Abstract
In this paper, we shall present some reverse arithmetic-geometric mean operator inequalities for unital positive linear maps. These inequalities improve some corresponding results due to Xue (J. Inequal. Appl. 2017:283, 2017).
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1 Introduction
Let m, \({m^{\prime}} \), \(m_{1}^{\prime}\), \(m_{2}^{\prime}\), \(m_{3}^{\prime}\), M, \({M^{\prime}}\), \(M_{1}^{\prime}\), \(M_{2}^{\prime}\) and \(M_{3}^{\prime}\) be scalars, I be the identity operator and the other capital letters be used to represent general elements of the \(C^{*}\)-algebra \(\mathcal{B}(\mathcal {H})\) of all bounded linear operators acting on a Hilbert space\((\mathcal{H},\langle\cdot,\cdot\rangle)\). The operator norm is denoted by \(\|\cdot\|\). An operator A is said to be positive if \(\langle Ax,x\rangle\geq0\) for all \(x \in\mathcal{H}\) and we write it as \(A\geq0\), it is said to be strictly positive if \(\langle Ax,x\rangle>0\) for all \(x \in {\mathcal{H}\setminus\{0\}}\) and we write it as \(A>0\). A linear map Φ is positive if \(\varPhi(A)\geq0\) whenever \(A\geq0\). It is said to be unital if \(\varPhi(I)=I\). For \(A, B>0\) the μ-weighted arithmetic mean and μ-weighted geometric mean of A and B are defined, respectively, by
where \(\mu\in[0,1]\), when \(\mu=1/2\), we write \(A\nabla B\) and \(A\sharp B \) for brevity for \(A\nabla_{1/2}B\) and \(A \sharp_{1/2}B\), respectively.
For \(0< m\leq A,B\leq M\), Tominaga [2] proved that the following operator reverse AM-GM inequality holds:
where \(S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}\) is called Specht’s ratio with \(h =\frac{M}{m}\).
The inequality (1.1) can be regarded as a counterpart of the following AM-GM inequality:
Lin [3, (3.3)] observed that
where \(K(h)=\frac{(h+1)^{2}}{4h} \) and \(h=\frac{M}{m}\). The constant \(K(t,2)=\frac{(t+1)^{2}}{4t}\) (\(t>0\)) is called the Kantorovich constant, which is simply represented by \(K(t)\) satisfying the following properties:
\(K(1,2)=1\),
\(K(t,2)= K(\frac{1}{t} ,2)\geq1\) (\(t>0\)),
\(K(t,2)\) is monotone increasing on \([1,\infty)\) and monotone decreasing on \((0,1]\).
By inequalities (1.1) and (1.3), we have
Because Φ is order preserving, (1.4) implies that
For a positive linear map Φ and \(A,B\geq0\). Ando [4] has proved the following inequality:
Then, by (1.5) and (1.6), we have
The studies of squaring operator inequalities start with [3, 5] and continued by a number of authors [6–10]. Lin [3] revealed that inequalities (1.5) and (1.7) can be squared as follows:
Recently, Xue [1] proved that if \(\sqrt{\frac{M}{m}}\leq2.314 \), then the following refinement of the inequality(1.4) holds:
Inspired by Lin’s idea [3] , Xue [1] also proved that if \(0< m\leq A, B\leq M\) and \(\sqrt{\frac{M}{m}}\leq2.314 \), then
and
inequalities (1.12) and (1.13) are refinements of the inequalities (1.8) and (1.9), respectively.
Moreover, she proved Lin’s conjecture [3] as follows:
Recently, Ali et al. obtained more refinements of the results presented by Xue [1] by using the relation (1.2), for comprehensive study, the reader is referred to [11]. In this article, in Sect. 2, we shall refine the inequalities (1.10)–(1.15), when \(\sqrt{\frac{M}{m}}\leq 2.314\), with the help of the Kantorovich constant.
2 Main results
We begin this section with the following lemmas.
Lemma 2.1
([12])
Let\(A,B >0\). Then the following norm inequality holds:
Remark 2.2
Lemma 2.1 is proved by Bhatia and Kittaneh in [12] for the finite dimensional case. However, all technical results used to prove this result for operator norm are also true for the infinite dimensional case. Here also, we mention that if \(A,B\) are compact operators, then a stronger result can be found in [13].
Lemma 2.3
([14])
Let\(A>0\). Then, for every positive unital linear mapΦ,
Lemma 2.4
([15])
Suppose that two operatorsA, Band positive real numbersm, \({m^{\prime}}\), M, \({M^{\prime}}\)satisfy either of the following conditions:
- (1)
\(0< m\leq A\leq m^{\prime}< M^{\prime}\leq B\leq M\),
- (2)
\(0< m\leq B\leq m^{\prime}< M^{\prime}\leq A\leq M\).
Then
for all\(\mu\in[0,1]\), \(r=\min[\mu, 1-\mu]\), \(h=\frac{M}{m}\)and\(h^{\prime}=\frac{{M^{\prime}}}{m^{\prime}}\).
Now, we prove the first main result in the following theorem.
Theorem 2.5
Let\(0< m\leq M\)and\(\sqrt{\frac{M}{m}}\leq2.314 \), we have
- (1)
If\(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then
$$\begin{aligned} \biggl(\frac{A+B}{2} \biggr)^{2}\leq\frac{K(h)}{K(h_{1}^{\prime})}(A\sharp B)^{2}, \end{aligned}$$(2.4)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{1}^{\prime})=\frac{(h_{1}^{\prime}+1)^{2}}{4h_{1}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{1}^{\prime}=\frac {M_{1}^{\prime}}{m_{1}^{\prime}}\).
- (2)
If\(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \biggl(\frac{A+B}{2} \biggr)^{2}\leq\frac{K(h)}{K(h_{2}^{\prime})}(A\sharp B)^{2}, \end{aligned}$$(2.5)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{2}^{\prime})= \frac{(h_{2}^{\prime}+1)^{2}}{4h^{\prime}_{2}}\), \(h = \frac{M}{m}\)and\(h_{2}^{\prime}= \frac {M_{2}^{\prime}}{m_{2}^{\prime}}\).
- (3)
If\(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then
$$\begin{aligned} \biggl(\frac{A+B}{2} \biggr)^{2}\leq\frac{K(h)}{K(h_{3}^{\prime})}(A\sharp B)^{2} , \end{aligned}$$(2.6)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{3}^{\prime})= \frac{(h_{3}^{\prime}+1)^{2}}{4h_{3}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{3}^{\prime}= \frac {M+m}{2m_{3}^{\prime}}\).
- (4)
If\(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \biggl(\frac{A+B}{2} \biggr)^{2}\leq\frac{K(h)}{K(h_{4}^{\prime})}(A\sharp B)^{2} , \end{aligned}$$(2.7)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{4}^{\prime})= \frac{(h_{4}^{\prime}+1)^{2}}{4h^{\prime}_{4}}\), \(h = \frac{M}{m}\)and\(h_{4}^{\prime}= \frac {2M_{3}^{\prime}}{M+m}\).
Proof
The operator inequality (2.4) is equivalent to
If \(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), we get
and
Compute
That is,
Since \(1\leq\sqrt{\frac{M}{m}}\leq2.314\), it follows that
It is easy to see that \(\frac{ (\frac{M+m}{2}+m )^{2}}{4\frac{M+m}{2}.m}\leq\frac {M+m}{2\sqrt{M m}}\) is equivalent to (2.10).
Thus
If \(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), we get
Similarly, we have
Since \(\frac{(\frac{M+m}{2}+M)^{2}}{4\frac{M+m}{2}.M}\leq\frac{(\frac {M+m}{2}+m)^{2}}{4\frac{M+m}{2}.m}\leq\frac{M+m}{2\sqrt{M m}}\), so (2.13) becomes
If \(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then we compute
so we have
If \(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), similarly, by (2.1), (2.3), (2.8) and (2.12), we have
This completes the proof. □
Remark 2.6
Because \(\frac{K(h)}{K(h_{1}^{\prime})}< K(h)\), \(\frac {K(h)}{K(h_{2}^{\prime})}< K(h)\), \(\frac{K(h)}{K(h_{3}^{\prime})}< K(h)\) and \(\frac{K(h)}{K(h_{4}^{\prime})}< K(h)\), so Theorem 2.5 is a refinement of the inequality (1.11).
Remark 2.7
Since \(t^{p}\) is operator monotone function for \(0\leq p\leq1\), so, by taking power \(\frac{1}{2}\) both sides of (2.4), (2.5), (2.6) and (2.7), respectively, we can easily get a refinement of the inequality (1.10) for the condition \(\sqrt{\frac{M}{m}}\leq2.314 \).
Theorem 2.8
Let\(0< m\leq M\)and\(\sqrt{\frac{M}{m}}\leq 2.314\), we have
- (1)
If\(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{1}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$(2.15)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{1}^{\prime})=\frac{(h_{1}^{\prime}+1)^{2}}{4h_{1}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{1}^{\prime}=\frac {M_{1}^{\prime}}{m_{1}^{\prime}}\).
- (2)
If\(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{2}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$(2.16)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{2}^{\prime})= \frac{(h_{2}^{\prime}+1)^{2}}{4h^{\prime}_{2}}\), \(h = \frac{M}{m}\)and\(h_{2}^{\prime}= \frac {M_{2}^{\prime}}{m_{2}^{\prime}}\).
- (3)
If\(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{3}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$(2.17)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{3}^{\prime})= \frac{(h_{3}^{\prime}+1)^{2}}{4h_{3}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{3}^{\prime}= \frac {M+m}{2m_{3}^{\prime}}\).
- (4)
If\(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{4}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$(2.18)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{4}^{\prime})= \frac{(h_{4}^{\prime}+1)^{2}}{4h^{\prime}_{4}}\), \(h = \frac{M}{m}\)and\(h_{4}^{\prime}= \frac {2M_{3}^{\prime}}{M+m}\).
Proof
Inequality (2.15) is equivalent to
If \(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then we compute
that is,
By \(1\leq\sqrt{\frac{M}{m}}\leq2.314\) and (2.10), we have
If \(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), similarly, by (2.1), (2.2), (2.3), (2.11), (2.12), \(\frac{(\frac{M+m}{2}+M)^{2}}{4\frac{M+m}{2}M}\leq\frac{(\frac {M+m}{2}+m)^{2}}{4\frac{M+m}{2}m}\) and (2.10), we obtain
If \(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then we compute
that is, we have
If \(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), similarly, by (2.1), (2.2), (2.3), (2.8) and (2.12), we have
It completes the proof. □
Remark 2.9
Obviously, Theorem 2.8 is a refinement of (1.12).
Theorem 2.10
Let\(0< m\leq M\)and\(\sqrt{\frac{M}{m}}\leq 2.314\), we have
- (1)
If\(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{1}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi(B)\bigr)^{2}, \end{aligned}$$(2.22)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{1}^{\prime})= \frac{(h_{1}^{\prime}+1)^{2}}{4h_{1}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{1}^{\prime}= \frac {M_{1}^{\prime}}{m_{1}^{\prime}}\).
- (2)
If\(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{2}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi(B)\bigr)^{2}, \end{aligned}$$(2.23)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{2}^{\prime})= \frac{(h_{2}^{\prime}+1)^{2}}{4h^{\prime}_{2}}\), \(h = \frac{M}{m}\)and\(h_{2}^{\prime}= \frac {M_{2}^{\prime}}{m_{2}^{\prime}}\).
- (3)
If\(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{3}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi(B)\bigr)^{2}, \end{aligned}$$(2.24)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{3}^{\prime})= \frac{(h_{3}^{\prime}+1)^{2}}{4h_{3}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{3}^{\prime}= \frac {M+m}{2m_{3}^{\prime}}\).
- (4)
If\(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{K(h)}{K(h_{4}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi(B)\bigr)^{2}, \end{aligned}$$(2.25)where\(K(h)= \frac{(h+1)^{2}}{4h}\), \(K(h_{4}^{\prime})= \frac{(h_{4}^{\prime}+1)^{2}}{4h^{\prime}_{4}}\), \(h = \frac{M}{m}\)and\(h_{4}^{\prime}= \frac {2M_{3}^{\prime}}{M+m}\).
Proof
Inequality (2.22) is equivalent to
If \(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then we compute
that is,
By \(1\leq\sqrt{\frac{M}{m}}\leq2.314\) and (2.10), we have
Since \(\frac{(\frac{M+m}{2}+M)^{2}}{4\frac{M+m}{2}M}\leq\frac{(\frac {M+m}{2}+m)^{2}}{4\frac{M+m}{2}m}\), by 2nd case \(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), we can easily obtain the inequality (2.23).
If \(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then we compute
thus, we have
The proof of (2.25) is similar to (2.24), we omit the details.
This completes the proof. □
Remark 2.11
Clearly Theorem 2.10 is a refinement of (1.13).
By (1.3) and Theorem 2.8, we obtain the following refinement of inequality (1.14).
Corollary 2.12
Let\(0< m\leq M\)and\(\sqrt{\frac{M}{m}}\leq 2.314\), we have
- (1)
If\(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{1}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}\), \(K(h_{1}^{\prime})= \frac{(h_{1}^{\prime}+1)^{2}}{4h^{\prime}_{1}}\), \(h = \frac{M}{ m}\)and\(h_{1}^{\prime}= \frac{M_{1}^{\prime}}{m_{1}^{\prime}}\).
- (2)
If\(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{2}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}\), \(K(h_{2}^{\prime})= \frac{(h_{2}^{\prime}+1)^{2}}{4h_{2}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{2}^{\prime}= \frac{M_{2}^{\prime}}{m_{2}^{\prime}}\).
- (3)
If\(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{3}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}\), \(K(h_{3}^{\prime})= \frac{(h_{3}^{\prime}+1)^{2}}{4h_{3}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{3}^{\prime}= \frac{M+m}{2m_{3}^{\prime}}\).
- (4)
If\(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{4}^{\prime})}\varPhi ^{2}(A\sharp B), \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e \log h^{\frac{1}{h-1}}}\), \(K(h_{4}^{\prime})= \frac{(h_{4}^{\prime}+1)^{2}}{4h^{\prime}_{4}}\), \(h = \frac{M}{m}\)and\(h_{4}^{\prime}= \frac{2M_{3}^{\prime}}{M+m}\).
By (1.3) and Theorem 2.10, we obtain the following refinement of the inequality (1.15).
Corollary 2.13
Let\(0< m\leq M\)and\(\sqrt{\frac{M}{m}}\leq2.314\), we have
- (1)
If\(0< m\leq A\leq m_{1}^{\prime}< M_{1}^{\prime}\leq B\leq\frac{M+m}{2}\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{1}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi( B)\bigr)^{2}, \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e \log h^{\frac{1}{h-1}}}\), \(K(h_{1}^{\prime})= \frac{(h_{1}^{\prime}+1)^{2}}{4h^{\prime}_{1}}\), \(h = \frac{M}{ m}\)and\(h_{1}^{\prime}= \frac{M_{1}^{\prime}}{m_{1}^{\prime}}\).
- (2)
If\(0<\frac{M+m}{2}\leq A\leq m_{2}^{\prime}<M_{2}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{2}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi( B)\bigr)^{2}, \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e \log h^{\frac{1}{h-1}}}\), \(K(h_{2}^{\prime})= \frac{(h_{2}^{\prime}+1)^{2}}{4h_{2}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{2}^{\prime}= \frac{M_{2}^{\prime}}{m_{2}^{\prime}}\).
- (3)
If\(0< m\leq A\leq m_{3}^{\prime}<\frac{M+m}{2}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{3}^{\prime})}\bigl( \varPhi (A)\sharp\varPhi(B)\bigr)^{2}, \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e \log h^{\frac{1}{h-1}}}\), \(K(h_{3}^{\prime})= \frac{(h_{3}^{\prime}+1)^{2}}{4h_{3}^{\prime}}\), \(h = \frac{M}{m}\)and\(h_{3}^{\prime}= \frac{M+m}{2m_{3}^{\prime}}\).
- (4)
If\(0< m\leq A\leq\frac{M+m}{2}<M_{3}^{\prime}\leq B\leq M\), then
$$\begin{aligned} \varPhi^{2} \biggl(\frac{A+B}{2} \biggr)\leq\frac{S^{2}(h)}{K(h_{4}^{\prime})}(\varPhi (A)\sharp\varPhi( B)^{2}, \end{aligned}$$where\(S(h)=\frac{h^{\frac{1}{h-1}}}{e \log h^{\frac{1}{h-1}}}\), \(K(h_{4}^{\prime})= \frac{(h_{4}^{\prime}+1)^{2}}{4h^{\prime}_{4}}\), \(h = \frac{M}{m}\)and\(h_{4}^{\prime}= \frac{2M_{3}^{\prime}}{M+m}\).
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Karim, S., Ali, I. & Mushtaq, M. Improvements of operator reverse AM-GM inequality involving positive linear maps. J Inequal Appl 2020, 73 (2020). https://doi.org/10.1186/s13660-020-02337-5
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DOI: https://doi.org/10.1186/s13660-020-02337-5