1 Introduction

Let \(M_{n} ( C )\) be the space of \(n\times n\) complex matrices. Let \(\Vert \cdot \Vert \) denote any unitarily invariant norm on \(M_{n} ( C )\). For \(A \in M_{n} ( C ) \), the Frobenius norm of A is defined by \(\Vert A \Vert _{F} = \sqrt {\operatorname{tr} ( {A^{*} A} )}\), where \(\operatorname{tr} ( X )\) is the trace of X. It is known that the Frobenius norm is unitarily invariant.

Let \(A,B \in M_{n} ( C )\) be positive semidefinite. Bhatia and Kittaneh proved in [1] that

$$ \bigl\Vert {A^{1/2} B^{1/2} } \bigr\Vert \le \biggl\Vert {\frac{{A + B}}{2}} \biggr\Vert , $$
(1.1)

which is known as the arithmetic–geometric mean inequality for unitarily invariant norms.

Let \(A,X,B\in M_{n} ( C ) \) and suppose that A and B are positive semidefinite. Bhatia and Davis proved in [2] that

$$ \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert \le \biggl\Vert {\frac{{AX + XB}}{2}} \biggr\Vert , $$
(1.2)

which is a generalization of inequality (1.1).

Recently, Kittaneh and Manasrah [3] showed a refinement of inequality (1.2) for the Frobenius norm, which can be stated as follows:

$$ \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} + \frac{1}{2} \bigl( {\sqrt{ \Vert {AX} \Vert _{F} } - \sqrt{ \Vert {XB} \Vert _{F} } } \bigr)^{2} \le \biggl\Vert {\frac{{AX + XB}}{2}} \biggr\Vert _{F} . $$
(1.3)

The authors of [4] and [5] gave some generalizations of inequality (1.3).

Let \(A,X,B\in M_{n} ( C )\) such that A and B are self-adjoint. Jocić and Kittaneh proved in [6] that for \(n \in N\) and \(j = 1, \ldots,n\),

$$ \bigl\Vert {A^{n + j} XB^{n - j + 1} - A^{n - j + 1} XB^{n + j} } \bigr\Vert \le \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} } \bigr\Vert . $$
(1.4)

Bhatia gave a simple proof of inequality (1.4) in [7]. For more information on inequalities of unitarily invariant norms, the reader is referred to [8–12] and the references therein.

In this short note, we first present a new generalization of inequality (1.3). After that, as an application of inequality (1.3), we show a refinement of inequality (1.4) for the Frobenius norm.

2 Main results

In this section, we show the main results of this paper. To do this, we need the following lemmas.

Lemma 2.1

([11])

Let \(A,X,B\in M_{n} ( C )\). If \(\alpha \in [ {0,1} ]\), then

$$ \bigl\Vert {A^{*} XB} \bigr\Vert ^{2} \le \bigl\Vert {\alpha AA^{*} X + ( {1 - \alpha} )XBB^{*} } \bigr\Vert \bigl\Vert { ( {1 - \alpha} )AA^{*} X + \alpha XBB^{*} } \bigr\Vert . $$
(2.1)

Lemma 2.2

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

$$\bigl\Vert {A^{*} XB} \bigr\Vert _{F} + \frac{1}{2} \bigl( { \sqrt{ \bigl\Vert {AA^{*} X} \bigr\Vert _{F} } - \sqrt{ \bigl\Vert {XBB^{*} } \bigr\Vert _{F} } } \bigr)^{2} \le \biggl\Vert {\frac{{AA^{*} X + XBB^{*} }}{2}} \biggr\Vert _{F}. $$

Proof

By the polar decomposition of matrices and the properties of unitary invariant norms, we know that inequality (1.3) is equivalent to Lemma 2.2. This completes the proof. □

Theorem 2.1

Let \(A,X,B\in M_{n} ( C )\), \(\alpha \in [ {0,1} ]\) such that A, B are positive semidefinite. Then

$$ \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} + \frac{1}{2} \bigl( {\sqrt {C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \le \biggl\Vert {\frac{{AX + XB}}{2}} \biggr\Vert _{F} , $$
(2.2)

where

$$C ( \alpha ) = \bigl\Vert {\alpha AX + ( {1 - \alpha} )XB} \bigr\Vert _{F} ,\qquad C ( {1 - \alpha} ) = \bigl\Vert { ( {1 - \alpha} )AX + \alpha XB} \bigr\Vert _{F}. $$

Proof

By definition of the Frobenius norm, we have

$$\begin{aligned}& \begin{aligned} C^{2} ( \alpha ) &= \bigl\Vert {\alpha AX + ( {1 - \alpha } )XB} \bigr\Vert _{F}^{2} \\ &= \alpha^{2} \Vert {AX} \Vert _{F}^{2} + ( {1 - \alpha} )^{2} \Vert {XB} \Vert _{F}^{2} + 2\alpha ( {1 - \alpha} ) \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F}^{2}, \end{aligned} \\& \begin{aligned} C^{2} ( {1 - \alpha} ) &= \bigl\Vert { ( {1 - \alpha} )AX + \alpha XB} \bigr\Vert _{F}^{2} \\ &= ( {1 - \alpha} )^{2} \Vert {AX} \Vert _{F}^{2} + \alpha ^{2} \Vert {XB} \Vert _{F}^{2} + 2 \alpha ( {1 - \alpha} ) \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F}^{2}, \end{aligned} \end{aligned}$$

and so

$$ \Vert {AX} \Vert _{F}^{2} + \Vert {XB} \Vert _{F}^{2} - C^{2} ( \alpha ) - C^{2} ( {1 - \alpha} ) = 2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2} . $$
(2.3)

It follows from (2.1) and (2.3) that

$$\begin{aligned}& \Vert {AX + XB} \Vert _{F}^{2} - \bigl( { \bigl( { \sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} + 2 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} } \bigr)^{2} \\& \qquad {} - 2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2} \\& \quad = \Vert {AX} \Vert _{F}^{2} + \Vert {XB} \Vert _{F}^{2} + 2 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F}^{2} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F}^{2} \\& \qquad {} - 2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2}- \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4} \\& \qquad {} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \quad = \Vert {AX} \Vert _{F}^{2} + \Vert {XB} \Vert _{F}^{2}- 2 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F}^{2} \\& \qquad {} - 2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2}- \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4} \\& \qquad {} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \quad \ge \Vert {AX} \Vert _{F}^{2} + \Vert {XB} \Vert _{F}^{2} - 2C ( \alpha )C ( {1 - \alpha} ) - \bigl( { \sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4} \\& \qquad {} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2}-2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2} \\& \quad = C^{2} ( \alpha ) + C^{2} ( {1 - \alpha} ) - 2C ( \alpha )C ( {1 - \alpha} ) - \bigl( {\sqrt {C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4} \\& \qquad {} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \quad = \bigl( {C ( \alpha ) - C ( {1 - \alpha} )} \bigr)^{2} - \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4} \\& \qquad {} - 4 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \quad = \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \bigl( {\sqrt{C ( \alpha )} + \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \qquad {} - \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{4}- 2 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \\& \quad = 4 \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} \bigl( {\sqrt{C ( \alpha )C ( {1 - \alpha} )} - \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} } \bigr) \\& \quad \ge 0. \end{aligned}$$

That is,

$$\begin{aligned}& \bigl( {2 \bigl\Vert {A^{1/2} XB^{1/2} } \bigr\Vert _{F} + \bigl( {\sqrt{C ( \alpha )} - \sqrt{C ( {1 - \alpha} )} } \bigr)^{2} } \bigr)^{2} + 2\alpha ( {1 - \alpha} ) \Vert {AX - XB} \Vert _{F}^{2} \\& \quad \le \Vert {AX + XB} \Vert _{F}^{2}, \end{aligned}$$

which implies inequality (2.2). This completes the proof. □

Remark 2.1

Putting \(\alpha = 0\) or \(\alpha = 1\) in inequality (2.2), we can obtain inequality (1.3). So, inequality (2.2) is a generalization of inequality (1.3).

Next, we will show a refinement of inequality (1.4).

Theorem 2.2

Let \(A,X,B\in M_{n} ( C )\) such that A and B are self-adjoint. If \(n \in N\) and \(j = 1, \ldots,n\), then we have

$$\bigl\Vert {A^{n + j} XB^{n - j + 1} - A^{n - j + 1} XB^{n + j} } \bigr\Vert _{F} +K^{2} ( {n,j} ) \le \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} } \bigr\Vert _{F} , $$

where

$$\begin{aligned} K ( {n,j} ) =& \sqrt{ \bigl\Vert {A^{2} \bigl( {A^{n + j - 1} XB^{n - j} - A^{n - j} XB^{n + j - 1} } \bigr)} \bigr\Vert _{F} } \\ &{}- \sqrt{ \bigl\Vert { \bigl( {A^{n + j - 1} XB^{n - j} - A^{n - j} XB^{n + j - 1} } \bigr)B^{2} } \bigr\Vert _{F} }. \end{aligned}$$

Proof

We prove it by induction. For \(j=1\) and any positive integer n, by Lemma 2.2 and the triangle inequality for unitary invariant norms, we have

$$\begin{aligned}& \bigl\Vert {A^{n + 1} XB^{n} - A^{n} XB^{n + 1} } \bigr\Vert _{F} + \frac{1}{2}K^{2} ( {n,1} ) \\& \quad = \bigl\Vert {A \bigl( {A^{n} XB^{n - 1} - A^{n - 1} XB^{n} } \bigr)B} \bigr\Vert _{F}+ \frac{1}{2}K^{2} ( {n,1} ) \\& \quad \le \frac{1}{2} \bigl\Vert {A^{2} \bigl( {A^{n} XB^{n - 1} - A^{n - 1} XB^{n} } \bigr) + \bigl( {A^{n} XB^{n - 1} - A^{n - 1} XB^{n} } \bigr)B^{2} } \bigr\Vert _{F} \\& \quad = \frac{1}{2} \bigl\Vert {A^{n + 2} XB^{n - 1} - A^{n - 1} XB^{n + 2} + A^{n} XB^{n + 1} - A^{n + 1} XB^{n} } \bigr\Vert _{F} \\& \quad \le \frac{1}{2} \bigl\Vert {A^{n + 2} XB^{n - 1} - A^{n - 1} XB^{n + 2} } \bigr\Vert _{F} + \frac{1}{2} \bigl\Vert {A^{n + 1} XB^{n} - A^{n} XB^{n + 1} } \bigr\Vert _{F}, \end{aligned}$$

which is equivalent to

$$\bigl\Vert {A^{n + 1} XB^{n} - A^{n} XB^{n + 1} } \bigr\Vert _{F} + K^{2} ( {n,1} ) \le \bigl\Vert {A^{n + 2} XB^{n - 1} - A^{n - 1} XB^{n + 2} } \bigr\Vert _{F}. $$

Now, suppose that Theorem 2.2 has been proved for \(j-1\). By Lemma 2.2, the triangle inequality for unitary invariant norms, and induction hypothesis, we have

$$\begin{aligned}& \bigl\Vert {A^{n + j} XB^{n - j + 1} - A^{n - j + 1} XB^{n + j} } \bigr\Vert _{F} + \frac{1}{2}K^{2} ( {n,j} ) \\& \quad = \bigl\Vert {A \bigl( {A^{n + j - 1} XB^{n - j} - A^{n - j} XB^{n + j - 1} } \bigr)B} \bigr\Vert _{F} + \frac{1}{2}K^{2} ( {n,j} ) \\& \quad \le \frac{1}{2} \bigl\Vert {A^{2} \bigl( {A^{n + j - 1} XB^{n - j} - A^{n - j} XB^{n + j - 1} } \bigr) + \bigl( {A^{n + j - 1} XB^{n - j} - A^{n - j} XB^{n + j - 1} } \bigr)B^{2} } \bigr\Vert _{F} \\& \quad = \frac{1}{2} \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} + A^{n + j - 1} XB^{n - ( {j - 1} ) + 1} - A^{n - ( {j - 1} ) + 1} XB^{n + j - 1} } \bigr\Vert _{F} \\& \quad \le \frac{1}{2} \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} } \bigr\Vert _{F} + \frac{1}{2} \bigl\Vert {A^{n + j - 1} XB^{n - ( {j - 1} ) + 1} - A^{n - ( {j - 1} ) + 1} XB^{n + j - 1} } \bigr\Vert _{F} \\& \quad \le \frac{1}{2} \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} } \bigr\Vert _{F} + \frac{1}{2} \bigl\Vert {A^{n + j} XB^{n - j + 1} - A^{n - j + 1} XB^{n + j} } \bigr\Vert _{F}, \end{aligned}$$

which is equivalent to

$$\bigl\Vert {A^{n + j} XB^{n - j + 1} - A^{n - j + 1} XB^{n + j} } \bigr\Vert _{F} +K^{2} ( {n,j} ) \le \bigl\Vert {A^{n + j + 1} XB^{n - j} - A^{n - j} XB^{n + j + 1} } \bigr\Vert _{F}. $$

This completes the proof. □