Let $ M_{m, n} $ be the space of $ m\times n $ complex matrices and $ M_n=M_{n, n} $. Let $ \left\| \cdot \right\| $ denote any unitarily invariant norm on $ M_n $, if $ ||UAV||=||A|| $ for all $ A\in M_n $ and for all unitary matrices $ U, V\in M_n $. The $ A> 0 $ is used to mean that $ A $ is a positive definite matrix. The Hilbert-Schmidt norm of $ A=(a_{ij})\in M_n $ is denoted by

$ \|A\|_2=\left(\sum\limits_{i, j=1}^n |a_{ij}|^2\right)^{\frac{1}{2}}. $

Let $ A, B\in M_n $ be positive definite and $ 0\leq v\leq 1 $, the weighted geometric mean of the matrices $ A $ and $ B $ is defined as follows:

$ A\sharp_vB=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^vA^{\frac{1}{2}}, $

for $ v = \frac{1}{2} $, we denote the geometric mean by $ A\sharp B $.

Kittaneh and Manasrah ^[1] proved that if $ A, B\in M_n $ are positive definite and $ 0\leq v\leq 1, $ then

$ \begin{equation} \begin{array}{lll} &2r_0(A+B-2A\sharp B)+A\sharp_vB+A\sharp_{1-v}B\\ &\leq A+B\\ &\leq 2s_0(A+B-2A\sharp B)+A\sharp_vB+A\sharp_{1-v}B, \end{array} \end{equation} $

(1.1)

where $ r_{0}=\min\{v, 1-v\}, s_{0}=\max\{v, 1-v\} $.

In 2018, Liu and Yang ^[2] refined the inequalities (1.1) as follows:

$ \begin{equation} \begin{array}{lll} &2r_0(A+B-2A\sharp B)+A\sharp_vB+A\sharp_{1-v}B\\ &\leq A+B\\ &\leq \alpha(v)(A+B-2A\sharp B)+A\sharp_vB+A\sharp_{1-v}B\\ &\leq 2s_0(A+B-2A\sharp B)+A\sharp_vB+A\sharp_{1-v}B, \end{array} \end{equation} $

(1.2)

where $ \alpha(v)=\frac{3}{2}-2(v-v^2). $

Let $ A, B, X \in M_n $ such that $ A $ and $ B $ are positive definite. Bhatia and Davis ^[3] proved that if $ 0\le v\le 1 $, then

$ 2||A^{\frac{1}{2}}XB^{\frac{1}{2}}||\leq ||A^{v}XB^{1-v}+A^{1-v}XB^{v}||\leq ||AX+XB||, $

where the second inequality is known as Heinz inequality.

He and Zou ^[4] showed if $ 0\le v\le 1 $, then

$ \begin{equation} ||AX+XB||_2^2\leq ||A^{v}XB^{1-v}+A^{1-v}XB^{v}||_2^2+2s_0||AX-XB||_2^2, \end{equation} $

(1.3)

where $ s_{0}=\max\{v, 1-v\}. $ Kittaneh and Manasrah ^[5] showed if $ 0\le v\le 1 $, then

$ \begin{equation} ||A^{v}XB^{1-v}+A^{1-v}XB^{v}||_2^2+2r_0||AX-XB||_2^2\leq ||AX+XB||_2^2, \end{equation} $

(1.4)

where $ r_{0}=\min\{v, 1-v\} $, inequality (1.4) is the inverse of inequality (1.3).

In 2018, Liu and Yang ^[2] refined inequality (1.3) as follows:

$ \begin{equation} ||AX+XB||_2^2\leq ||A^{v}XB^{1-v}+A^{1-v}XB^{v}||_2^2+\alpha(v)||AX-XB||_2^2, \end{equation} $

(1.5)

where $ \alpha(v)=\frac{3}{2}-2(v-v^2) $.

Recently, many interesting articles have been devoted to study the unitarily invariant norm inequalities for matrices, see ^[6-8] and references therein.

In this paper, we first give two scalar inequalities. By using scalar inequalities, we improve inequalities (1.2) and (1.5).

In the following, we give two scalar inequalities which will turn out to be useful in the proof of our results.

Theorem 2.1 Let $ a, b>0 $, $ 0\leq v\leq 1 $, then

$ \begin{equation} a+b\leq a^{v}b^{1-v}+a^{1-v}b^{v}+\gamma(v)(\sqrt{a}-\sqrt{b})^2, \end{equation} $

(2.1)

where $ \gamma(v)=\frac{5}{4}-(v-v^2) $.

Proof To prove inequality (2.1), we only need prove that the following inequality

$ (1-\gamma(v))(a+b)+2\gamma(v)\sqrt{ab}\leq a^{v}b^{1-v}+a^{1-v}b^{v}. $

Let $ a=e^x, b=e^y $, by the definition of the hyperbolic function, we have

$ \begin{equation} (v-v^{2}-\frac{1}{4})cosh(\frac{x-y}{2})+(\frac{5}{4}-(v-v^2))\leq cosh((1-2v)(\frac{x-y}{2})). \end{equation} $

(2.2)

Let $ z=\frac{x-y}{2} $, by the series expansion of the hyperbolic $ coshz $ function, we know that inequality (2.2) is equivalent to

$ \begin{equation} (v-v^{2}-\frac{1}{4})(1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots)+(\frac{5}{4}-(v-v^2)) \leq 1+\frac{(1-2v)^2z^2}{2!}+\frac{(1-2v)^4z^4}{4!}+\cdots. \end{equation} $

(2.3)

For $ 0\leq v\leq 1 $, it is easy to know that inequality (2.3) holds.

This completes the proof.

Corollary 2.2 Let $ a, b>0 $, $ 0\leq v\leq 1 $, then

$ \begin{equation} (a+b)^2\leq (a^{v}b^{1-v}+a^{1-v}b^{v})^2+\gamma(v)(a-b)^2, \end{equation} $

(2.4)

where $ \gamma(v)=\frac{5}{4}-(v-v^2) $.

Proof By inequality (2.1), we have

$ \begin{array}{lll} (\sqrt{a}+\sqrt{b})^2-(a^\frac{v}{2}b^\frac{1-v}{2}+a^\frac{1-v}{2}b^\frac{v}{2})^2&=& a+b-(a^{v}b^{1-v}+a^{1-v}b^{v})\\ \\ &\le& (\frac{5}{4}-(v-v^2))(\sqrt{a}-\sqrt{b})^2, \end{array} $

hence

$ (a+b)^2\leq (a^{v}b^{1-v}+a^{1-v}b^{v})^2+(\frac{5}{4}-(v-v^2))(a-b)^2. $

This completes the proof.

Theorem 2.3 Let $ A, B\in M_n $ be positive definite. Then

$ \begin{equation} \begin{array}{lll} &2r_{0}(A+B-2A\sharp B)+A\sharp_{v}B+A\sharp_{1-v}B\\ \\ &\leq A+B\\ \\ &\leq \gamma(v)(A+B-2A\sharp B)+A\sharp_{v}B+A\sharp_{1-v}B\\ \\ &\leq \alpha(v)(A+B-2A\sharp B)+A\sharp_{v}B+A\sharp_{1-v}B, \end{array} \end{equation} $

(2.5)

where $ v \in [0, 1] $, $ r_{0}=\min\{v, 1-v\}, \gamma(v)=\frac{5}{4}-(v-v^2), \alpha(v)=\frac{3}{2}-2(v-v^2) $.

Proof By inequalities (1.2), we know that the first inequality of (2.5) holds. For the second inequality of (2.5). Since $ T\in M_n $ is positive definite, it follows by the spectral theorem that there exists unitary matrix $ U\in M_n $ such that

$ T=UPU^*, $

where $ P=diag(\lambda_1, \lambda_2, \cdots, \lambda_n), \lambda_j> 0, 1\leq j\leq n $. For $ a > 0, b = 1 $, by inequality (2.1), we have

$ a+1\leq a^v+b^{1-v}+\gamma(v)(\sqrt{a}-1)^2, $

and so

$ \begin{equation} P+I\leq P^v+P^{1-v}+\gamma(v)(\sqrt{P}-I)^2. \end{equation} $

(2.6)

Multiplying the left and right sides of the inequality of (2.6) by $ U $ and $ U^* $, we have

$ T+I\leq T^v+T^{1-v}+\gamma(v)(\sqrt{T}-I)^2, $

let $ T=A^{-\frac{1}{2}}BA^{-\frac{1}{2}} $, the second inequality of (2.5) holds. For $ 0\leq v\leq 1 $, it easy to know that

$ \alpha(v)-\gamma(v)=\frac{1}{4}-(v-v^2)=(v-\frac{1}{2})^2\geq 0. $

Therefore, Theorem 2.3 is a refinement of the inequalities (1.2).

This completes the proof.

Theorem 2.4 Let $ A, B, X\in M_n $ such that $ A, B $ are positive definite. Then

$ \begin{equation} ||AX+XB||^2_2\leq ||A^vXB^{1-v}+A^{1-v}XB^v||^2_2+\gamma(v)||AX-XB||^2_2, \end{equation} $

(2.7)

where $ v \in [0, 1] $, $ \gamma(v)=\frac{5}{4}-(v-v^2). $

Proof Since every positive definite matrix is unitarily diagonalizable, it follows that there exist unitary matrices $ U, V\in M_n $ such that

$ A=UP_1U^*, B=VP_2V^*, $

where $ P_1=diag(\lambda_1, \lambda_2, \cdots, \lambda_n), P_2=diag(\mu_1, \mu_2, \cdots, \mu_n), \lambda_j, \mu_j> 0, 1\leq j\leq n $.

Let $ C=U^*XV=(c_{ij}) $, then

$ \begin{array}{lll} A^vXB^{1-v}+A^{1-v}XB^v&=(UP_1U^*)^vX(VP_2V^*)^{1-v}+(UP_1U^*)^{1-v}X(VP_2V^*)^v\\ & = UP^v_1(U^*XV)P^{1-v}_2V^*+UP^{1-v}_1(U^*XV)P^v_2V^*\\ & = U(P^v_1CP^{1-v}_2+P^{1-v}_1CP^v_2)V^* , \end{array} $

and

$ \begin{array}{lll} ||A^vXB^{1-v}+A^{1-v}XB^v||^2_2&=||P^v_1CP^{1-v}_2+P^{1-v}_1CP^v_2||^2_2\\ & = \sum\limits_{i, j=1}^n(\lambda^v_i\mu^{1-v}_j+\lambda^{1-v}_i\mu^v_j)^2|c_{ij}|^2. \end{array} $

Using the same method, we have

$ ||AX+XB||^2_2= \sum\limits_{i, j=1}^n(\lambda_i+\mu_j)^2|c_{ij}|^2, $

$ ||AX-XB||^2_2= \sum\limits_{i, j=1}^n(\lambda_i-\mu_j)^2|c_{ij}|^2. $

By inequality (2.4), we obtain

$ \sum\limits_{i, j=1}^n(\lambda^v_i\mu^{1-v}_j+\lambda^{1-v}_i\mu^v_j)^2|c_{ij}|^2+\gamma(v)\sum\limits_{i, j=1}^n(\lambda_i-\mu_j)^2|c_{ij}|^2\geq \sum\limits_{i, j=1}^n(\lambda_i+\mu_j)^2|c_{ij}|^2. $

Therefore, inequality (2.7) holds, it is a refinement of the inequality (1.5).

This completes the proof.