1 Introduction and Preliminaries

Let \({\mathbb B}({\mathscr {H}})\) denote the \(C^*\)-algebra of all bounded linear operators on a complex Hilbert space \({\mathscr {H}}\) with the identity I. An operator \(A\in {\mathbb B}({\mathscr {H}})\) is called positive if \(\langle Ax,x\rangle \ge 0\) for all \(x\in {\mathscr {H} }\), and we then write \(A\ge 0\). We write \(A>0\) if A is a positive invertible operator. The set of all positive invertible operators is denoted by \({\mathbb B}({\mathscr {H}})_+\). For self-adjoint operators \(A, B\in {\mathbb B}({\mathscr {H}})\), we say \(B\ge A\) if \(B-A\ge 0\). The Gelfand map f to f(A) is an isometric \(*\)-isomorphism between the \(C^*\)-algebra \(C(\mathrm {sp}(A))\) of a complex-valued continuous functions on the spectrum \(\mathrm {sp}(A)\) of a self-adjoint operator A and the \(C^*\)-algebra generated by I and A. If \(f, g\in C(\mathrm{sp}(A))\), then \(f(t)\ge g(t)\,\,(t\in \mathrm{sp}(A))\) implies that \(f(A)\ge g(A)\).

It is known that the Hadamard product can be presented by filtering the tensor product \(A \otimes B\) through a positive linear map. In fact,\(A\circ B=U^*(A\otimes B)U\), where \(U:{\mathscr {H}}\rightarrow {\mathscr {H}}\otimes {\mathscr {H}}\) is the isometry defined by \(Ue_j=e_j\otimes e_j\), where \((e_j)\) is an orthonormal basis of the Hilbert space \({\mathscr {H}}\); see [7].

For \(A, B\in {\mathbb {B}}({\mathscr {H}})_+\), the operator geometric mean \(A\sharp B\) is defined by \(A\sharp B=A^{\frac{1}{2}}\left( A^{\frac{-1}{2}}BA^{\frac{-1}{2}}\right) ^{\frac{1}{2}}A^{\frac{1}{2}}\). For \(\alpha \in (0,1)\), the operator-weighted geometric mean is defined by

$$\begin{aligned} A\sharp _\alpha B=A^{\frac{1}{2}}\left( A^{\frac{-1}{2}}BA^{\frac{-1}{2}}\right) ^{\alpha }A^{\frac{1}{2}}. \end{aligned}$$

Callebaut [4] showed the following refinement of the Cauchy–Schwarz inequality

$$\begin{aligned} \left( \sum _{j=1}^n x_j^{\frac{1}{2}}y_j^{\frac{1}{2}}\right) ^2&\le \sum _{j=1}^n x_j^{\frac{1+s}{2}}y_j^{\frac{1-s}{2}}\sum _{j=1}^n x_j^{\frac{1-s}{2}}y_j^{\frac{1+s}{2}}\nonumber \\&\le \sum _{j=1}^n x_j^{\frac{1+t}{2}}y_j^{\frac{1-t}{2}}\sum _{j=1}^n x_j^{\frac{1-t}{2}}y_j^{\frac{1+t}{2}}\nonumber \\&\le \left( \sum _{j=1}^n x_j\right) \left( \sum _{j=1}^ny_j\right) , \end{aligned}$$
(1.1)

where \( x_j, y_j \,\,(1\le j\le n)\) are positive real numbers and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). This is indeed an extension of the Cauchy–Schwarz inequality.

Wada [10] gave an operator version of the Callebaut inequality by showing that if \(A, B\in {\mathbb B}({\mathscr {H}})_+\), then

$$\begin{aligned} (A\sharp B)\otimes (A\sharp B)&\le \frac{1}{2}\left\{ (A\sharp _\alpha B)\otimes (A\sharp _{1-\alpha } B)+(A\sharp _{1-\alpha } B)\otimes (A\sharp _{\alpha } B)\right\} \\ {}&\le \frac{1}{2}\left\{ (A\otimes B)+(B\otimes A)\right\} , \end{aligned}$$

where \(\alpha \in [0,1]\). In [6], the authors showed another operator version of the Callebaut inequality:

$$\begin{aligned} \sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)&\le \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\nonumber \\&\le \sum _{j=1}^n(A_j\sharp _tB_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j)\nonumber \\ {}&\le \left( \sum _{j=1}^nA_j\right) \circ \left( \sum _{j=1}^nB_j\right) , \end{aligned}$$
(1.2)

where \(A_j, B_j\in {{\mathbb {B}}}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\).

In [3], the authors presented the following refinement of inequality (1.2) as follows:

$$\begin{aligned} \sum _{j=1}^n(A_j\sharp _{s}B_j)&\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\nonumber \\&\le \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\nonumber \\&\quad +\left( \frac{t-s}{s-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \nonumber \\&\le \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j), \end{aligned}$$
(1.3)

in which \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\).

There have been obtained several Cauchy–Schwarz type inequalities for Hilbert space operators and matrices; see [1, 2, 5, 8] and references therein.

In this paper, we present some refinements and reverses of the Callebaut inequality involving the weighted geometric mean and Hadamard product of Hilbert space operators.

2 Further Refienements of the Callebaut Inequality Involving Hadamard Product

The Kontorovich constant is

$$\begin{aligned} K(t,2)=\frac{(t+1)^2}{4t}\qquad (t>0). \end{aligned}$$

The classical Young inequality states that

$$\begin{aligned} a^\nu b^{1-\nu }\le \nu a+(1-\nu )b, \end{aligned}$$

where \(a,b\ge 0\) and \(\nu \in [0,1]\). Recently, Zuo et. al. [11] showed an improvement of the Young inequality as follows:

$$\begin{aligned} K\left( \sqrt{\frac{a}{b}},2\right) ^{r}a^\nu b^{1-\nu }\le \nu a+(1-\nu )b, \end{aligned}$$

where \(a,b>0\), \(\nu \in [0,1]\), \(r=\min \left\{ \nu ,1-\nu \right\} \). Applying this inequality, J. Wu and J. Zhao [9] showed the following refinement of the Young inequality:

$$\begin{aligned} K\left( \sqrt{\frac{a}{b}},2\right) ^{r{^\prime }}a^\nu b^{1-\nu }+r\left( \sqrt{a}-\sqrt{b}\right) ^2\le \nu a+(1-\nu )b, \end{aligned}$$
(2.1)

where \(a,b>0\), \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \), \(r=\min \left\{ \nu ,1-\nu \right\} \) and \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \). Using (2.1), we get the following lemmas.

Lemma 2.1

Let \(a,b>0\) and \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \). Then

$$\begin{aligned} K\left( \sqrt{\frac{a}{b}},2\right) ^{r{^\prime }}\left( a^\nu b^{1-\nu }+a^{1-\nu }b^{\nu }\right) +2r\left( \sqrt{a}-\sqrt{b}\right) ^2\le a+b, \end{aligned}$$
(2.2)

where \(r=\min \left\{ \nu ,1-\nu \right\} \) and \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \).

Lemma 2.2

Let \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned} K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }}&(A^{s}\otimes B^{1-s}+A^{1-s}\otimes B^{s})\nonumber \\&\,\,\,\,+\left( \frac{t-s}{t-1/2}\right) \left( A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}-2(A^{\frac{1}{2}}\otimes B^{\frac{1}{2}})\right) \nonumber \\&\le A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}, \end{aligned}$$
(2.3)

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Proof

Let \(a>0\). If we replace b by \(a^{-1}\) and take \(\nu =\frac{1-\mu }{2}\) (2.2), then we get

$$\begin{aligned} K(a,2)^{r{^\prime }}\left( a^\mu +a^{-\mu }\right) +(1-\mu )\left( a+a^{-1}-2\right) \le a+a^{-1}, \end{aligned}$$
(2.4)

in which \(\mu \in (0,1]\) and \(r{^\prime }=\min \left\{ 1-\mu ,\mu \right\} \). Let us fix positive real numbers \(\alpha ,\beta \) such that \(\beta <\alpha \). It follows from \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) that \(I\le h=\left( \frac{M}{m}\right) ^\alpha \le A^\alpha \otimes B^{-\alpha }\le h{^\prime }=\left( \frac{M{^\prime }}{m{^\prime }}\right) ^\alpha \) and \(\mathrm {sp}(A^\alpha \otimes B^{-\alpha })\subseteq [h,h{^\prime }]\subseteq (1,+\infty )\). Since the Kontorovich constant \(\frac{(a+1)^2}{4a}\) is an increasing function on \((1,+\infty )\), by (2.4), we have

$$\begin{aligned} K\left( \frac{M^\alpha }{m^\alpha },2\right) ^{r{^\prime }}\left( a^\mu +a^{-\mu }\right) +(1-\mu )\left( a+a^{-1}-2\right) \le a+a^{-1}, \end{aligned}$$

where \(\mu \in (0,1]\) and \(r{^\prime }=\min \left\{ 1-\mu ,\mu \right\} \). Using the functional calculus, if we replace a by the operator \(A^\alpha \otimes B^{-\alpha }\) and \(\mu \) by \(\frac{\beta }{\alpha }\) we have

$$\begin{aligned}&K\left( \frac{M^\alpha }{m^\alpha },2\right) ^{r{^\prime }}\left( A^\beta \otimes B^{-\beta }+A^{-\beta }\otimes B^{\beta }\right) \nonumber \\&\qquad +\left( 1-\frac{\beta }{\alpha }\right) \left( A^\alpha \otimes B^{-\alpha }+A^{-\alpha }\otimes B^{\alpha }-2I\right) \nonumber \\&\quad \le A^\alpha \otimes B^{-\alpha }+A^{-\alpha }\otimes B^{\alpha }, \end{aligned}$$
(2.5)

where \(r{^\prime }=\min \left\{ 1-\frac{\beta }{\alpha },\frac{\beta }{\alpha }\right\} \). Multiplying, we both sides of (2.5) by \(A^\frac{1}{2}\otimes B^\frac{1}{2}\), we obtain

$$\begin{aligned}&K\left( \frac{M^{\alpha }}{m^{\alpha }},2\right) ^{r{^\prime }}\left( A^{1+\beta }\otimes B^{1-\beta }+A^{1-\beta }\otimes B^{1+\beta }\right) \nonumber \\&\qquad +\left( 1-\frac{\beta }{\alpha }\right) \left( A^{1+\alpha }\otimes B^{1-\alpha }+A^{1-\alpha }\otimes B^{1+\alpha }-2(A\otimes B)\right) \nonumber \\&\quad \le A^{1+\alpha }\otimes B^{1-\alpha }+A^{1-\alpha }\otimes B^{1+\alpha }. \end{aligned}$$
(2.6)

Now, if we replace \(\alpha , \beta , A, B\) by \(2t-1, 2s-1, A^\frac{1}{2}, B^\frac{1}{2}\), respectively, in (2.6), we obtain

$$\begin{aligned}&K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime } }(A^{s}\otimes B^{1-s}+A^{1-s}\otimes B^{s})\\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}-2(A^{\frac{1}{2}}\otimes B^{\frac{1}{2}})\right) \\&\quad \le A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}\, \end{aligned}$$

for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \). \(\square \)

We are ready to prove the first result of this section.

Theorem 2.3

Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }} \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\nonumber \\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \nonumber \\&\quad \le \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j), \end{aligned}$$
(2.7)

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Proof

Put \(C_j=A_j^{-{1\over 2}}B_jA_j^{-{1\over 2}}\,\,(1\le j\le n)\). By inequality (2.3), we get

$$\begin{aligned}&K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }}\left( C_j^{s}\otimes C_i^{1-s}+C_j^{1-s}\otimes C_i^{s}\right) \nonumber \\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( C_j^{t}\otimes C_i^{1-t}+C_j^{1-t}\otimes C_i^{t}-2\big (C_j^{\frac{1}{2}}\otimes C_i^{\frac{1}{2}}\big )\right) \nonumber \\&\quad \le C_j^{t}\otimes C_i^{1-t}+C_j^{1-t}\otimes C_i^{t}\qquad (1\le i,j\le n). \end{aligned}$$
(2.8)

Multiplying both sides of (2.8) by \(A_j^{\frac{1}{2}}\otimes A_i^{\frac{1}{2}}\), we get

$$\begin{aligned}&K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }}\left( (A_j\sharp _{s} B_j)\otimes (A_i\sharp _{1-s} B_i)+(A_j\sharp _{1-s} B_j)\otimes (A_i\sharp _{s}B_i)\right) \nonumber \\&\qquad +\left( \frac{t-s}{t-1/2}\right) \Big ((A_j\sharp _{t} B_j)\otimes (A_i\sharp _{1-t} B_i)+(A_j\sharp _{1-t} B_j)\nonumber \\&\qquad \otimes (A_i\sharp _{t} B_i)-2(A_j\sharp B_j)\otimes (A_i\sharp B_i)\Big )\nonumber \\&\quad \le (A_j\sharp _{t}B_j)\otimes (A_i\sharp _{1-t} B_i)+(A_j\sharp _{1-t}B_j)\otimes (A_i\sharp _{t}B_i)\, \end{aligned}$$
(2.9)

for all \(1\le i,j\le n\). Therefore

$$\begin{aligned}&K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }}\left( \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\right) \\&\,+\left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t}B_j) -\Big (\sum _{j=1}^nA_j\sharp B_j\Big )\circ \Big (\sum _{j=1}^n A_j\sharp B_j\Big )\right) \\&\quad =\frac{1}{2}\sum _{i,j=1}^n\Big [K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }}\Big ((A_j\sharp _{s} B_j)\circ (A_i\sharp _{1-s} B_i)+(A_j\sharp _{1-s} B_j)\circ (A_i\sharp _{s}B_i)\Big )\nonumber \\&\qquad +\left( \frac{t-s}{t-1/2}\right) \Big ((A_j\sharp _{t} B_j)\circ (A_i\sharp _{1-t} B_i)+(A_j\sharp _{1-t} B_j)\nonumber \\&\qquad \circ (A_i\sharp _{t} B_i)-2(A_j\sharp B_j)\circ (A_i\sharp B_i)\Big )\Big ]\\&\le \frac{1}{2}\sum _{i,j=1}^n\left( (A_j\sharp _{t}B_j)\circ (A_i\sharp _{1-t} B_i)+(A_j\sharp _{1-t}B_j)\circ (A_i\sharp _{t}B_i)\right) \ \, (\text {by inequality (2.9)})\\&\quad =\sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j)\,. \end{aligned}$$

Remark 2.4

It follows from

$$\begin{aligned} \left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \ge 0, \end{aligned}$$

where \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\), either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(K(t,2)=\frac{(t+1)^2}{4t}\ge 1\,\,(t>0)\) that inequality (2.7) is a refinement of the second inequality of inequalities (1.2) and (1.3).

We conclude an application of Theorem 2.3 for numerical cases which is a refinement of inequality (1.1).

Corollary 2.5

Let \(0<m{^\prime } \le y_j\le m <M \le x_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\). Then

$$\begin{aligned}&\sum _{j=1}^nx_j^{{1+s}\over 2}y_j^{{1-s\over 2}} \sum _{j=1}^nx_j^{{1-s}\over 2}y_j^{{1+s\over 2}}\\&\quad \le K\left( \frac{M^{t}}{m^{t}},2\right) ^{r{^\prime }} \sum _{j=1}^nx_j^{{1+s}\over 2}y_j^{{1-s\over 2}} \sum _{j=1}^nx_j^{{1-s}\over 2}y_j^{{1+s\over 2}}\\&\qquad +\left( \frac{t-s}{t}\right) \left( \sum _{j=1}^nx_j^{{1+t}\over 2}y_j^{{1-t\over 2}} \sum _{j=1}^nx_j^{{1-t}\over 2}y_j^{{1+t\over 2}} -\left( \sum _{j=1}^nx_j^{1\over 2}y_j^{1\over 2}\right) ^2 \right) \\&\quad \le \sum _{j=1}^nx_j^{{1+t}\over 2}y_j^{{1-t\over 2}} \sum _{j=1}^nx_j^{{1-t}\over 2}y_j^{{1+t\over 2}}, \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t},\frac{s}{t}\right\} \).

Proof

If we put \(A_j=x_j, B_j=y_j\,\,(1\le j\le n)\) in Theorem 2.3 and inequality (1.2), then we get

$$\begin{aligned}&\sum _{j=1}^nx_j^{{1-s}}y_j^{{s}} \sum _{j=1}^nx_j^{{s}}y_j^{{1-s}}\\&\quad \le K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{r{^\prime }} \sum _{j=1}^nx_j^{{1-s}}y_j^{{s}} \sum _{j=1}^nx_j^{{s}}y_j^{{1-s}}\\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^nx_j^{{1-t}}y_j^{{t}} \sum _{j=1}^nx_j^{{t}}y_j^{{1-t}} -\left( \sum _{j=1}^nx_j^{1\over 2}y_j^{1\over 2}\right) ^2 \right) \\ \quad&\le \sum _{j=1}^nx_j^{{1-t}}y_j^{{t}} \sum _{j=1}^nx_j^{{t}}y_j^{{1-t}}, \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \) and either \(0\le t\le s<\frac{1}{2}\) or \(1\ge t\ge s>{\frac{1}{2}}\). Now if we replace s by \({s+1\over 2}\) and t by \({t+1\over 2}\), respectively, where either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\), then we obtain the desired inequalities. \(\square \)

Lemma 2.6

Let \(a,b>0\) and \(\nu \in (0,1)\). Then

$$\begin{aligned}&a^\nu b^{1-\nu }+a^{1-\nu } b^\nu +2r(\sqrt{a}-\sqrt{b})^2\\&\qquad +r{^\prime }\left( 2\sqrt{ab}+a+b-2a^{\frac{1}{4}}b^{\frac{3}{4}}-2a^{\frac{3}{4}}b^{\frac{1}{4}}\right) \le a+b, \end{aligned}$$

where \(r=\min \{\nu ,1-\nu \}\) and \(r{^\prime }=\min \{2r,1-2r\}\).

Proof

Let \(a,b>0\), \(\nu \in (0,1)\), \(r=\min \{\nu ,1-\nu \}\) and \(r{^\prime }=\min \{2r,1-2r\}\). Applying [12, Lemma1], we have the inequalities

$$\begin{aligned} a^{1-\nu } b^\nu +\nu (\sqrt{a}-\sqrt{b})^2+r{^\prime }(\root 4 \of {ab}-\sqrt{a})^2 \le (1-\nu ) a+\nu b, \end{aligned}$$

where \(0<\nu \le \frac{1}{2}\) and

$$\begin{aligned} a^{1-\nu } b^\nu +(1-\nu )(\sqrt{a}-\sqrt{b})^2+r{^\prime }(\root 4 \of {ab}-\sqrt{b})^2 \le (1-\nu ) a+\nu b, \end{aligned}$$

where \(\frac{1}{2}<\nu <1\). Summing these inequalities, we get the desired result. \(\square \)

Lemma 2.7

Let \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&(A^{s}\otimes B^{1-s}+A^{1-s}\otimes B^{s})\\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( A^{s}\otimes B^{1-s}+A^{1-s}\otimes B^{s}-2(A^{\frac{1}{2}}\otimes B^{\frac{1}{2}})\right) \\&\qquad +r{^\prime }\Big (A^{t}\otimes B^{1-s}+A^{1-s}\otimes B^{s}+2(A^{\frac{1}{2}}\otimes B^{\frac{1}{2}})-2A^{\frac{1+2s}{4}}\\&\qquad \otimes B^{\frac{3-2s}{4}}-2A^{\frac{3-2s}{4}}\otimes B^{\frac{1+2s}{4}}\Big ) \\&\quad \le A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}, \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Using Lemma 2.7 and the same argument in the proof of Lemma 2.2, we get another refinement of inequality (1.2).

Theorem 2.8

Let \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&\sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\nonumber \\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \nonumber \\&\qquad +r{^\prime }\Big (\sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) +\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\nonumber \\&\qquad -2\sum _{j=1}^n(A_j\sharp _{\frac{3-2s}{4}}B_j)\circ \sum _{j=1}^n(A_j\sharp _{\frac{1+2s}{4}}B_j)\Big )\nonumber \\&\quad \le \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j), \end{aligned}$$
(2.10)

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Remark 2.9

If \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\), then

$$\begin{aligned}&\sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) +\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\\&\qquad -2\sum _{j=1}^n(A_j\sharp _{\frac{3-2s}{4}}B_j)\circ \sum _{j=1}^n(A_j\sharp _{\frac{1+2s}{4}}B_j) \end{aligned}$$

and

$$\begin{aligned} \left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \end{aligned}$$

are positive operators. Hence, inequality (2.10) is another refinement of the second inequality of inequalities (1.2) and (1.3).

If we put \(B_j=I\,\,(1\le j\le n)\) in Theorem 2.8, then we get the next result.

Corollary 2.10

Let \(A_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&\sum _{j=1}^nA_j^{1-s}\circ \sum _{j=1}^nA_j^s\\&\qquad +\left( \frac{t-s}{t-1/2}\right) \left( \sum _{j=1}^nA_j^{1-s}\circ \sum _{j=1}^nA_j^s -\sum _{j=1}^nA_j^{\frac{1}{2}}\circ \sum _{j=1}^nA_j^{\frac{1}{2}}\right) \\&\qquad +r{^\prime }\Big (\sum _{j=1}^nA_j^{1-s}\circ \sum _{j=1}^nA_j^s +\sum _{j=1}^nA_j^{\frac{1}{2}}\circ \sum _{j=1}^nA_j^{\frac{1}{2}}\\&\qquad -2\sum _{j=1}^nA_j^{\frac{1+2s}{4}}\circ \sum _{j=1}^nA_j^{\frac{3-2s}{4}}\Big )\\&\quad \le \sum _{j=1}^nA_j^{1-t}\circ \sum _{j=1}^nA_j^t, \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

If in Theorem 2.8 we replace \(A_j, B_j, s, t\), by \(x_j, y_j, {s+1\over 2}, {t+1\over 2}\,\,(1\le j\le n)\), respectively, then we obtain another refinement of inequality (1.1).

Corollary 2.11

Let \(x_j,y_j\,\,(1\le j\le n)\) be positive numbers and either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\). Then

$$\begin{aligned}&\sum _{j=1}^nx_j^{1+s\over 2}y_j^{1-s\over 2} \sum _{j=1}^nx_j^{1-s\over 2}y_j^{1+s\over 2}\\&\qquad +\left( \frac{t-s}{t}\right) \left( \sum _{j=1}^nx_j^{1+s\over 2}y_j^{1-s\over 2} \sum _{j=1}^nx_j^{1-s\over 2}y_j^{1+s\over 2} -\left( \sum _{j=1}^nx_j^{1\over 2}y_j^{1\over 2}\right) ^2\right) \\&\qquad +r{^\prime }\Big (\sum _{j=1}^nx_j^{1+s\over 2}y_j^{1-s\over 2} \sum _{j=1}^nx_j^{1-s\over 2}y_j^{1+s\over 2} +\left( \sum _{j=1}^nx_j^{1\over 2}y_j^{1\over 2}\right) ^2\\&\qquad -2\sum _{j=1}^nx_j^{2+s\over 4}y_j^{2-s\over 4} \sum _{j=1}^nx_j^{2-s\over 4}y_j^{2+s\over 4}\Big )\\&\quad \le \sum _{j=1}^nx_j^{1+t\over 2}y_j^{1-t\over 2} \sum _{j=1}^nx_j^{1-t\over 2}y_j^{1+t\over 2}, \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t},\frac{s-1}{t-1}\right\} \).

3 Some Reverses of the Callebaut Type Inequality

In [9], the authors showed a reverse of the Young inequality as follows:

$$\begin{aligned} \nu a+(1-\nu )b\le K\left( \sqrt{\frac{a}{b}},2\right) ^{-r{^\prime }}a^\nu b^{1-\nu }+s\left( \sqrt{a}-\sqrt{b}\right) ^2, \end{aligned}$$
(3.1)

in which \(a,b>0\), \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \), \(r=\min \left\{ \nu ,1-\nu \right\} \), \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \) and \(s=\max \left\{ \nu ,1-\nu \right\} \). Applying (3.1), we have the next result.

Lemma 3.1

Let \(a,b>0\) and \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \). Then

$$\begin{aligned} a+b\le K\left( \sqrt{\frac{a}{b}},2\right) ^{-r{^\prime }}\left( a^\nu b^{1-\nu }+a^{1-\nu }b^{\nu }\right) +2s\left( \sqrt{a}-\sqrt{b}\right) ^2, \end{aligned}$$

where \(r=\min \left\{ \nu ,1-\nu \right\} \), \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \) and \(s=\max \left\{ \nu ,1-\nu \right\} \). In particular, If\(\nu \in [0,\frac{1}{2})\), then

$$\begin{aligned} a+a^{-1}\le K(a,2)^{-r{^\prime }}\left( a^{1-2\nu }+a^{-(1-2\nu )}\right) +2(1-\nu )\left( a^\frac{1}{2}-a^\frac{-1}{2}\right) ^2. \end{aligned}$$
(3.2)

Now, utilizing inequality (3.2) and the same argument in the proof of Lemma 2.2, we can accomplish the corresponding result:

Lemma 3.2

Let \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}\le K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{-r{^\prime }} (A^{s}\otimes B^{1-s}+A^{1-s}\otimes B^{s})\\&\qquad +\left( \frac{s-1/2}{t-1/2}\right) \left( A^{t}\otimes B^{1-t}+A^{1-t}\otimes B^{t}-2(A^{\frac{1}{2}}\otimes B^{\frac{1}{2}})\right) , \end{aligned}$$

where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

As a consequence of Lemma 3.2, we have the following result:

Theorem 3.3

Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\). Then

$$\begin{aligned}&\sum _{j=1}^n(A_j\sharp _{t}B_j)\!\circ \!\sum _{j=1}^n(A_j\sharp _{1-t} B_j)\!\le \!K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{-r{^\prime }} \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\\&\qquad +\left( \frac{s-1/2}{t-1/2}\right) \left( \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t}B_j) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) \end{aligned}$$

for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Remark 3.4

If we put \(t=1\) and \(1\ge s>{\frac{1}{2}}\) in Theorem 3.3, then we get a reverse of the third inequality of (1.2):

$$\begin{aligned}&\left( \sum _{j=1}^nA_j \right) \circ \left( \sum _{j=1}^n B_j\right) \le K\left( \frac{M}{m},2\right) ^{-r{^\prime }} \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\\&\qquad +\left( {2s-1}\right) \left( \left( \sum _{j=1}^nA_j \right) \circ \left( \sum _{j=1}^n B_j\right) -\sum _{j=1}^n(A_j\sharp B_j)\circ \sum _{j=1}^n(A_j\sharp B_j)\right) , \end{aligned}$$

where \(r{^\prime }=\min \left\{ {2-2s},{2s-1}\right\} \).

If we replace \(A_j, B_j, s, t\), by \(x_j, y_j, {s+1\over 2}, {t+1\over 2}\,\,(1\le j\le n)\) in Theorem 3.3, respectively, then we obtain a reverse of the second inequality of (1.1).

Corollary 3.5

Let \(0<m{^\prime } \le y_j\le m <M \le x_j\le M{^\prime }\,\,(1\le j\le n)\). Then

$$\begin{aligned}&\sum _{j=1}^nx_j^{1+t\over 2}y_j^{1-t\over 2} \sum _{j=1}^nx_j^{1-t\over 2}y_j^{1+t\over 2}\le K\left( \frac{M^{2t-1}}{m^{2t-1}},2\right) ^{-r{^\prime }} \sum _{j=1}^nx_j^{1+s\over 2}y_j^{1-s\over 2} \sum _{j=1}^nx_j^{1-s\over 2}y_j^{1+s\over 2}\\&\qquad +\left( \frac{s-1/2}{t-1/2}\right) \left( \sum _{j=1}^nx_j^{1+t\over 2}y_j^{1-t\over 2} \sum _{j=1}^nx_j^{1-t\over 2}y_j^{1+t\over 2} -\left( \sum _{j=1}^nx_j^{1\over 2}y_j^{1\over 2}\right) ^2\right) \end{aligned}$$

for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Proposition 3.6

Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then

$$\begin{aligned}&K\left( {h^{2t-1}},2\right) ^{r{^\prime }} \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j) +\left( \frac{t-s}{t-1/2}\right) \left( \sqrt{h}- \sqrt{\frac{1}{h}}\right) ^2\\&\quad \le \sum _{j=1}^n(A_j\sharp _{t}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-t} B_j)\\&\quad \le K\left( {h^{2t-1}},2\right) ^{-r{^\prime }} \sum _{j=1}^n(A_j\sharp _{s}B_j)\circ \sum _{j=1}^n(A_j\sharp _{1-s}B_j)\\&\qquad +\left( \frac{s-1/2}{t-1/2}\right) \left( \sqrt{h{^\prime }}-\sqrt{\frac{1}{h{^\prime }}}\right) ^2, \end{aligned}$$

where \(h=\frac{M}{m}\), \(h{^\prime }=\frac{M{^\prime }}{m{^\prime }}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).

Proof

Since the function \(f(a)=a-\frac{1}{a}\) is increasing on \((0,\infty )\), we have

$$\begin{aligned} \left( \sqrt{h}-\sqrt{\frac{1}{h}}\right) ^2 \le \left( \sqrt{ a}-\sqrt{\frac{1}{a}}\right) ^2\le \left( \sqrt{h{^\prime }}-\sqrt{\frac{1}{h{^\prime }}}\right) ^2\,\,\,(h\le a\le h{^\prime }). \end{aligned}$$

Applying inequalities (2.2), (3.2) and the same argument in the proof of Theorem 2.3, we get the desired result.