1 Introduction

In this sequel, we use the standard notation \(M_{n}, M_{n}^{+}\) and \(M_{n}^{++}\) for the algebra of all \(n\times n\) complex matrices, the cone of positive (or positive semidefinite) matrix and that of strictly positive matrices in \(M_{n}\), respectively. Matrices and their inequalities have attracted researchers working in functional analysis. These inequalities have been studied in different approaches among which unitarily invariant norms inequalities are most popular. Recall that a unitarily invariant norm is a norm \(\Vert \cdot \Vert \) defined on \(M_{n}\) satisfying the property \(\Vert UAV\Vert =\Vert A\Vert \) for all \(A\in M_{n}\) and unitaries \(U,V\in M_{n}\). The absolute value of a matrix \(A = (a_{ij})\) is defined by \(|A| = (A^{*}A)^{1/2}\). The motivation behind this work starts with some crucial inequalities which will be presented as follows.

The classical arithmetic-geometric mean inequality [1] states that for \(A,B\in M_{n}^{+}\) and \(X\in M_{n}\),

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert \le \frac{1}{2}\Vert AX+XB\Vert . \end{aligned}$$
(1)

Heinz inequality [1] is a refinement of inequality (1) which states that

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert \le \Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert \le \Vert \frac{AX+XB}{2}\Vert \end{aligned}$$
(2)

hold for \(A,B\in M_{n}^{+}, X\in M_{n}\) and \(0\le t\le 1\).

A general form of Cauchy-Schwartz inequality [2] states that for \(A,B\in M_{n}^{+}, X\in M_{n}\) and \(r>0\),

$$\begin{aligned} \Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\le \Vert |AX|^{r}\Vert \Vert |XB|^{r}\Vert . \end{aligned}$$
(3)

We remark that the above inequalities have been studied deeply in the literature. We refer the reader to [3,4,5] as samples of recent work treating such inequalities and their variants.

Motivated by Bhatia and Bourin [2, 6], here we define two functions f and h for a given unitarily invariant norm \(\Vert \cdot \Vert \),

$$\begin{aligned} f(t)=\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert \quad \mathrm{and}\quad h(t)=\Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert ^{2}, \end{aligned}$$

where \(A,B\in M_{n}^{+}\) and \(X\in M_{n}\). The above functions f and h are convex on [0,1] and attain their minimum at \(t=\frac{1}{2}\). In this article, we utilize convexity of these functions to obtain refinements of arithmetic-geometric mean, Heinz and Cauchy-Schwartz inequalities. The following convex function inequalities are also essential to our results.

Hermite-Hadaward inequality [7] states that for every real-valued convex function g on the interval [a, b], we have

$$\begin{aligned} g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\int _{a}^{b}g(t)dt \le \frac{g(a)+g(b)}{2}. \end{aligned}$$

In 2010, EL Farissi [8] refined Hermite-Hadaward inequality as follows

$$\begin{aligned} g\left(\frac{a+b}{2}\right)\le l(\lambda )\le \frac{1}{b-a}\int _{a}^{b}g(t)dt \le L(\lambda )\le \frac{g(a)+g(b)}{2} \end{aligned}$$

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

$$\begin{aligned} l(\lambda )=\lambda g\left(\frac{\lambda b+(2-\lambda )a}{2}\right)+(1-\lambda )g\left(\frac{(1+\lambda )b+(1-\lambda )a}{2}\right) \end{aligned}$$

and

$$\begin{aligned} L(\lambda )=\frac{1}{2}(g(\lambda b+(1-\lambda )a)+\lambda g(a)+(1-\lambda )g(b)). \end{aligned}$$

A few years later, Abbas and Mourad [9] got that

$$\begin{aligned} g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\int _{a}^{b}g(t)dt&\le \frac{1}{4n}\left[(2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b)\right]\\&\le \frac{g(a)+g(b)}{2}. \end{aligned}$$

The following lemma combining Farissi and Abbas’ results will be essential for our main results. The main results in this paper, Theorems 1, 2 and 3, are obtained by applying some refinements of Hermite-Hadaward inequalities on the convex functions f and h using the same method from Kittaneh [10].

Lemma 1

Let g be a real-valued convex function which is convex on the interval [a,b]. Then for any positive integer n, we have

$$\begin{aligned} g\left(\frac{a+b}{2}\right)&\le \frac{1}{n}g\left(\frac{b+(2n-1)a}{2n}\right)+\left(1-\frac{1}{n}\right)g\left(\frac{(n+1)b+(n-1)a}{2n}\right)\\&\le \frac{1}{b-a}\int _{a}^{b}g(t)dt\\&\le \frac{1}{4n}\left[(2n-1)g(a)+2g\left(\frac{a+b}{2}\right)+(2n-1)g(b)\right]\\&\le \frac{g(a)+g(b)}{2}. \end{aligned}$$

Recently, Chen, Chen and Gao [11] obtained the following refinements of Hermite-Hadaward inequality.

Lemma 2

Let \(m, n: [a,b]\rightarrow [0,+\infty )\) be convex functions and meet \([m(a)-m(b)]\cdot [n(a)-n(b)]\le 0\). Then for all \(\lambda \in [0,1]\), we have

$$\begin{aligned} \frac{1}{b-a}\int _{a}^{b}m(t)n(t)dt\le L'(\lambda )\le \frac{1}{3}M(a,b)+\frac{1}{6}N(a,b) \end{aligned}$$

and

$$\begin{aligned} 2m\left(\frac{a+b}{2}\right)n\left(\frac{a+b}{2}\right)-\frac{1}{6}M(a,b)-\frac{1}{3}N(a,b)\le l'(\lambda )\le \frac{1}{b-a}\int _{a}^{b}m(t)n(t)dt \end{aligned}$$

where

$$\begin{aligned} M(a,b)= & {} m(a)n(a)+m(b)n(b), N(a,b)=m(a)n(b)+m(b)n(a), \\ L'(\lambda )= & {} \frac{\lambda }{3}m(a)n(a)+\frac{1-\lambda }{3}m(b)n(b)+\frac{\lambda }{3}m(\lambda b+(1-\lambda )a)n(\lambda b+(1-\lambda )a)\\&\quad&+\frac{\lambda }{6}m(\lambda b+(1-\lambda )a)[\lambda n(a)+(1-\lambda )n(b)]\\&\quad&+\frac{\lambda }{6}n(\lambda b+(1-\lambda )a)[\lambda n(b)+(1-\lambda )n(a)] \end{aligned}$$

and

$$\begin{aligned}&l'(\lambda )=2\lambda m\left(\frac{(2-\lambda )a+\lambda b}{2}\right)n\left(\frac{(2-\lambda )a+\lambda b}{2}\right)-\frac{1+3\lambda -3\lambda ^{2}}{6}M(a,b)\\&-\frac{2-3\lambda +3\lambda ^{2}}{6}N(a,b) +2(1-\lambda )m\left(\frac{(1-\lambda )a+(1+\lambda ) b}{2}\right)n\left(\frac{(1-\lambda )a+(1+\lambda ) b}{2}\right). \end{aligned}$$

The organization of this article will be as follows. In the following, we mainly present some unitarily invariant norm inequalities for matrix means which are refinements of arithmetic-geometric mean, Heinz and Cauchy-Schwartz inequalities utilizing Lemmas 1 and 2.

2 Unitarily invariant norm inequalities

Now we are in a position to begin our main results.

Applying Lemma 1 to the convex function h(t) on the interval \([\mu ,1-\mu ]\) when \(0\le \mu <\frac{1}{2}\) and on the interval \([1-\mu ,\mu ]\) when \(\frac{1}{2}<\mu \le 1\), we obtain the following refinement of arithmetic-geometric mean and Heinz inequalities.

Theorem 1

If \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu \le 1\), then for unitarily invariant norm \(\Vert \cdot \Vert \),

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert \frac{A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}+A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}}{2}\Vert ^{2}\\&\quad&+(1-\frac{1}{n})\Vert \frac{A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}+A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}}{2}\Vert ^{2}\\\le & {} \frac{1}{|2\mu -1|}|\int _{\mu }^{1-\mu }\Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert ^{2}dt| \quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}+\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right] \\\le & {} \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2} \end{aligned}$$

hold for any positive integer n.

Proof

Assume that \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu < \frac{1}{2}\), then it follows by Lemma 1 that

$$\begin{aligned} h\left(\frac{\mu +1-\mu }{2}\right)\le & {} \frac{1}{n}h\left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)h\left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }h(t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{4n}\left[(2n-1)h(\mu )+2h\left(\frac{\mu +1-\mu }{2}\right)+(2n-1)h(1-\mu )\right]\\\le & {} \frac{h(\mu )+h(1-\mu )}{2}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} h\left(\frac{1}{2}\right)\le & {} \frac{1}{n}h\left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)h\left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }h(t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{4n}\left[(2n-1)h(\mu )+2h\left(\frac{1}{2}\right)+(2n-1)h(1-\mu )\right]\\\le & {} \frac{h(\mu )+h(1-\mu )}{2}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert \frac{A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}+A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}}{2}\Vert ^{2}\\&\quad&+(1-\frac{1}{n})\Vert \frac{A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}+A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}}{2}\Vert ^{2}\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }\Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert ^{2}dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}+\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right]\\\le & {} \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}. \end{aligned}$$
(4)

On the other hand, if \(\frac{1}{2}<\mu \le 1\), then it follows by symmetry (i.e., by applying the above inequality (4) to \(1-\mu \)) that

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert \frac{A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}+A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}}{2}\Vert ^{2}\\&\quad&+(1-\frac{1}{n})\Vert \frac{A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}+A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}}{2}\Vert ^{2}\\\le & {} \frac{1}{2\mu -1}\int _{1-\mu }^{\mu }\Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert ^{2}dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}+\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right]\\\le & {} \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}. \end{aligned}$$
(5)

We complete the proof of Theorem 1 by combining the inequalities (4) and (5). \(\square \)

Following the same logic of Theorem 1 and applying Lemma 1 to the function f(t) on the interval \([\mu ,1-\mu ]\) when \(0\le \mu <\frac{1}{2}\) and on the interval \([1-\mu ,\mu ]\) when \(\frac{1}{2}<\mu \le 1\), we have the following refinement of Cauchy-Schwartz inequality.

Theorem 2

If \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu \le 1\), then for unitarily invariant norm \(\Vert \cdot \Vert \),

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}\Vert \Vert A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}\Vert \\&\quad&+(1-\frac{1}{n})\Vert A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}\Vert \Vert A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}\Vert \\\le & {} \frac{1}{|1-2\mu |}|\int _{\mu }^{1-\mu }\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert dt|\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert +\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right]\\\le & {} \Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert \end{aligned}$$

hold for any positive integer n.

Proof

Assume that \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu < \frac{1}{2}\), then it follows by Lemma 1 that

$$\begin{aligned} f\left(\frac{\mu +1-\mu }{2}\right)&\le \frac{1}{n}f\left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)f\left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\&\le \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }f(t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\&\le \frac{1}{4n}\left[(2n-1)f(\mu )+2f\left(\frac{\mu +1-\mu }{2}\right)+(2n-1)f(1-\mu )\right]\\&\le \frac{f(\mu )+f(1-\mu )}{2}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} f\left(\frac{1}{2}\right)\le & {} \frac{1}{n}f\left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)f\left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }f(t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{4n}\left[(2n-1)f(\mu )+2f\left(\frac{1}{2}\right)+(2n-1)f(1-\mu )\right]\\\le & {} \frac{f(\mu )+f(1-\mu )}{2}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}\Vert \Vert A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}\Vert \\&\quad&+(1-\frac{1}{n})\Vert A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}\Vert \Vert A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}\Vert \\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert +\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right]\\\le & {} \Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert . \end{aligned}$$
(6)

On the other hand, if \(\frac{1}{2}<\mu \le 1\), then it follows by symmetry (i.e., by applying the above inequality (6) to \(1-\mu \)) that

$$\begin{aligned} \Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\le & {} \frac{1}{n}\Vert A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}\Vert \Vert A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}\Vert \\&\quad&+(1-\frac{1}{n})\Vert A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}\Vert \Vert A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}\Vert \\\le & {} \frac{1}{2\mu -1}\int _{1-\mu }^{\mu }\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert +\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{2}\right]\\\le & {} \Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert . \end{aligned}$$
(7)

We complete the proof of Theorem 2 by combining the inequalities (6) and (7). \(\square \)

Next, for every positive real number r, we consider the function

$$\begin{aligned} \phi (t)=\Vert |A^{t}XB^{1-t}|^{r}\Vert \cdot \Vert |A^{1-t}XB^{t}|^{r}\Vert \end{aligned}$$

which is convex on [0,1] and attains its minimum at \(t =\frac{1}{2}\) obtained by Hiai and Zhan [12].

Applying Lemma 1 to the function \(\phi (t)\) on the interval \([\mu ,1-\mu ]\) when \(0\le \mu <\frac{1}{2}\) and on the interval \([1-\mu ,\mu ]\) when \(\frac{1}{2}<\mu \le 1\), then we have the following refinement of general Cauchy-Schwartz inequality.

Theorem 3

If \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu \le 1\), then for unitarily invariant norm \(\Vert \cdot \Vert \),

$$\begin{aligned} \Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\le & {} \frac{1}{n}\Vert |A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \\&\quad&+(1-\frac{1}{n})\Vert |A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}|^{r}\Vert \\\le & {} \frac{1}{|1-2\mu |}|\int _{\mu }^{1-\mu }\Vert |A^{t}XB^{1-t}|^{r}\Vert \cdot \Vert |A^{1-t}XB^{t}|^{r}\Vert dt| \quad \left(\mu \ne \frac{1}{2}\right) \\\le & {} \frac{1}{2n}\left[ (2n-1)\Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert +\Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\right] \\\le & {} \Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert \end{aligned}$$

hold for any positive integer n.

Proof

Assume that \(A,B\in M_{n}^{+}\), \(X\in M_{n}\) and \(0\le \mu < \frac{1}{2}\), then it follows by Lemma 1 that

$$\begin{aligned} \phi \left(\frac{\mu +1-\mu }{2}\right)\le & {} \frac{1}{n}\phi \left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)\phi \left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }\phi (t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le &{} \frac{1}{4n}\left[(2n-1)\phi (\mu )+2\phi \left(\frac{\mu +1-\mu }{2}\right)+(2n-1)\phi (1-\mu )\right]\\\le & {} \frac{\phi (\mu )+\phi (1-\mu )}{2}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \phi \left(\frac{1}{2}\right)\le & {} \frac{1}{n}\phi \left(\frac{1-\mu +(2n-1)\mu }{2n}\right)+\left(1-\frac{1}{n}\right)\phi \left(\frac{(n+1)(1-\mu )+(n-1)\mu }{2n}\right)\\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }\phi (t)dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{4n}\left[(2n-1)\phi (\mu )+2\phi (\frac{1}{2})+(2n-1)\phi (1-\mu )\right]\\\le & {} \frac{\phi (\mu )+\phi (1-\mu )}{2}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\le & {} \frac{1}{n}\Vert |A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \\&\quad&+(1-\frac{1}{n})\Vert |A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{(n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}|^{r}\Vert \\\le & {} \frac{1}{1-2\mu }\int _{\mu }^{1-\mu }\Vert |A^{t}XB^{1-t}|^{r}\Vert \cdot \Vert |A^{1-t}XB^{t}|^{r}\Vert dt\quad \left(\mu \ne \frac{1}{2}\right)\\\le & {} \frac{1}{2n}\left[(2n-1)\Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert +\Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\right]\\\le & {} \Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert . \end{aligned}$$
(8)

On the other hand, if \(\frac{1}{2}<\mu \le 1\), then it follows by symmetry (i.e., by applying the above inequality (8) to \(1-\mu \)) that

$$\begin{aligned} \Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\le & {} \frac{1}{n}\Vert |A^{\frac{1+(2n-2)\mu }{2n}}XB^{1-\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{1+(2n-2)\mu }{2n}}XB^{\frac{1+(2n-2)\mu }{2n}}|^{r}\Vert \\&\quad&+(1-\frac{1}{n})\Vert |A^{\frac{n+1-2\mu }{2n}}XB^{1-\frac{n+1-2\mu }{2n}}|^{r}\Vert \cdot \Vert |A^{1-\frac{n+1-2\mu }{2n}}XB^{\frac{n+1-2\mu }{2n}}|^{r}\Vert \\\le & {} \frac{1}{2\mu -1}\int _{1-\mu }^{\mu }\Vert |A^{t}XB^{1-t}|^{r}\Vert \cdot \Vert |A^{1-t}XB^{t}|^{r}\Vert dt \quad \left(\mu \ne \frac{1}{2}\right) \\\le & {} \frac{1}{2n}\left[(2n-1)\Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert +\Vert |A^{\frac{1}{2}}XB^{\frac{1}{2}}|^{r}\Vert ^{2}\right] \\\le & {} \Vert |A^{\mu }XB^{1-\mu }|^{r}\Vert \cdot \Vert |A^{1-\mu }XB^{\mu }|^{r}\Vert . \end{aligned}$$
(9)

We complete the proof of Theorem 3 by combining the inequalities (8) and (9). \(\square \)

In view of the fact that the functions f(t) and h(t) are symmetric, we have

$$\begin{aligned}&\quad |f(\mu )-f(1-\mu )|\cdot |h(\mu )-h(1-\mu )|\\&=|\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert -\Vert A^{1-\mu }XB^{\mu }\Vert \Vert A^{\mu }XB^{1-\mu }\Vert |\cdot \\&\quad&\quad |\Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2} -\Vert \frac{A^{1-\mu }XB^{\mu }+A^{\mu }XB^{1-\mu }}{2}\Vert ^{2} |\ \\&=0. \end{aligned}$$

We can have the following result by applying Lemma 2 to function

$$\begin{aligned} f(t)\cdot h(t)=\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}. \end{aligned}$$

Corollary 1

For \(0\le \mu \le 1\) and all \(\lambda \in [0,1]\), we have

$$\begin{aligned}&2\Vert A^{\frac{1}{2}}XB^{\frac{1}{2}}\Vert ^{4}-\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2} \\&\le l'(\lambda )\\&\le \frac{1}{|1-2\mu |}|\int _{\mu }^{1-\mu }\Vert A^{t}XB^{1-t}\Vert \Vert A^{1-t}XB^{t}\Vert \Vert \frac{A^{t}XB^{1-t}+A^{1-t}XB^{t}}{2}\Vert ^{2} dt|\quad \left(\mu \ne \frac{1}{2}\right)\\&\le L'(\lambda )\\&\le \Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}, \end{aligned}$$

where

$$\begin{aligned} L'(\lambda )= & {} \frac{1}{3}\Vert A^{1-\mu }XB^{\mu }\Vert \Vert A^{\mu }XB^{1-\mu }\Vert \Vert \frac{A^{1-\mu }XB^{\mu }+A^{\mu }XB^{1-\mu }}{2}\Vert ^{2}\\&\quad&+\frac{\lambda }{3}\Vert A^{\lambda +\mu -2\lambda \mu }XB^{1-(\lambda +\mu -2\lambda \mu )}\Vert \Vert A^{1-(\lambda +\mu -2\lambda \mu ) }XB^{\lambda +\mu -2\lambda \mu }\Vert \cdot \\&\quad&\Vert \frac{A^{\lambda +\mu -2\lambda \mu }XB^{1-(\lambda +\mu -2\lambda \mu )}+A^{1-(\lambda +\mu -2\lambda \mu )}XB^{\lambda +\mu -2\lambda \mu }}{2}\Vert ^{2}\\&\quad&+\frac{\lambda }{6}\Vert A^{\lambda +\mu -2\lambda \mu }XB^{1-(\lambda +\mu -2\lambda \mu )}\Vert \Vert A^{1-(\lambda +\mu -2\lambda \mu )}XB^{\lambda +\mu -2\lambda \mu }\Vert \cdot \\&\Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2}\\&\quad&+\frac{\lambda }{6}\Vert \frac{A^{\lambda +\mu -2\lambda \mu }XB^{1-(\lambda +\mu -2\lambda \mu )}+A^{1-(\lambda +\mu -2\lambda \mu )}XB^{\lambda +\mu -2\lambda \mu }}{2}\Vert ^{2}\cdot \\&\Vert \frac{A^{1-\mu }XB^{\mu }+A^{\mu }XB^{1-\mu }}{2}\Vert ^{2} \end{aligned}$$

and

$$\begin{aligned} l'(\lambda )= & {} 2\lambda \Vert A^{\frac{\lambda +2\mu -2\lambda \mu }{2}}XB^{1-\frac{\lambda +2\mu -2\lambda \mu }{2}}\Vert \Vert A^{1-\frac{\lambda +2\mu -2\lambda \mu }{2}}XB^{\frac{\lambda +2\mu -2\lambda \mu }{2}}\Vert \cdot \\&\Vert \frac{A^{\frac{\lambda +2\mu -2\lambda \mu }{2}}XB^{1-\frac{\lambda +2\mu -2\lambda \mu }{2}}+A^{1-\frac{\lambda +2\mu -2\lambda \mu }{2}}XB^{\frac{\lambda +2\mu -2\lambda \mu }{2}}}{2}\Vert ^{2}\\&-\Vert A^{\mu }XB^{1-\mu }\Vert \Vert A^{1-\mu }XB^{\mu }\Vert \Vert \frac{A^{\mu }XB^{1-\mu }+A^{1-\mu }XB^{\mu }}{2}\Vert ^{2} \\&+2(1-\lambda )\Vert A^{\frac{\lambda +1-2\lambda \mu }{2}}XB^{1-\frac{\lambda +1-2\lambda \mu }{2}}\Vert \Vert A^{1-\frac{\lambda +1-2\lambda \mu }{2}}XB^{\frac{\lambda +1-2\lambda \mu }{2}}\Vert \cdot \\&\Vert \frac{A^{\frac{\lambda +1-2\lambda \mu }{2}}XB^{1-\frac{\lambda +1-2\lambda \mu }{2}}+A^{1-\frac{\lambda +1-2\lambda \mu }{2}}XB^{\frac{\lambda +1-2\lambda \mu }{2}}}{2}\Vert ^{2}. \end{aligned}$$

Here we remark that \(|f(\mu )-f(1-\mu )|\cdot |\phi (\mu )-\phi (1-\mu )|=0\) and \(|h(\mu )-h(1-\mu )|\cdot |\phi (\mu )-\phi (1-\mu )|=0\) for \(0\le \mu \le 1\). Hence, results similar to Corollary 1 can be obtained by using \(f\cdot \phi \) and \(h\cdot \phi \).