1 Introduction

Let \(\mathbb {B} ( \mathbb {H} )\) denote the space of all bounded linear operators on a complex separable Hilbert space \(\mathbb {H}\) and let \(\mathbb {K} ( \mathbb {H} )\) denote the two-sided ideal of compact operators in \(\mathbb {B} ( \mathbb {H} )\). For \(A\in \mathbb {K} ( \mathbb {H} )\), the singular values of A denoted by \(s_{_{1}}(A),s_{2}(A),...\) are the eigenvalues of the positive operator \(\left| A\right| =\left( A^{*}A\right) ^{1/2}\), which is denoted by \(\left| A\right| \ge 0\), enumerated as \(s_{_{1}}(A)\ge s_{2}(A)\ge \cdots\) and repeated according to multiplicity. Properties of singular values where \(A,B\in \mathbb {K} ( \mathbb {H} )\) are listed below:

  1. (a)
    $$\begin{aligned} s_{j}(A)=s_{j}(A^{*})=s_{j}(\left| A\right| )=s_{j}(\left| A^{*}\right| ) \end{aligned}$$
    (1)

    for \(j=1,2,...\)

  2. (b)

    If \(A,B\ge 0\) and \(A\le B\), then

    $$\begin{aligned} s_{j}(A)\le s_{j}(B) \end{aligned}$$
    (2)

    for \(j=1,2,...\) This fact follows by applying Weyl’s monotonicity principle (see, e.g., [7, p. 63] or [10, p. 26]). Moreover, \(s_{j}(A)\le s_{j}(B)\) if and only if \(s_{j}(A\oplus A)\le s_{j}(B\oplus B)\) for \(j=1,2,...\). Here, we use the direct sum notation \(A\oplus B\) for the block-diagonal operator \(\left[ \begin{array}{cc} A &{} 0 \\ 0 &{} B \end{array} \right]\) defined on \(\mathbb {H} \oplus \mathbb {H}\).

  3. (c)
    $$\begin{aligned} s_{j}\left[ \begin{array}{cc} A &{} 0 \\ 0 &{} B \end{array} \right] =s_{j}\left[ \begin{array}{cc} 0 &{} B \\ A &{} 0 \end{array} \right] \end{aligned}$$
    (3)

    for \(j=1,2,...\), and they consist of those of A together with those of B.

Some related inequalities with our study are summarized below where \(A,B,X,Y\in \mathbb {K} ( \mathbb {H} )\):

Bhatia and Kittaneh proved in [8] that if A is self-adjoint, \(B\ge 0\) and \(\ \pm A\le B,\) then

$$\begin{aligned} s_{j}(A)\le s_{j}(B\oplus B) \end{aligned}$$
(4)

for \(j=1,2,...\)

Audeh and Kittaneh obtained in [6] an equivalent inequality of (4):

If \(\left[ \begin{array}{cc} A &{} B \\ B^{*} &{} C \end{array} \right] \ge 0,\) then

$$\begin{aligned} s_{j}(B)\le s_{j}(A\oplus C) \end{aligned}$$
(5)

for \(\ j=1,2,...\)

Bhatia and Kittaneh in [9] obtained the arithmetic-geometric mean inequality of singular values,

$$\begin{aligned} 2s_{j}(AB^{*})\le s_{j}(A^{*}A+B^{*}B) \end{aligned}$$
(6)

for \(\ j=1,2,...\) Zhan proved in [12] that if \(A,B\ge 0\), then

$$\begin{aligned} s_{j}(A-B)\le s_{j}(A\oplus B) \end{aligned}$$
(7)

for \(\ j=1,2,...\) Hirzallah in [11] generalized inequality (6):

$$\begin{aligned} \sqrt{2}s_{j}\left( \left| A_{1}A_{2}^{*}+A_{3}A_{4}^{*}\right| ^{1/2}\right) \le s_{j}\left( \left[ \begin{array}{cc} A_{1} &{} A_{3} \\ A_{2} &{} A_{4} \end{array} \right] \right) \end{aligned}$$
(8)

for \(j=1,2,...\) Audeh in [4] gave another generalization of inequality (6):

$$\begin{aligned} s_{j}(AXY^{*}B^{*})\le \frac{1}{2}s_{j}\left( X^{*}\left| A\right| ^{2}X+Y^{*}\left| B\right| ^{2}Y\right) \end{aligned}$$
(9)

for \(j=1,2,...\) Moreover, it has been shown in the same paper that if \(X_{i},Y_{i}\ge 0\), \(i=1,2,...,n\). Then

$$\begin{aligned} 2s_{j}\left( \sum \limits _{i=1}^{n}A_{i}X_{i}^{1/2}Y_{i}^{1/2}B_{i}^{*}\right) \le s_{j}^{2}\left( W\right) \end{aligned}$$
(10)

for \(j=1,2,...,\) where \(W=\left[ \begin{array}{cccc} A_{1}X_{1}^{1/2} &{} A_{2}X_{2}^{1/2} &{} ... &{} A_{n}X_{n}^{1/2} \\ B_{1}Y_{1}^{1/2} &{} B_{2}Y_{2}^{1/2} &{} ... &{} B_{n}Y_{n}^{1/2} \end{array} \right]\). Several results are demonstrated as special cases for this inequality, some of these results are summarized below:

(i) Let \(X,Y\ge 0\). Then

$$\begin{aligned} 2s_{j}\left( AX^{1/2}Y^{1/2}B^{*}+BX^{1/2}Y^{1/2}A^{*}\right) \le s_{j}^{2}\left( \left[ \begin{array}{cc} AX^{1/2} &{} BX^{1/2} \\ BY^{1/2} &{} AY^{1/2} \end{array} \right] \right) \end{aligned}$$
(11)

for \(j=1,2,...\) In particular, replacing Y by X in inequality (11), leads to the following inequality:

$$\begin{aligned} 2s_{j}\left( AXB^{*}+BXA^{*}\right) \le s_{j}^{2}\left( \left[ \begin{array}{cc} AX^{1/2} &{} BX^{1/2} \\ BX^{1/2} &{} AX^{1/2} \end{array} \right] \right) \end{aligned}$$
(12)

for \(j=1,2,...\)

(ii) Let \(A,B,X\ge 0\). Then

$$\begin{aligned} s_{j}\left( A^{1/2}XA^{1/2}+B^{1/2}XB^{1/2}\right) \le s_{j}\left( (P+\left| Q^{*}\right| \right) \oplus \left( R+\left| Q\right| \right) \end{aligned}$$
(13)

for \(j=1,2,...\), where \(P=X^{1/2}AX^{1/2}\), \(Q=X^{1/2}A^{1/2}B^{1/2}X^{1/2}\) , and \(R=X^{1/2}BX^{1/2}\). Let \(X=I\), we have

$$\begin{aligned} s_{j}\left( A+B\right) \le s_{j}\left( \left( A+\left| B^{1/2}A^{1/2}\right| \right) \oplus \left( B+\left| A^{1/2}B^{1/2}\right| \right) \right) \end{aligned}$$
(14)

for \(j=1,2,...\)

(iii) Let \(X_{1},X_{2},Y_{1},Y_{2}\ge 0\). Then

$$\begin{aligned} 2s_{j}\left( E-F\right) \le s_{j}\left( \left( H+\left| L^{*}\right| \right) \oplus \left( K+\left| L\right| \right) \right) \end{aligned}$$
(15)

for \(j=1,2,...\),where \(E=AX_{1}^{1/2}Y_{1}^{1/2}A^{*}\), \(F=BX_{2}^{1/2}Y_{2}^{1/2}B^{*}\), \(H=X_{1}^{1/2}A^{*}AX_{1}^{1/2}+Y_{1}^{1/2}A^{*}AY_{1}^{1/2}\), \(L=X_{1}^{1/2}A^{*}BX_{2}^{1/2}-Y_{1}^{1/2}A^{*}BY_{2}^{1/2}\), and \(K=X_{2}^{1/2}B^{*}BX_{2}^{1/2}+Y_{2}^{1/2}B^{*}BY_{2}^{1/2}\). For recent studies about generalizations and applications for singular value inequalities, we refer the reader to [1,2,3,4,5,6].

In Sect. 2, we provide generalizations of the inequalities (6)–(15).

2 Singular value inequalities for compact operators

The following lemma is well-known.

Lemma 2.1

Let A be self-adjoint. Then

$$\begin{aligned} \pm A\le \left| A\right| \text {.} \end{aligned}$$
(16)

We are ready to state the first main result in this section.

Theorem 2.2

Let\(\ A_{1},A_{2},B_{1},B_{2},X_{1},X_{2},Y_{1},Y_{2}\in \mathbb {K} ( \mathbb {H} )\). Then

$$\begin{aligned} 2s_{j}(A_{1}^{*}X_{1}^{*}X_{2}A_{2}+B_{1}^{*}Y_{1}^{*}Y_{2}B_{2})\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(17)

for \(j=1,2,...\) where

$$\begin{aligned}&L=X_{1}A_{1}A_{1}^{*}X_{1}^{*}+X_{2}A_{2}A_{2}^{*}X_{2}^{*} \text {,} \\&M=Y_{1}B_{1}B_{1}^{*}Y_{1}^{*}+Y_{2}B_{2}B_{2}^{*}Y_{2}^{*} \end{aligned}$$

and

$$\begin{aligned} N=Y_{1}B_{1}A_{1}^{*}X_{1}^{*}+Y_{2}B_{2}A_{2}^{*}X_{2}^{*} \text {.} \end{aligned}$$

Proof

In what follows in this proof, let

$$\begin{aligned} E=\left[ \begin{array}{cc} X_{1}A_{1}A_{1}^{*}X_{1}^{*}+X_{2}A_{2}A_{2}^{*}X_{2}^{*} &{} 0 \\ 0 &{} Y_{1}B_{1}B_{1}^{*}Y_{1}^{*}+Y_{2}B_{2}B_{2}^{*}Y_{2}^{*} \end{array} \right] \end{aligned}$$

and

$$\begin{aligned} F=\left[ \begin{array}{cc} 0 &{} X_{1}A_{1}B_{1}^{*}Y_{1}^{*}+X_{2}A_{2}B_{2}^{*}Y_{2}^{*}\\ Y_{1}B_{1}A_{1}^{*}X_{1}^{*}+Y_{2}B_{2}A_{2}^{*}X_{2}^{*} &{} 0 \end{array} \right] \text {.} \end{aligned}$$

Let \(S=\left[ \begin{array}{cc} X_{1}A_{1} &{} 0 \\ Y_{1}B_{1} &{} 0 \end{array} \right]\) and \(T=\left[ \begin{array}{cc} X_{2}B_{2} &{} 0 \\ Y_{2}B_{2} &{} 0 \end{array} \right]\). Note that

$$\begin{aligned} S^{*}T=A_{1}^{*}X_{1}^{*}X_{2}A_{2}+B_{1}^{*}Y_{1}^{*}Y_{2}B_{2} \end{aligned}$$

and

$$\begin{aligned} SS^{*}+TT^{*}=C+D \end{aligned}$$

where

$$\begin{aligned} C=\left[ \begin{array}{cc} X_{1}A_{1}A_{1}^{*}X_{1}^{*} &{} X_{1}A_{1}B_{1}^{*}Y_{1}^{*}\\ Y_{1}B_{1}A_{1}^{*}X_{1}^{*} &{} Y_{1}B_{1}B_{1}^{*}Y_{1}^{*} \end{array} \right] \end{aligned}$$

and

$$\begin{aligned} D=\left[ \begin{array}{cc} X_{2}A_{2}A_{2}^{*}X_{2}^{*} &{} X_{2}A_{2}B_{2}^{*}Y_{2}^{*}\\ Y_{2}B_{2}A_{2}^{*}X_{2}^{*} &{} Y_{2}B_{2}B_{2}^{*}Y_{2}^{*} \end{array} \right] \text {.} \end{aligned}$$

Apply inequality (6) for the operator matrices S and T, we get

$$\begin{aligned}&2s_{j}(A_{1}^{*}X_{1}^{*}X_{2}A_{2}+B_{1}^{*}Y_{1}^{*}Y_{2}B_{2}) \le s_{j}(C+D) \\&\quad = s_{j}(E+F) \\&\quad = s_{j}\left[ \begin{array}{cc} L &{} N^{*} \\ N &{} M \end{array} \right] \\&\quad =s_{j}\left( \left[ \begin{array}{cc} L &{} 0 \\ 0 &{} M \end{array} \right] +\left[ \begin{array}{cc} 0 &{} N^{*} \\ N &{} 0 \end{array} \right] \right) \\&\quad \le s_{j}\left( \left[ \begin{array}{cc} L &{} 0 \\ 0 &{} M \end{array} \right] +\left[ \begin{array}{cc} \left| N\right| &{} 0 \\ 0 &{} \left| N^{*}\right| \end{array} \right] \right) , \\&\quad \text {(By applying inequalities (1.2) and (2.1)).} \\&\quad = s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \text {.} \end{aligned}$$

Thus inequality (17) has thus been substantiated. \(\square\)

In the following, we will see some special cases of inequality (17).

Remark 2.3

Letting \(X_{1}=X_{2}=I\), \(B_{1}=B_{2}=Y_{1}=Y_{2}=0\) in inequality (17), we give inequality (6).

Remark 2.4

Letting \(A_{1}=A_{2}=A^{1/2}\), \(B_{1}=-B_{2}=B^{1/2}\), \(X_{1}=X_{2}=Y_{1}=Y_{2}=I\) in inequality (17), implies inequality (7).

Remark 2.5

Letting \(B_{1}=B_{2}=Y_{1}=Y_{2}=0\) in inequality (17), leads to inequality (9).

Remark 2.6

Letting \(A_{1}=A_{2}=A^{1/2}\), \(B_{1}=B_{2}=B^{1/2}\), and \(X_{1}=X_{2}=Y_{1}=Y_{2}=I\) in inequality (17), one can get inequality (14).

In the following, we will present special case of inequality (17) which in turns a generalization of several known results.

Corollary 2.7

Let\(\ A_{1},A_{2},B_{1},B_{2},X\in \mathbb {K} ( \mathbb {H} )\) such that \(X\ge 0\). Then

$$\begin{aligned} 2s_{j}(A_{1}^{*}XA_{2}+B_{1}^{*}XB_{2})\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(18)

for \(j=1,2,...\) where

$$\begin{aligned}&L=X^{1/2}\left( A_{1}A_{1}^{*}+A_{2}A_{2}^{*}\right) X^{1/2}\text {,}\\&M=X^{1/2}\left( B_{1}B_{1}^{*}+B_{2}B_{2}^{*}\right) X^{1/2} \end{aligned}$$

and

$$\begin{aligned} N=X^{1/2}B_{1}A_{1}^{*}X^{1/2}+X^{1/2}B_{2}A_{2}^{*}X^{1/2}\text {.} \end{aligned}$$

Proof

Letting \(X_{1}=X_{2}=Y_{1}=Y_{2}=X^{1/2}\) in inequality (17), we give inequality (18). \(\square\)

Example

Let \(A_{1}=B_{1}=\left[ \begin{array}{cc} i &{} 0 \\ 0 &{} i \end{array} \right]\), \(A_{2}=B_{2}=\left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} -1 \end{array} \right]\), and \(X=\left[ \begin{array}{cc} 4 &{} 0 \\ 0 &{} 1 \end{array} \right]\). Then \(2s_{j}(A_{1}^{*}XA_{2}+B_{1}^{*}XB_{2})=16,4,0,0\) for \(j=1,2,3,4,\) and \(s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) =16,16,4,4\) for \(j=1,2,3,4.\)

For \(A,B,X\in \mathbb {B} ( \mathbb {H} )\), an operator of the form \(AX-XA\) is called a commutator and an operator of the form \(AX+XA\) is called anticommutator. Now we are ready to state the following generalization of singular value inequality for anticommutators.

Corollary 2.8

Let\(\ A,B,X\in \mathbb {K} ( \mathbb {H} )\) such that \(X\ge 0\). Then

$$\begin{aligned} 2s_{j}(AXB+BXA)\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(19)

for \(j=1,2,...\) where

$$\begin{aligned}&L=X^{1/2}\left( A^{*}A+BB^{*}\right) X^{1/2}\text {,} \\&M=X^{1/2}\left( AA^{*}+B^{*}B\right) X^{1/2} \end{aligned}$$

and

$$\begin{aligned} N=X^{1/2}B^{*}AX^{1/2}+X^{1/2}AB^{*}X^{1/2}\text {.} \end{aligned}$$

Proof

Let \(A_{1}^{*}=B_{2}=A\) and \(A_{2}=B_{1}^{*}=B\) in inequality (18), we give inequality (19). \(\square\)

Example

Let \(A=\left[ \begin{array}{cc} 0 &{} i \\ i &{} 0 \end{array} \right] ,\) \(B=\left[ \begin{array}{cc} 0 &{} 2i \\ i &{} 0 \end{array} \right] ,\) and \(X=\left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} 4 \end{array} \right]\). Then \(2s_{j}(AXB+BXA)=24,6,0,0\) for \(j=1,2,3,4,\) and \(s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) =32,20,8,5\) for \(j=1,2,3,4.\)

A remarkable inequality for singular value inequalities of anticommutators is now ready to present.

Corollary 2.9

Let\(\ A,B\in \mathbb {K} ( \mathbb {H} )\). Then

$$\begin{aligned} 2s_{j}(AB+BA)\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(20)

for \(j=1,2,...\) where

$$\begin{aligned}&L=\left( A^{*}A+BB^{*}\right) \text {,}\\&M=\left( AA^{*}+B^{*}B\right) \end{aligned}$$

and

$$\begin{aligned} N=B^{*}A+AB^{*}\text {.} \end{aligned}$$

Proof

Letting \(X=I\) in inequality (19), we give inequality (20 ). \(\square\)

Corollary 2.10

Let\(\ A_{1},A_{2},B_{1},B_{2},X_{1},X_{2},Y_{1},Y_{2}\in \mathbb {K} ( \mathbb {H} )\). Then

$$\begin{aligned} 2s_{j}(A_{1}^{*}X_{1}^{*}X_{2}A_{2}-B_{1}^{*}Y_{1}^{*}Y_{2}B_{2})\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(21)

for \(j=1,2,...\) where

$$\begin{aligned}&L=X_{1}A_{1}A_{1}^{*}X_{1}^{*}+X_{2}A_{2}A_{2}^{*}X_{2}^{*} \text {,}\\&M=Y_{1}B_{1}B_{1}^{*}Y_{1}^{*}+Y_{2}B_{2}B_{2}^{*}Y_{2}^{*} \end{aligned}$$

and

$$\begin{aligned} N=Y_{1}B_{1}A_{1}^{*}X_{1}^{*}-Y_{2}B_{2}A_{2}^{*}X_{2}^{*} \text {.} \end{aligned}$$

Proof

Substituting \(B_{2}\) by \(-B_{2}\) in inequality (17), we give inequality (21). \(\square\)

We will present the following inequality which extends singular value inequality of commutators.

Corollary 2.11

Let\(\ A_{1},A_{2},B_{1},B_{2}\in \mathbb {K} ( \mathbb {H} )\). Then

$$\begin{aligned} 2s_{j}(A_{1}^{*}A_{2}-B_{1}^{*}B_{2})\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(22)

for \(j=1,2,...\) where

$$\begin{aligned}&L=A_{1}A_{1}^{*}+A_{2}A_{2}^{*}\text {,}\\&M=B_{1}B_{1}^{*}+B_{2}B_{2}^{*} \end{aligned}$$

and

$$\begin{aligned} N=B_{1}A_{1}^{*}-B_{2}A_{2}^{*}\text {.} \end{aligned}$$

Proof

Letting \(X_{1}=X_{2}=Y_{1}=Y_{2}=I\) in inequality (21), we give inequality (22). \(\square\)

Now we state the singular value inequality of commutators.

Corollary 2.12

Let\(\ A,B\in \mathbb {K} ( \mathbb {H} )\). Then

$$\begin{aligned} 2s_{j}(AB-BA)\le s_{j}\left( \left( L+\left| N\right| \right) \oplus \left( M+\left| N^{*}\right| \right) \right) \end{aligned}$$
(23)

for \(j=1,2,...\) where

$$\begin{aligned}&L=A^{*}A+BB^{*}\text {,}\\&M=AA^{*}+B^{*}B \end{aligned}$$

and

$$\begin{aligned} N=B^{*}A-AB^{*}\text {.} \end{aligned}$$

Proof

Letting \(A_{1}^{*}=B_{2}=A\) and \(A_{2}=B_{1}^{*}=B\) in inequality (22), we give inequality (23). \(\square\)

We are ready to state the second general result of this section.

Theorem 2.13

Let \(A_{i},B_{i},X_{i},Y_{i}\in \mathbb {K} ( \mathbb {H} )\), \(i=1,2,3,4\). Then

$$\begin{aligned} 2s_{j}\left( K+L\right) \le s_{j}\left( \left( O+\left| T\right| \right) \oplus \left( V+\left| T^{*}\right| \right) \right) \end{aligned}$$
(24)

where

$$\begin{aligned}&K=A_{1}^{*}X_{1}^{*}X_{3}A_{3}+A_{2}^{*}X_{2}^{*}X_{4}A_{4} \text {,}\\&L=B_{1}^{*}Y_{1}^{*}Y_{3}B_{3}+B_{2}^{*}Y_{2}^{*}Y_{4}B_{4} \text {,}\\&O=\left[ \begin{array}{cc} X_{1}A_{1}A_{1}^{*}X_{1}^{*}+X_{3}A_{3}A_{3}^{*}X_{3}^{*} &{} X_{1}A_{1}A_{2}^{*}X_{2}^{*}+X_{3}A_{3}A_{4}^{*}X_{4}^{*} \\ X_{2}A_{2}A_{1}^{*}X_{1}^{*}+X_{4}A_{4}A_{3}^{*}X_{3}^{*} &{} X_{2}A_{2}A_{2}^{*}X_{2}^{*}+X_{4}A_{4}A_{4}^{*}X_{4}^{*} \end{array} \right] \text {,}\\&V=\left[ \begin{array}{cc} Y_{1}B_{1}B_{1}^{*}Y_{1}^{*}+Y_{3}B_{3}B_{3}^{*}Y_{3} &{} Y_{1}B_{1}B_{2}^{*}Y_{2}^{*}+Y_{3}B_{3}B_{4}^{*}Y_{4}^{*} \\ Y_{2}B_{2}B_{1}^{*}Y_{1}^{*}+Y_{4}B_{4}B_{3}^{*}Y_{3}^{*} &{} Y_{2}B_{2}B_{2}^{*}Y_{2}^{*}+Y_{4}B_{4}B_{4}^{*}Y_{4}^{*} \end{array} \right] \end{aligned}$$

and

$$\begin{aligned} T=\left[ \begin{array}{cc} Y_{1}B_{1}A_{1}^{*}X_{1}^{*}+Y_{3}B_{3}A_{3}^{*}X_{3} &{} Y_{1}B_{1}A_{2}^{*}X_{2}^{*}+Y_{3}B_{3}A_{4}^{*}X_{4}^{*} \\ Y_{2}B_{2}A_{1}^{*}X_{1}^{*}+Y_{4}B_{4}A_{3}^{*}X_{3}^{*} &{} Y_{2}B_{2}A_{2}^{*}X_{2}^{*}+Y_{4}B_{4}A_{4}^{*}X_{4}^{*} \end{array} \right] \end{aligned}$$

for \(j=1,2,...\)

Proof

On \(\oplus _{j=1}^{2}H\), let \(C_{1}=\left[ \begin{array}{cc} A_{1} &{} 0 \\ A_{2} &{} 0 \end{array} \right]\), \(C_{2}=\left[ \begin{array}{cc} A_{3} &{} 0 \\ A_{4} &{} 0 \end{array} \right]\), \(D_{1}=\left[ \begin{array}{cc} B_{1} &{} 0 \\ B_{2} &{} 0 \end{array} \right] ,\) \(D_{2}=\left[ \begin{array}{cc} B_{3} &{} 0 \\ B_{4} &{} 0 \end{array} \right]\), \(S_{1}=\left[ \begin{array}{cc} X_{1} &{} 0 \\ 0 &{} X_{2} \end{array} \right]\), \(S_{2}=\left[ \begin{array}{cc} X_{3} &{} 0 \\ 0 &{} X_{4} \end{array} \right]\), \(T_{1}=\left[ \begin{array}{cc} Y_{1} &{} 0 \\ 0 &{} Y_{2} \end{array} \right]\), and \(T_{2}=\left[ \begin{array}{cc} Y_{3} &{} 0 \\ 0 &{} Y_{4} \end{array} \right]\). It follows that \(C_{1}^{*}S_{1}^{*}S_{2}C_{2}+D_{1}^{*}T_{1}^{*}T_{2}D_{2}=K+L\), \(S_{1}C_{1}C_{1}^{*}S_{1}^{*}+S_{2}C_{2}C_{2}^{*}S_{2}^{*}=O\), \(T_{1}D_{1}D_{1}^{*}T_{1}^{*}+T_{2}D_{2}D_{2}^{*}T_{2}^{*}=V\) and \(T_{1}D_{1}C_{1}^{*}S_{1}^{*}+T_{2}D_{2}C_{2}^{*}S_{2}^{*}=T\) . Substitute the operators \(A_{1},A_{2},B_{1},B_{2},X_{1},X_{2},Y_{1}\) and \(Y_{2}\) by \(C_{1},C_{2},D_{1},D_{2},S_{1},S_{2},T_{1}\) and \(T_{2}\), respective‘ly, in inequality (17), we give inequality (24). \(\square\)

Inequality (24) is an extension of several known results, some of them are listed below.

Remark 2.14

Letting \(B_{i}=0\) for \(i=1,2,3,4\), \(X_{1}=X_{1}^{1/2}\), \(X_{2}=X_{2}^{1/2}\), \(X_{3}=Y_{1}^{1/2}\), \(X_{4}=Y_{2}^{1/2}\) in inequality (24), we give inequality (10) for \(n=2\).

Remark 2.15

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\), \(A_{1}=A_{4}=A\), \(A_{2}=A_{3}=B\), \(X_{1}=X_{2}=X^{1/2}\) and \(X_{3}=X_{4}=Y^{1/2\text { }}\)in inequality (24), one can get inequality (11).

Remark 2.16

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\), \(A_{1}=A_{4}=A\), \(A_{2}=A_{3}=B\), \(X_{1}=X_{2}=X_{3}=X_{4}=X^{1/2}\) in inequality (24), leads to inequality (12).

Remark 2.17

Letting \(B_{i}=Y_{i}=0\) and \(X_{i}=I\) for \(i=1,2,3,4\) in inequality (24), implies inequality (8).

Remark 2.18

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\) and \(A_{i}=0\) for \(i=2,4\) in inequality (24), we have inequality (9).

Remark 2.19

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\), \(A_{i}=0\) for \(i=2,4\), and \(X_{1}=X_{3}=I\) in inequality (24), we get inequality (6).

We will give a special case of inequality (24) which is a generalization of inequality (13).

Corollary 2.20

Let \(A,B,X,Y\in \mathbb {K} ( \mathbb {H} )\ge 0\). Then

$$\begin{aligned} s_{j}\left( A^{1/2}XA^{1/2}+B^{1/2}YB^{1/2}\right) \le s_{j}\left( (P+\left| Q^{*}\right| \right) \oplus \left( R+\left| Q\right| \right) \end{aligned}$$
(25)

for \(j=1,2,...\) where \(P=X^{1/2}AX^{1/2}\), \(Q=X^{1/2}A^{1/2}B^{1/2}Y^{1/2}\) and \(R=Y^{1/2}BY^{1/2}\). In particular, letting \(Y=X\) in inequality (25), we give

$$\begin{aligned} s_{j}\left( A^{1/2}XA^{1/2}+B^{1/2}XB^{1/2}\right) \le s_{j}\left( (P+\left| T^{*}\right| \right) \oplus \left( S+\left| T\right| \right) \end{aligned}$$

for \(j=1,2,...\), where \(P=X^{1/2}AX^{1/2}\), \(T=X^{1/2}A^{1/2}B^{1/2}X^{1/2}\) and \(S=X^{1/2}BX^{1/2}\). Moreover, letting \(X=I\) in inequality (13), we give inequality (14).

Proof

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\), \(A_{1}=A_{3}=A^{1/2},\) \(A_{2}=A_{4}=B^{1/2},\) \(X_{1}=X_{3}=X^{1/2}\) and \(X_{2}=X_{4}=Y^{1/2}\) in inequality (24), leads to

$$\begin{aligned}&2s_{j}(A^{1/2}XA^{1/2}+B^{1/2}YB^{1/2}) \nonumber \\&\quad \le 2s_{j}\left( \left[ \begin{array}{cc} X^{1/2}AX^{1/2} &{} X^{1/2}A^{1/2}B^{1/2}Y^{1/2} \\ Y^{1/2}B^{1/2}A^{1/2}X^{1/2} &{} Y^{1/2}BY^{1/2} \end{array} \right] \right) \nonumber \\&\quad =2s_{j}\left( \left[ \begin{array}{cc} P &{} Q \\ Q^{*} &{} R \end{array} \right] \right) \nonumber \\&\quad =2s_{j}\left( \left[ \begin{array}{cc} P &{} 0 \\ 0 &{} R \end{array} \right] +\left[ \begin{array}{cc} 0 &{} Q \\ Q^{*} &{} 0 \end{array} \right] \right) . \end{aligned}$$
(26)

But

$$\begin{aligned} \left[ \begin{array}{cc} 0 &{} Q \\ Q^{*} &{} 0 \end{array} \right] \le \left| \left[ \begin{array}{cc} 0 &{} Q \\ Q^{*} &{} 0 \end{array} \right] \right| =\left[ \begin{array}{cc} \left| Q^{*}\right| &{} 0 \\ 0 &{} \left| Q\right| \end{array} \right] \end{aligned}$$

combining this with inequality (26), one can get

$$\begin{aligned}&s_{j}(A^{1/2}XA^{1/2}+B^{1/2}YB^{1/2}) \nonumber \\&\quad \le s_{j}\left( \left[ \begin{array}{cc} P &{} 0 \\ 0 &{} R \end{array} \right] +\left[ \begin{array}{cc} \left| Q^{*}\right| &{} 0 \\ 0 &{} \left| Q\right| \end{array} \right] \right) \nonumber \\&\quad = s_{j}\left( (P+\left| Q^{*}\right| \right) \oplus \left( R+\left| Q\right| \right) . \end{aligned}$$
(27)

\(\square\)

In the following, we will give another special case of inequality (24) which has been proved in [13].

Corollary 2.21

Let \(A,B,X_{1},X_{2},Y_{1},Y_{2}\in \mathbb {K} ( \mathbb {H} )\) such that\(\ X_{1},X_{2},Y_{1},Y_{2}\ge 0\). Then

$$\begin{aligned} 2s_{j}\left( E-F\right) \le s_{j}\left( \left( H+\left| L^{*}\right| \right) \oplus \left( K+\left| L\right| \right) \right) \end{aligned}$$

for \(j=1,2,...\), where \(E=AX_{1}^{1/2}Y_{1}^{1/2}A^{*}\), \(F=BX_{2}^{1/2}Y_{2}^{1/2}B^{*}\), \(H=X_{1}^{1/2}A^{*}AX_{1}^{1/2}+Y_{1}^{1/2}A^{*}AY_{1}^{1/2}\), \(L=X_{1}^{1/2}A^{*}BX_{2}^{1/2}-Y_{1}^{1/2}A^{*}BY_{2}^{1/2}\) and \(K=X_{2}^{1/2}B^{*}BX_{2}^{1/2}+Y_{2}^{1/2}B^{*}BY_{2}^{1/2}\).

Proof

Letting \(B_{i}=Y_{i}=0\) for \(i=1,2,3,4\), \(A_{1}=A_{3}=A^{*}\), \(A_{2}=-A_{4}=B^{*}\), \(X_{1}=X_{1}^{1/2}\), \(X_{2}=X_{2}^{1/2}\), \(X_{3}=Y_{1}^{1/2}\) and \(X_{4}=Y_{2}^{1/2}\) in inequality (24), we give

$$\begin{aligned}&2s_{j}\left( AX_{1}^{1/2}Y_{1}^{1/2}A^{*}-BX_{2}^{1/2}Y_{2}^{1/2}B^{*}\right) \\&\quad \le s_{j}\left( \left[ \begin{array}{cc} H &{} L \\ L^{*} &{} K \end{array} \right] \right) \\&\quad =s_{j}\left( \left[ \begin{array}{cc} H &{} 0 \\ 0 &{} K \end{array} \right] +\left[ \begin{array}{cc} 0 &{} L \\ L^{*} &{} 0 \end{array} \right] \right) \\&\quad \le s_{j}\left( \left[ \begin{array}{cc} H &{} 0 \\ 0 &{} K \end{array} \right] +\left[ \begin{array}{cc} \left| L^{*}\right| &{} 0 \\ 0 &{} \left| L\right| \end{array} \right] \right) \\&\quad = s_{j}\left( \left( H+\left| L^{*}\right| \right) \oplus \left( K+\left| L\right| \right) \right) \text {,} \end{aligned}$$

which is exactly inequality (15). \(\square\)

3 Singular value inequalities for matrices

Let \(\mathbb {M} _{n}\) be the space of all \(n\times n\) complex matrices In this section, attractive generalizations of inequalities (6) and (7) for matrices are proved.

Bhatia and Kittaneh in [8] proved that if \(A,B\in \mathbb {M} _{n}\) and \(Q=AA^{*}+BB^{*}\), then

$$\begin{aligned} s_{j}(AB^{*}+BA^{*})\le s_{j}(Q\oplus Q) \end{aligned}$$
(28)

for\(\ j=1,2,...,n\). Among our results in this section, we obtained an inequality that is sharper than inequality (28).

Recently in [13] a new generalization of inequality (6) has been given: If \(A,B,X\in \mathbb {M} _{n}\) such that \(X\ge 0\), then

$$\begin{aligned} 2s_{j}(AXB^{*})\le s_{j}\left[ \left( \left| A\right| ^{2}+\left| B\right| ^{2}\right) ^{1/2}X\left( \left| A\right| ^{2}+\left| B\right| ^{2}\right) ^{1/2}\right] \end{aligned}$$
(29)

for \(j=1,2,...,n\). In this section, we have established singular value inequality that is equivalent to inequality (29). Several relevant singular value inequalities are also given.

We start this section with the following lemmas.

Lemma 3.1

Let A be self-adjoint matrix. Then

$$\begin{aligned} \pm A\le \left| A\right| \end{aligned}$$
(30)

Lemma 3.2

Let \(A,B,X\in \mathbb {M} _{n}\) such that \(X\ge 0\). Then

$$\begin{aligned} s_{j}(AXB^{*})\le \frac{1}{2}s_{j}\left( X^{1/2}\left| A\right| ^{2}X^{1/2}+X^{1/2}\left| B\right| ^{2}X^{1/2}\right) \end{aligned}$$
(31)

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

Proof

Inequality (31) is a direct consequence of inequality (6) by substituting \(A=AX^{1/2}\) and \(B=BX^{1/2}\). \(\square\)

Corollary 3.3

Let\(\ A,B,X\ge 0\). Then

$$\begin{aligned} s_{j}(A^{1/2}XB^{1/2})\le \frac{1}{2}s_{j}\left( X^{1/2}AX^{1/2}+X^{1/2}BX^{1/2}\right) \end{aligned}$$
(32)

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

Proof

Inequality (32) is followed from Lemma 3.2 by substituting \(A=A^{1/2}\) and \(B=B^{1/2}.\) \(\square\)

Now, we can present the first result of this section, which is an impressive generalization of arithmetic–geometric mean inequality.

Theorem 3.4

Let \(A_{i},B_{i},X_{i}\in \mathbb {M} _{n}\) such that \(X_{i}\ge 0\) for \(i=1,2,...,n\),

$$\begin{aligned} K=\left[ \begin{array}{ccc} A_{1}^{*}A_{1}+B_{1}^{*}B_{1} &{} \cdots &{} A_{1}^{*}A_{n}+B_{1}^{*}B_{n} \\ A_{2}^{*}A_{1}+B_{2}^{*}B_{1} &{} \cdots &{} A_{2}^{*}A_{n}+B_{2}^{*}B_{n} \\ \vdots &{} \ddots &{} \vdots \\ A_{n}^{*}A_{1}+B_{n}^{*}B_{1} &{} \cdots &{} A_{n}^{*}A_{n}+B_{n}^{*}B_{n} \end{array} \right] \end{aligned}$$

and

$$\begin{aligned} X=\left[ \begin{array}{cccc} X_{1} &{} 0 &{} \ldots &{} 0 \\ 0 &{} X_{2} &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} X_{n} \end{array} \right] \text {,} \end{aligned}$$

then

$$\begin{aligned} 2s_{j}\left( \sum \limits _{i=1}^{n}A_{i}X_{i}B_{i}^{*}\right) \le s_{j}\left( K^{1/2}XK^{1/2}\right) \end{aligned}$$
(33)

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

Proof

Replace \(A=\left[ \begin{array}{cccc} A_{1} &{} A_{2} &{} \ldots &{} A_{n} \\ 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 0 \end{array} \right] ,\) \(B=\left[ \begin{array}{cccc} B_{1} &{} B_{2} &{} \ldots &{} B_{n} \\ 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 0 \end{array} \right]\) and \(X=\left[ \begin{array}{cccc} X_{1} &{} 0 &{} \ldots &{} 0 \\ 0 &{} X_{2} &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} X_{n} \end{array} \right]\) in inequality (29), we get inequality (33). \(\square\)

Remark 3.5

Substituting \(A_{i}=B_{i}=X_{i}=0\) for \(i=2,3,...,n\) in inequality (33), leads to inequality (29), the way to show that inequalities (29) and (33) are equivalent.

Corollary 3.6

Let \(A_{i},B_{i},X_{i}\in \mathbb {M} _{n}\) such that \(X_{i}\ge 0\) for \(i=1,2\),

$$\begin{aligned} L=\left[ \begin{array}{cc} A_{1}^{*}A_{1}+B_{1}^{*}B_{1} &{} A_{1}^{*}A_{2}+B_{1}^{*}B_{2} \\ A_{2}^{*}A_{1}+B_{2}^{*}B_{1} &{} A_{2}^{*}A_{2}+B_{2}^{*}B_{2} \end{array} \right] \end{aligned}$$

and

$$\begin{aligned} X=\left[ \begin{array}{cc} X_{1} &{} 0 \\ 0 &{} X_{2} \end{array} \right] \text {.} \end{aligned}$$

Then

$$\begin{aligned} 2s_{j}\left( \sum \limits _{i=1}^{2}A_{i}X_{i}B_{i}^{*}\right) \le s_{j}\left( L^{1/2}XL^{1/2}\right) \text {.} \end{aligned}$$
(34)

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

Proof

Specifies inequality (33) to \(n=2\), we give inequality (34). \(\square\)

Remark 3.7

Substituting \(X_{1}=X_{2}=I\) in inequality (34), we give inequality (8).

Depending on inequality (33), we now present our next inequality.

Corollary 3.8

Let \(A,B,X_{1},X_{2}\) \(\in \mathbb {M} _{n}\) such that \(X_{1},X_{2}\ge 0\), where

$$\begin{aligned} L=\left[ \begin{array}{cc} A^{*}A+B^{*}B &{} A^{*}B+B^{*}A \\ A^{*}B+B^{*}A &{} A^{*}A+B^{*}B \end{array} \right] \text { and }X=\left[ \begin{array}{cc} X_{1} &{} 0 \\ 0 &{} X_{2} \end{array} \right] \text {.} \end{aligned}$$

Then

$$\begin{aligned} 2s_{j}\left( AX_{1}B^{*}+BX_{2}A^{*}\right) \le s_{j}\left( L^{1/2}XL^{1/2}\right) \end{aligned}$$
(35)

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

Proof

Inequality (35) follows by substituting \(n=2,\) \(A_{1}=B_{2}=A\) and \(A_{2}=B_{1}=B\) in inequality (33). \(\square\)

Depending on inequality (35), we now present our next result, which is a refinement of inequality (28).

Corollary 3.9

Let \(A,B\in \mathbb {M} _{n}\), \(Q_{1}=AA^{*}+BB^{*}+AB^{*}+BA^{*}\), \(Q_{2}=AA^{*}+BB^{*}-AB^{*}-BA^{*}.\) Then

$$\begin{aligned} 2s_{j}\left( AB^{*}+BA^{*}\right) \le s_{j}(Q_{1}\oplus Q_{2}) \end{aligned}$$
(36)

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

Proof

Substituting \(X=I\) in inequality (35), we give

$$\begin{aligned} 2s_{j}\left( AB^{*}+BA^{*}\right)\le & {} s_{j}\left( \left[ \begin{array}{cc} A^{*}A+B^{*}B &{} A^{*}B+B^{*}A \\ B^{*}A+A^{*}B &{} A^{*}A+B^{*}B \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{cc} A^{*} &{} B^{*} \\ B^{*} &{} A^{*} \end{array} \right] \left[ \begin{array}{cc} A &{} B \\ B &{} A \end{array} \right] \right) \\= & {} s_{j}^{2}\left( \left[ \begin{array}{cc} A &{} B \\ B &{} A \end{array} \right] \right) \\= & {} s_{j}^{2}\left( \left[ \begin{array}{cc} A^{*} &{} B^{*} \\ B^{*} &{} A^{*} \end{array} \right] \right) \\= & {} s_{j}^{2}\left( \left[ \begin{array}{cc} A^{*}+B^{*} &{} 0 \\ 0 &{} A^{*}-B^{*} \end{array} \right] \right) \\&\text {(Since unitarily equivalent matrices } \\&\text {have the same singular values)} \\= & {} s_{j}\left( \left[ \begin{array}{cc} Q_{1} &{} 0 \\ 0 &{} Q_{2} \end{array} \right] \right) \text {,} \end{aligned}$$

which is precisely inequality (36). \(\square\)

Remark 3.10

In view of the fact that \(\pm \left( AB^{*}+BA^{*}\right) \le AA^{*}+BB^{*}\) and Weyl’s monotonicity principle, one can see that inequality (36) is sharper than inequality (28).

Depending on inequality (35), we now present our next result, which is another refinement of inequality (28).

Corollary 3.11

Let \(A,B\in \mathbb {M} _{n},\) \(Q=A^{*}A+B^{*}B+\left| A^{*}B+B^{*}A\right|\). Then

$$\begin{aligned} 2s_{j}\left( AB^{*}+BA^{*}\right) \le s_{j}(Q\oplus Q) \end{aligned}$$
(37)

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

Proof

Throughout this proof let \(T=\left[ \begin{array}{cc} A^{*}A+B^{*}B &{} 0 \\ 0 &{} A^{*}A+B^{*}B \end{array} \right]\), \(Z=\left[ \begin{array}{cc} 0 &{} A^{*}B+B^{*}A \\ B^{*}A+A^{*}B &{} 0 \end{array} \right]\). Substituting \(X=I\) in inequality (35), we give

$$\begin{aligned} 2s_{j}\left( AB^{*}+BA^{*}\right)\le & {} s_{j}\left( \left[ \begin{array}{cc} A^{*}A+B^{*}B &{} A^{*}B+B^{*}A \\ B^{*}A+A^{*}B &{} A^{*}A+B^{*}B \end{array} \right] \right) \\= & {} s_{j}\left( T+S\right) \\\le & {} s_{j}\left( \left| \left( T+S\right) \right| \right) \\\le & {} s_{j}\left( \left| T\right| +\left| S\right| \right) \\= & {} s_{j}\left( T+\left[ \begin{array}{cc} \left| A^{*}B+B^{*}A\right| &{} 0 \\ 0 &{} \left| A^{*}B+B^{*}A\right| \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{cc} Q &{} 0 \\ 0 &{} Q \end{array} \right] \right) \text {.} \end{aligned}$$

which is precisely inequality (37). \(\square\)

Remark 3.12

By the fact that \(\left| A^{*}B+B^{*}A\right| \le A^{*}A+B^{*}B\) and by applying Weyl’s monotonicity principle, one can see that inequality (37) is sharper than inequality (28).

The following result is an application of inequality (33).

Corollary 3.13

Let A, B,  \(X_{1},X_{2}\in \mathbb {M} _{n}\ge 0\). Then

$$\begin{aligned} s_{j}\left( A^{1/2}X_{1}A^{1/2}+B^{1/2}X_{2}B^{1/2}\right) \le s_{j}(X^{1/2}JX^{1/2}) \end{aligned}$$
(38)

for \(j=1,2,...,n\) where

$$\begin{aligned} J=\left[ \begin{array}{cc} A &{} A^{1/2}B^{1/2} \\ B^{1/2}A^{1/2} &{} B \end{array} \right] \text { and }X=\left[ \begin{array}{cc} X_{1} &{} 0 \\ 0 &{} X_{2} \end{array} \right] \text {.} \end{aligned}$$

Proof

Substituting \(n=2,\) \(A_{1}=B_{1}=A^{1/2},\) \(A_{2}=B_{2}=B^{1/2}\) in inequality (33), leads to

$$\begin{aligned} 2s_{j}(A^{1/2}X_{1}A^{1/2}+B^{1/2}X_{2}B^{1/2})\le s_{j}\left( (2J)^{1/2}X(2J)^{1/2}\right) \text {,} \end{aligned}$$

which implies that

$$\begin{aligned} s_{j}\left( A^{1/2}X_{1}A^{1/2}+B^{1/2}X_{2}B^{1/2}\right)\le & {} s_{j}(J^{1/2}XJ^{1/2}) \\= & {} \lambda _{j}(J^{1/2}XJ^{1/2}) \\= & {} \lambda _{j}(JX) \\= & {} \lambda _{j}(X^{1/2}JX^{1/2}) \\= & {} s_{j}(X^{1/2}JX^{1/2})\text {,} \end{aligned}$$

as required. \(\square\)

Corollary 3.14

Let A, B,  \(X_{1},X_{2}\in \mathbb {M} _{n}\ge 0\), where

$$\begin{aligned}&S_{1}=X_{1}^{1/2}AX_{1}^{1/2}\text {, } S_{2}=X_{2}^{1/2}B^{1/2}A^{1/2}X_{1}^{1/2}\text {,} \\&T_{1}=X_{2}^{1/2}BX_{2}^{1/2}\text { and } T_{2}=X_{1}^{1/2}A^{1/2}B^{1/2}X_{2}^{1/2}\text {.} \end{aligned}$$

Then

$$\begin{aligned} 2s_{j}(A^{1/2}X_{1}A^{1/2}+B^{1/2}X_{2}B^{1/2})\le s_{j}(\left( S_{1}+\left| S_{2}\right| \right) \oplus \left( T_{1}+\left| T_{2}\right| \right) ) \end{aligned}$$
(39)

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

Proof

Spreading inequality (38), leads to

$$\begin{aligned} s_{j}\left( A^{1/2}X_{1}A^{1/2}+B^{1/2}X_{2}B^{1/2}\right)\le & {} s_{j}(X^{1/2}JX^{1/2}) \\= & {} s_{j}\left[ \begin{array}{cc} S_{1} &{} T_{2} \\ S_{2} &{} T_{1} \end{array} \right] \\= & {} s_{j}\left| \left[ \begin{array}{cc} S_{1} &{} T_{2} \\ S_{2} &{} T_{1} \end{array} \right] \right| \text { (Since }\left[ \begin{array}{cc} S_{1} &{} T_{2} \\ S_{2} &{} T_{1} \end{array} \right] \ge 0\text {)} \\= & {} s_{j}\left( \left| \left[ \begin{array}{cc} S_{1} &{} 0 \\ 0 &{} T_{1} \end{array} \right] +\left[ \begin{array}{cc} 0 &{} T_{2} \\ S_{2} &{} 0 \end{array} \right] \right| \right) \\\le & {} s_{j}\left( \left| \left[ \begin{array}{cc} S_{1} &{} 0 \\ 0 &{} T_{1} \end{array} \right] \right| +\left| \left[ \begin{array}{cc} 0 &{} T_{2} \\ S_{2} &{} 0 \end{array} \right] \right| \right) \\= & {} s_{j}\left( \left[ \begin{array}{cc} S_{1} &{} 0 \\ 0 &{} T_{1} \end{array} \right] +\left[ \begin{array}{cc} \left| S_{2}\right| &{} 0 \\ 0 &{} \left| T_{2}\right| \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{cc} S_{1}+\left| S_{2}\right| &{} 0 \\ 0 &{} T_{1}+\left| T_{2}\right| \end{array} \right] \right) \text {,} \end{aligned}$$

which is precisely inequality (39). \(\square\)

Remark 3.15

Substituting \(X_{1}=X_{2}=I\) in inequality (39), we give the following result which was proved in [5].

$$\begin{aligned} s_{j}(A+B)\le s_{j}\left( \left( A+\left| B^{1/2}A^{1/2}\right| \right) \oplus (B+\left| A^{1/2}B^{1/2}\right| \right) ) \end{aligned}$$

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

The following inequality is a generalization of inequality (7).

Corollary 3.16

Let A,  B,  \(X_{1},\) \(X_{2}\in \mathbb {M} _{n}\) such that \(X_{1},X_{2}\ge 0\). Then

$$\begin{aligned} s_{j}\left( AX_{1}A^{*}-BX_{2}B^{*}\right) \le s_{j}(X_{1}^{1/2}A^{*}AX_{1}^{1/2}\oplus X_{2}^{1/2}A^{*}AX_{2}^{1/2}) \end{aligned}$$
(40)

for \(j=1,2,...,n\). If \(A=B=I,\) we obtain inequality (7), and if \(X_{1}=X_{2}=I,\) then

$$\begin{aligned} s_{j}\left( AA^{*}-BB^{*}\right) \le s_{j}(A^{*}A\oplus B^{*}B) \end{aligned}$$
(41)

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

Proof

Substituting \(n=2,\) \(A_{1}=B_{1}=A,\) \(A_{2}=-B_{2}=B,\) in inequality (33), where \(Z=\left[ \begin{array}{cc} A^{*}A &{} 0 \\ 0 &{} B^{*}B \end{array} \right]\) and \(X=\left[ \begin{array}{cc} X_{1} &{} 0 \\ 0 &{} X_{2} \end{array} \right]\), leads to

$$\begin{aligned} 2s_{j}\left( AX_{1}A^{*}-BX_{2}B^{*}\right)\le & {} s_{j}\left( Z^{1/2}XZ^{1/2}\right) \\= & {} \lambda _{j}\left( Z^{1/2}XZ^{1/2}\right) \\= & {} \lambda _{j}\left( ZX\right) \\= & {} \lambda _{j}\left( X^{1/2}ZX^{1/2}\right) \\= & {} \lambda _{j}\left( X^{1/2}ZX^{1/2}\right) \\= & {} s_{j}\left( X^{1/2}ZX^{1/2}\right) \text {,} \end{aligned}$$

which is inequality (40). \(\square\)

By making use of inequality (40) incites, we here by present the following theorem which has been proven in completely different technique in [13].

Theorem 3.17

Let \(A,B,X\in \mathbb {M} _{n}\) such that \(X\ge 0\). Then

$$\begin{aligned} 2s_{j}(AXB^{*})\le s_{j}\left( (A^{*}A+B^{*}B)^{1/2}X(A^{*}A+B^{*}B)^{1/2}\right) \end{aligned}$$
(42)

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

Proof

Let \(C=\left[ \begin{array}{c} A \\ B \end{array} \right] ,\) \(D=\left[ \begin{array}{c} A \\ -B \end{array} \right] ,\) \(X_{1}=X_{2}=X\) , and \(W=X^{1/2}(A^{*}A+B^{*}B)X^{1/2}\). Then

$$\begin{aligned} CXC^{*}-DXD^{*}=\left[ \begin{array}{cc} 0 &{} 2AXB^{*} \\ 2BXA^{*} &{} 0 \end{array} \right] \text {,} \end{aligned}$$

and

$$\begin{aligned} X^{1/2}C^{*}CX^{1/2}\oplus X^{1/2}D^{*}DX^{1/2}=W\oplus W\text {,} \end{aligned}$$

Now, applying inequality (40), leads to

$$\begin{aligned} 2s_{j}\left[ \begin{array}{cc} BXA^{*} &{} 0 \\ 0 &{} AXB^{*} \end{array} \right] \le s_{j}\left( \left( W\right) \oplus \left( W\right) \right) \text {.} \end{aligned}$$

This gives

\(2s_{j}(AXB^{*})\le s_{j}\left( X^{1/2}(A^{*}A+B^{*}B)X^{1/2}\right)\) for \(j=1,2,...,n\). as required. \(\square\)