdots &{}0\\ 0&{}0&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}0 \end{array}} \right] , \end{aligned}$$

and

$$\begin{aligned} p{X^{\frac{1}{2}}}{\left| A \right| ^2}{X^{\frac{1}{2}}} + (1 - p){Y^{\frac{1}{2}}}{\left| B \right| ^2}{Y^{\frac{1}{2}}}=\left[ {\begin{array}{*{20}{c}} {{M_{11}}}&{}{{M_{12}}}&{} \cdots &{}{{M_{1n}}}\\ {{M_{21}}}&{}{{M_{22}}}&{} \cdots &{}{{M_{2n}}}\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ {{M_{n1}}}&{}{{M_{n2}}}&{} \cdots &{}{{M_{nn}}} \end{array}} \right] =L^{*}L, \end{aligned}$$

where \(L=\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A_1}{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} {A_2}{X_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{p} {A_n}{X_n}^{\frac{1}{2}}}\\ {\sqrt{1 - p} {B_1}{Y_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} {B_2}{Y_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{1 - p} {B_n}{Y_n}^{\frac{1}{2}}} \end{array}} \right]\).

Thus,

$$\begin{aligned}&p{X^{\frac{1}{2}}}{\left| A \right| ^2}{X^{\frac{1}{2}}} + (1 - p){Y^{\frac{1}{2}}}{\left| B \right| ^2}{Y^{\frac{1}{2}}} \\= & {} {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A_1}{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} {A_2}{X_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{p} {A_n}{X_n}^{\frac{1}{2}}}\\ {\sqrt{1 - p} {B_1}{Y_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} {B_2}{Y_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{1 - p} {B_n}{Y_n}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2}=W(p). \end{aligned}$$

Similarly,

$$\begin{aligned}&(1 - p){X^{\frac{1}{2}}}{\left| A \right| ^2}{X^{\frac{1}{2}}} + p{Y^{\frac{1}{2}}}{\left| B \right| ^2}{Y^{\frac{1}{2}}} \\= & {} {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} {A_1}{X_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} {A_2}{X_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{1 - p} {A_n}{X_n}^{\frac{1}{2}}}\\ {\sqrt{p} {B_1}{Y_1}^{\frac{1}{2}}}&{}{\sqrt{p} {B_2}{Y_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{p} {B_n}{Y_n}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2}=W(1-p). \end{aligned}$$

Our desired result follows from applying inequality (16) to A, B, X and Y.

Corollary 2

Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) such that X and Y are positive semidefinite. Then,

$$\begin{aligned} {\left| \left| \left| A{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^*} + B{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{A^*} \right| \right| \right| }^{2}\le & {} {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} A{X^{\frac{1}{2}}}}&{}{\sqrt{p} B{X^{\frac{1}{2}}}}\\ {\sqrt{1 - p} B{Y^{\frac{1}{2}}}}&{}{\sqrt{1 - p} A{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} A{X^{\frac{1}{2}}}}&{}{\sqrt{1 - p} B{X^{\frac{1}{2}}}}\\ {\sqrt{p} B{Y^{\frac{1}{2}}}}&{}{\sqrt{p} A{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| }, \end{aligned}$$
(19)

for all \(p\in [0,1]\).

Proof

Let \(n=2, A_{1}=B_{2}=A, A_{2}=B_{1}=B, X_{1}=X_{2}=X\) and \(Y_{1}=Y_{2}=Y\) in inequality (18) to get the result.

Remark 1

Letting \(Y=X\) in inequality (19) leads to the following inequality:

$$\begin{aligned} {\left| \left| \left| AXB^* + BXA^* \right| \right| \right| }^{2}\le & {} {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} A{X^{\frac{1}{2}}}}&{}{\sqrt{p} B{X^{\frac{1}{2}}}}\\ {\sqrt{1 - p} B{X^{\frac{1}{2}}}}&{}{\sqrt{1 - p} A{X^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} A{X^{\frac{1}{2}}}}&{}{\sqrt{1 - p} B{X^{\frac{1}{2}}}}\\ {\sqrt{p} B{X^{\frac{1}{2}}}}&{}{\sqrt{p} A{X^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| }. \end{aligned}$$
(20)

Replacing X by I in inequality (20) leads to the following inequality:

$$\begin{aligned} {\left| \left| \left| AB^* + BA^* \right| \right| \right| }^{2} \le {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} A}&{}{\sqrt{p} B}\\ {\sqrt{1 - p} B}&{}{\sqrt{1 - p} A} \end{array}} \right] } \right| ^2} \right| \right| \right| } {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} A}&{}{\sqrt{1 - p} B}\\ {\sqrt{p} B}&{}{\sqrt{p} A} \end{array}} \right] } \right| ^2} \right| \right| \right| }. \end{aligned}$$
(21)

Corollary 3

Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be positive semidefinite matrices. Then,

$$\begin{aligned} {\left| \left| \left| S+T \right| \right| \right| }^2\le & {} {\left| \left| \left| \left( {H(p) + \left| {{K^*}(p)} \right| } \right) \oplus \left( {L(p) + \left| {K(p)} \right| } \right) \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left( {H(1 - p) + \left| {{K^*}(1 - p)} \right| } \right) \oplus \left( {L(1 - p) + \left| {K(1 - p)} \right| } \right) \right| \right| \right| }, \end{aligned}$$
(22)

for all \(p \in [0,1]\), where \(S={A^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{A^{\frac{1}{2}}}\), \(T={B^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^{\frac{1}{2}}}\), \(H(p)=p{X^{\frac{1}{2}}}A{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}A{Y^{\frac{1}{2}}}\), \(L(p)=p{X^{\frac{1}{2}}}B{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}B{Y^{\frac{1}{2}}}\), and \(K(p)=p{X^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}\). In particular, if \(Y=X\), then

$$\begin{aligned} {\left| \left| \left| {A^{\frac{1}{2}}}X{A^{\frac{1}{2}}} + {B^{\frac{1}{2}}}X{B^{\frac{1}{2}}} \right| \right| \right| } \le {\left| \left| \left| \left( {M + \left| {{Q^*}} \right| } \right) \oplus \left( {N + \left| {Q} \right| } \right) \right| \right| \right| } \end{aligned}$$
(23)

where \(M = {X^{\frac{1}{2}}}A{X^{\frac{1}{2}}}\), \(N = {X^{\frac{1}{2}}}B{X^{\frac{1}{2}}}\), and \(Q = {X^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{X^{\frac{1}{2}}}\). Moreover, if \(X=I\), then

$$\begin{aligned} {\left| \left| \left| A+B \right| \right| \right| } \le {\left| \left| \left| (A+\left| {{B^{\frac{1}{2}}}{A^{\frac{1}{2}}}} \right| ) \oplus (B+\left| {{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}} \right| ) \right| \right| \right| }. \end{aligned}$$
(24)

Proof

Let \(n=2\), \(A_{1}=B_{1}=A^{\frac{1}{2}}\), \(A_{2}=B_{2}=B^{\frac{1}{2}}\) , \(X_{1}=X_{2}=X\), and \(Y_{1}=Y_{2}=Y\) in inequality (18), to get that

$$\begin{aligned} {\left| \left| \left| {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{A^{\frac{1}{2}}} + {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^{\frac{1}{2}}} \right| \right| \right| }^2\le & {} {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}}&{}{\sqrt{p} {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}}\\ {\sqrt{1 - p} {A^{\frac{1}{2}}}{Y^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}}\\ {\sqrt{p} {A^{\frac{1}{2}}}{Y^{\frac{1}{2}}}}&{}{\sqrt{p} {B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| }. \end{aligned}$$
(25)

But,

$$\begin{aligned}&{\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}}&{}{\sqrt{p} {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}}\\ {\sqrt{1 - p} {A^{\frac{1}{2}}}{Y^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\&\quad = {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {\sqrt{p} {X^{\frac{1}{2}}}{A^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {Y^{\frac{1}{2}}}{A^{\frac{1}{2}}}}\\ {\sqrt{p} {X^{\frac{1}{2}}}{B^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {Y^{\frac{1}{2}}}{B^{\frac{1}{2}}}} \end{array}} \right] {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}}&{}{\sqrt{p} {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}}\\ {\sqrt{1 - p} {A^{\frac{1}{2}}}{Y^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| \right| \right| } \nonumber \\&\quad = {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}{K(p)}\\ {{K^*}(p)}&{}{L(p)} \end{array}} \right] \right| \right| \right| }. \end{aligned}$$
(26)

Similarly,

$$\begin{aligned} {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}}&{}{\sqrt{1 - p} {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}}\\ {\sqrt{p} {A^{\frac{1}{2}}}{Y^{\frac{1}{2}}}}&{}{\sqrt{p} {B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}} \end{array}} \right] } \right| ^2} \right| \right| \right| }= {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(1 - p)}&{}{K(1 - p)}\\ {{K^*}(1 - p)}&{}{L(1 - p)} \end{array}} \right] \right| \right| \right| }. \end{aligned}$$
(27)

Combining inequality (25) with the equations (26) and (27) leads to

$$\begin{aligned} {\left| \left| \left| {A^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{A^{\frac{1}{2}}} + {B^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^{\frac{1}{2}}} \right| \right| \right| }^2\le & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}{K(p)}\\ {{K^*}(p)}&{}{L(p)} \end{array}} \right] \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(1 - p)}&{}{K(1 - p)}\\ {{K^*}(1 - p)}&{}{L(1 - p)} \end{array}} \right] \right| \right| \right| }. \end{aligned}$$
(28)

Using Lemmas 1 and 2, we can get that

$$\begin{aligned} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}{K(p)}\\ {{K^*}(p)}&{}{L(p)} \end{array}} \right] \right| \right| \right| }= & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}0\\ 0&{}{L(p)} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{}{K(p)}\\ {{K^*}(p)}&{}0 \end{array}} \right] \right| \right| \right| } \nonumber \\\le & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}0\\ 0&{}{L(p)} \end{array}} \right] + \left| {\left[ {\begin{array}{*{20}{c}} 0&{}{K(p)}\\ {{K^*}(p)}&{}0 \end{array}} \right] } \right| \right| \right| \right| } \nonumber \\= & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(p)}&{}0\\ 0&{}{L(p)} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\left| {{K^*}(p)} \right| }&{}0\\ 0&{}{\left| {K(p)} \right| } \end{array}} \right] \right| \right| \right| } \nonumber \\= & {} {\left| \left| \left| \left( {H(p) + \left| {{K^*}(p)} \right| } \right) \oplus \left( {L(p) + \left| {K(p)} \right| } \right) \right| \right| \right| }. \end{aligned}$$
(29)

Similarly, we can get that

$$\begin{aligned}&{\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {H(1 - p)}&{}{K(1 - p)}\\ {{K^*}(1 - p)}&{}{L(1 - p)} \end{array}} \right] \right| \right| \right| } \nonumber \\&\quad \le {\left| \left| \left| \left( {H(1 - p) + \left| {{K^*}(1 - p)} \right| } \right) \oplus \left( {L(1 - p) + \left| {K(1 - p)} \right| } \right) \right| \right| \right| }. \end{aligned}$$
(30)

Combining inequality (28) with inequalities (29) and (30) leads to the desired result.

Corollary 4

Let \(A, B, X_{1}, Y_{1}, X_{2}, Y_{2} \in {M_{n}(\mathbb {C})}\) such that \(X_{1}, Y_{1}, X_{2}\) and \(Y_{2}\) are positive semidefinite matrices. Then,

$$\begin{aligned} {\left| \left| \left| C-D \right| \right| \right| }^2\le & {} {\left| \left| \left| \left( {L(p) + \left| {{K^*}(p)} \right| } \right) \oplus \left( {H(p) + \left| {K(p)} \right| } \right) \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| \left( {L(1 - p) + \left| {{K^*}(1 - p)} \right| } \right) \oplus \left( {H(1 - p) + \left| {K(1 - p)} \right| } \right) \right| \right| \right| }, \end{aligned}$$
(31)

for all \(p \in [0,1]\), where \(C = A{X_{1}}^{\frac{1}{2}}{Y_{1}}^{\frac{1}{2}}{A^{*}}\), \(D = B{X_{2}}^{\frac{1}{2}}{Y_{2}}^{\frac{1}{2}}{B^{*}}\), \(L(p) = p{X_1}^{\frac{1}{2}}{\left| A \right| ^2} {X_1}^{\frac{1}{2}} + (1 - p){Y_1}^{\frac{1}{2}}{\left| A \right| ^2}{Y_1}^{\frac{1}{2}}\), \(H(p) = p{X_2}^{\frac{1}{2}}{\left| B \right| ^2}{X_2}^{\frac{1}{2}} + (1 - p){Y_2}^{\frac{1}{2}}{\left| B \right| ^2}{Y_2}^{\frac{1}{2}}\) and \(K(p) = p{X_1}^{\frac{1}{2}}{A^*}B{X_2}^{\frac{1}{2}} - (1 - p){Y_1}^{\frac{1}{2}}{A^*}B{Y_2}^{\frac{1}{2}}\).

Proof

Letting \(n=2\), \(A_{1}=B_{1}=A\), and \(A_{2}=-B_{2}=B\) in inequality (18) leads to

$$\begin{aligned} {\left| \left| \left| A{X_1}^{\frac{1}{2}}{Y_1}^{\frac{1}{2}}{A^*} - B{X_2}^{\frac{1}{2}}{Y_2}^{\frac{1}{2}}{B^*} \right| \right| \right| }^2\le & {} {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} A{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} B{X_2}^{\frac{1}{2}}}\\ {\sqrt{1 - p} A{Y_1}^{\frac{1}{2}}}&{}{ - \sqrt{1 - p} B{Y_2}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\&\quad \times {\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} A{X_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} B{X_2}^{\frac{1}{2}}}\\ {\sqrt{p} A{Y_1}^{\frac{1}{2}}}&{}{ - \sqrt{p} B{Y_2}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2} \right| \right| \right| }. \end{aligned}$$
(32)

But,

$$\begin{aligned}&{\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{p} A{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} B{X_2}^{\frac{1}{2}}}\\ {\sqrt{1 - p} A{Y_1}^{\frac{1}{2}}}&{}{ - \sqrt{1 - p} B{Y_2}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \nonumber \\= & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {\sqrt{p} {X_1}^{\frac{1}{2}}{A^*}}&{}{\sqrt{1 - p} {Y_1}^{\frac{1}{2}}{A^*}}\\ {\sqrt{p} {X_2}^{\frac{1}{2}}{B^*}}&{}{ - \sqrt{1 - p} {Y_2}^{\frac{1}{2}}{B^*}} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {\sqrt{p} A{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} B{X_2}^{\frac{1}{2}}}\\ {\sqrt{1 - p} A{Y_1}^{\frac{1}{2}}}&{}{ - \sqrt{1 - p} B{Y_2}^{\frac{1}{2}}} \end{array}} \right] \right| \right| \right| } \nonumber \\= & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {L(p)}&{}{K(p)}\\ {{K^*}(p)}&{}{H(p)} \end{array}} \right] \right| \right| \right| } = {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {L(p)}&{}0\\ 0&{}{H(p)} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{}{K(p)}\\ {{K^*}(p)}&{}0 \end{array}} \right] \right| \right| \right| } \nonumber \\\le & {} {\left| \left| \left| \left[ {\begin{array}{*{20}{c}} {L(p)}&{}0\\ 0&{}{H(p)} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\left| {{K^*}(p)} \right| }&{}0\\ 0&{}{\left| {K(p)} \right| } \end{array}} \right] \right| \right| \right| } \nonumber \\= & {} {\left| \left| \left| \left( {L(p) + \left| {{K^*}(p)} \right| } \right) \oplus \left( {H(p) + \left| {K(p)} \right| } \right) \right| \right| \right| }. \end{aligned}$$
(33)

Similarly, we can get that

$$\begin{aligned}&{\left| \left| \left| {\left| {\left[ {\begin{array}{*{20}{c}} {\sqrt{1 - p} A{X_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} B{X_2}^{\frac{1}{2}}}\\ {\sqrt{p} A{Y_1}^{\frac{1}{2}}}&{}{ - \sqrt{p} B{Y_2}^{\frac{1}{2}}} \end{array}} \right] } \right| ^2} \right| \right| \right| } \le \nonumber \\&\quad {\left| \left| \left| \left( {L(1 - p) + \left| {{K^*}(1 - p)} \right| } \right) \oplus \left( {H(1 - p) + \left| {K(1 - p)} \right| } \right) \right| \right| \right| }. \end{aligned}$$
(34)

Combining inequalities (33) and (34) with inequality (32) leads to our result.