Let $M_{n}$ denote the algebra of all $n\times n$ complex matrices, $M^{+}_{n}$ be the set of all the positive semidefinite matrices in $M_{n}.$ For two Hermitian matrices $A$ and $B$, $A\geq B$ means $A-B\in M^{+}_{n}, $ $A> B$ means $A-B\in M^{++}_{n}, $ where $M^{++}_{n}$ is the set of all the strictly positive matrices in $M_{n}.$ $I$ stands for the identity matrix. The Hilbert-Schmidt norm of $A=[a_{ij}]\in M_{n}$ is defined by $||A||_{2}=\sqrt{{\sum\limits_{i, j=1}^n}|a_{ij}|^{2}}.$ It is well-known that the Hilbert-Schmidt norm is unitarily invariant in the sense that $||UAV || = ||A||$ for all unitary matrices $U, V\in M_{n}.$ What's more, we use the following notions

$ A\nabla _{v} B= (1-v)A+vB, $

$ A\sharp _{v} B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{v}A^{\frac{1}{2}}, $

$ A!_{v} B= ((1-v)A^{-1}+vB^{-1})^{-1}, $

for $A, B\in M^{++}_{n}$ and $0\leq v\leq1.$ Usually we denote by $A \nabla B, $ $A\sharp B$ and $A!B$ for brevity respectively when $v=\frac{1}{2}.$

In this paper, we setting $a, b>0.$ As we all know, the scalar harmonic-geometric-arithmetic mean inequalities

$ a!_{v} b\leq a\sharp _{v} b \leq a\nabla _{v} b $

(1.1)

hold and the second inequality is called Young inequality. Similarly, we also have the related operator version

$ A!_{v} B\leq A\sharp _{v} B \leq A\nabla _{v} B $

(1.2)

for two strictly positive operators $A$ and $B$.

The first refinements of Young inequality is the squared version proved in [1]

$ (a^{v}b^{1-v})^{2}+\min \{ v, 1-v \} ^{2} (a-b)^{2} \leq (va+(1-v)b)^{2}. $

(1.3)

Later, the authors in [2] obtained the other interesting refinement

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

(1.4)

Then many results about Young inequalities presented in recent years. wa can see [3], [4] and [5] for some related results. Also, in [3] authors proved that

$ A\nabla _{v} B\geq A!_{v} B+2\min \{ v, 1-v \}(A\nabla _{v} B- A!_{v} B) $

(1.5)

for $A, B\in M^{++}_{n}$ and $0\leq v\leq1.$

Alzer [6] proved that

$ (\frac{\nu}{\tau})^{\lambda}\leq\frac{(a\nabla_{\nu}b)^{\lambda}-(a\sharp _{\nu} b)^{\lambda}}{(a\nabla_{\tau}b)^{\lambda}-(a\sharp _{\tau} b)^{\lambda}}\leq(\frac{1-\nu}{1-\tau})^{\lambda} $

(1.6)

for $0< \nu\leq\tau<1$ and $\lambda\geq1, $ which is a different form of $(1.5)$. By a similar technique, Liao [7] presented that

$ (\frac{\nu}{\tau})^{\lambda}\leq\frac{(a\nabla_{\nu}b)^{\lambda}-(a!_{\nu}b)^{\lambda}}{(a\nabla_{\tau}b)^{\lambda}-(a!_{\tau}b)^{\lambda}}\leq(\frac{1-\nu}{1-\tau})^{\lambda} $

(1.7)

for $0< \nu\leq\tau<1$ and $\lambda\geq1.$ Sababheh [8] generalized $(1.6)$ and $(1.7)$ by convexity of function $f$

$ (\frac{\nu}{\tau})^{\lambda}\leq\frac{((1-\nu)f(0)+\nu f(1))^{\lambda}-f^{\lambda}(\nu)}{((1-\tau)f(0)+\tau f(1))^{\lambda}-f^{\lambda}(\tau)}\leq(\frac{1-\nu}{1-\tau})^{\lambda}, $

(1.8)

where $0< \nu\leq\tau<1$ and $\lambda\geq1.$ In the same paper [8], it is proved that

$ (\frac{\nu}{\tau})^{\lambda}\leq\frac{(a\sharp _{\nu} b)^{\lambda}-(a!_{\nu}b)^{\lambda}}{(a\sharp _{\tau} b)^{\lambda}-(a!_{\tau}b)^{\lambda}}, $

(1.9)

where $0< \nu\leq\tau<1$ and $\lambda\geq1.$

Our main task of this paper is to improved $(1.7)$ for scalar and matrix under some conditions. The article is organized in the following way: in Section 2, new refinements of harmonic-arithmetic mean are presented for scalars. In Section 3, similar inequalities for operators will be presented. And the Hilbert-Schmidt norm and determinant inequalities will be presented in Section 4 and 5 respectively.

In this part, we firstly give an improved version of harmonic-arithmetic mean inequality for scalar. It is also the base of this paper.

Theorem 2.1   Let $\nu, \tau $ a and $b$ are real positive numbers with $0< \nu, \tau<1, $ then we have

$ \begin{eqnarray} \frac{a\nabla_{\nu}b-a!_{\nu}b}{a\nabla_{\tau}b-a!_{\tau}b}\leq\frac{\nu(1-\nu)}{\tau(1-\tau)}, \ \ for\ (b-a)(\tau-\nu)>0; \end{eqnarray} $

(2.1)

and

$ \begin{eqnarray} \frac{a\nabla_{\nu}b-a!_{\nu}b}{a\nabla_{\tau}b-a!_{\tau}b}\geq\frac{\nu(1-\nu)}{\tau(1-\tau)}, \ \ for\ (b-a)(\tau-\nu)<0. \end{eqnarray} $

(2.2)

Proof   Put $f(v)=\frac{1-v+vx-(1-v+vx^{-1})^{-1}}{v(1-v)}, $ then we have $f'(v)=\frac{1}{v^{2}(1-v)^{2}}h(x), $ where $h(x)=v(1-v)[x-1+(1-v+vx^{-1})^{-2}(\frac{1}{x}-1)]-(1-2v)[1-v+vx-(1-v+vx^{-1})^{-1}].$ By a carefully and directly computation, we can have $h'(x)=\frac{v^{2}}{((1-v)x+v)^{3}}g(x), $ where $g(x)=2(1-v)(1-x)-(1-v)x-v+((1-v)x+v)^{3}, $ so we can get $g'(x)=3(1-v)^{2}(x-1)((1-v)x+v+1)$ easily.

Now if $0<x\leq1, $ then $g'(x)\leq0, $ which means $g(x)\geq g(1)=0, $ and then $h'(x)\geq0;$ and if $1\leq x<\infty, $ then $g'(x)\geq0, $ which means $g(x)\geq g(1)=0, $ and then $h'(x)\geq0.$

That is to say that $h'(x)\geq0$ for all $x\in(0, \infty).$ Hence when $0<x\leq1, $ we have $h(x)\leq h(1)=0, $ and so $f'(v)\leq0, $ which means that $f(v)$ is decreasing on $(0, 1);$ and when $1\leq x<\infty, $ $h(x)\geq h(1)=0, $ and so $f'(v)\geq0, $ which means that $f(v)$ is increasing on $(0, 1).$ Put $x=\frac{b}{a}$, we can get our desired results easily.

Remark 2.2   Let $0<\nu\leq\tau<1$, then we have the following inequalities from (2.1),

$ \begin{eqnarray} \frac{a\nabla_{\nu}b-a!_{\nu}b}{a\nabla_{\tau}b-a!_{\tau}b}\leq\frac{\nu(1-\nu)}{\tau(1-\tau)}\leq\frac{1-\nu}{1-\tau}, \ \ for\ (b-a)>0. \end{eqnarray} $

(2.3)

On the other hand, when $0<\nu\leq\tau<1$ and $b-a<0$, we also have by(2.1),

$ \begin{eqnarray} \frac{a^{-1}\nabla_{\nu}b^{-1}-a^{-1}!_{\nu}b^{-1}}{a^{-1}\nabla_{\tau}b^{-1}-a^{-1}!_{\tau}b^{-1}}\leq\frac{\nu(1-\nu)}{\tau(1-\tau)}, \end{eqnarray} $

(2.4)

which implies

$ \begin{eqnarray} \frac{\nu}{\tau}&\leq&\frac{\nu(1-\nu)}{\tau(1-\tau)}\leq\frac{a\nabla_{\nu}b-a!_{\nu}b}{a\nabla_{\tau}b-a!_{\tau}b}\nonumber\\ &\leq&\frac{(a\nabla_{\nu}b)(a!_\nu b)}{(a\nabla_{\tau}b)(a!_\tau b)}\frac{\nu(1-\nu)}{\tau(1-\tau)} \ \ for\ (b-a)<0, \end{eqnarray} $

(2.5)

by (2.2). Therefore, it is clearly that (2.3) and (2.5) are sharper than (1.7) under some conditions for $\lambda=1.$ Next, we give a quadratic refinement of Theorem 2.1, which is better than (1.7) when $\lambda=2.$

Theorem 2.3   Let $\nu, \tau, a$ and $b$ are real numbers with $0< \nu, \tau<1, $ then we have

$ \begin{eqnarray} \frac{(a\nabla_{\nu}b)^{2}-(a!_{\nu}b)^{2}}{(a\nabla_{\tau}b)^{2}-(a!_{\tau}b)^{2}}\leq\frac{\nu(1-\nu)}{\tau(1-\tau)}, \ \ for\ (b-a)(\tau-\nu)>0; \end{eqnarray} $

(2.6)

and

$ \begin{eqnarray} \frac{(a\nabla_{\nu}b)^{2}-(a!_{\nu}b)^{2}}{(a\nabla_{\tau}b)^{2}-(a!_{\tau}b)^{2}}\geq\frac{\nu(1-\nu)}{\tau(1-\tau)}, \ \ for\ (b-a)(\tau-\nu)<0. \end{eqnarray} $

(2.7)

Proof   Put $f(v)=\frac{(1-v+vx)^{2}-(1-v+vx^{-1})^{-2}}{v(1-v)}, $ then we have $f'(v)=\frac{1}{v^{2}(1-v)^{2}}h(x), $ where

$ h(x)=2v(1-v)[(1-v+vx)(x-1)+(1-v+vx^{-1})^{-3}(\frac{1}{x}-1)]\\-(1-2v)[(1-v+vx)^{2}-(1-v+vx^{-1})^{-2}] $

by a directly computation, we have $h'(x)=\frac{2xv^{2}}{((1-v)x+v)^{4}}g(v), $ where $g(v)=3(1-v)(1-x)-(1-v)x-v+((1-v)x+v)^{4}.$ By $g'(v)=-4(1-x)^{2}(1-v)[((1-v)x+v)^{2}+1+(1-v)x+v]\leq0, $ so we have $g(v)\geq g(1)=0, $ which means $h'(x)\geq0.$ Now if $0<x\leq1, $ then $h(x)\leq h(1)=0, $ and so $f'(v)\leq0, $ which means that $f(v)$ is decreasing on $(0, 1).$ On the other hand, if $1\leq x<\infty, $ then $h(x)\geq h(1)=0, $ and so $f'(v)\geq0, $ which means that $f(v)$ is increasing on $(0, 1).$ Put $x=\frac{b}{a}, $ we can get our desired results directly.

In this section, we will give some refinements of harmonic-arithmetic mean for operators, which are based on the inequalities $(2.1)$ and $(2.2)$.

Lemma3.1   Let $X\in M_{n}$ be self-adjoint and let $f$ and $g$ be continuous real functions such that $f(t)\geq g(t)$ for all $t\in Sp(X)$ (the Spectrum of $X$). Then $f(X)\geq g(X)$.

For more details about this property, readers can refer to [9].

Theorem 3.2   Let $A, B\in M^{++}_{n}$ and $0< \nu, \tau<1, $ then

$ \begin{eqnarray} \tau(1-\tau)(A\nabla _{\nu} B-A!_{\nu} B)\leq\nu(1-\nu)(A\nabla _{\tau} B-A!_{\tau} B) \end{eqnarray} $

(3.1)

for $(B-A)(\tau-\nu)\geq0;$and

$ \begin{eqnarray} \tau(1-\tau)(A\nabla _{\nu} B-A!_{\nu} B)\geq\nu(1-\nu)(A\nabla _{\tau} B-A!_{\tau} B) \end{eqnarray} $

(3.2)

for $(B-A)(\tau-\nu)\leq0.$

Proof   Let $a=1$ in $(2.1), $ for $(b-1)(\tau-v)\geq0, $ we have

$ \begin{eqnarray}\label{3.3} &&\tau(1-\tau)[1-\nu+\nu b-(1-\nu+\nu b^{-1})^{-1}]\nonumber\\ &&\leq\nu(1-\nu)[1-\tau+\tau b-(1-\tau+\tau b^{-1})^{-1}]. \end{eqnarray} $

(3.3)

We may assume $0<\tau<v<1, $ and $0<b\leq1$. For $(B-A)(\tau-\nu)\geq0$, we have $A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\leq I.$ The operator $X=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}$ has a positive Spectrum. By Lemma3.1 and (3.3), we have

$ \begin{eqnarray} &&\tau(1-\tau)[1-\nu+\nu X-(1-\nu+\nu X^{-1})^{-1}]\nonumber\\ &&\leq\nu(1-\nu)[1-\tau+\tau X-(1-\tau+\tau X^{-1})^{-1}]. \end{eqnarray} $

(3.4)

Multiplying (3.4) by $A^{\frac{1}{2}}$ on the both sides, we can get the desired inequality (3.1).

Using the same technique, we can get (3.2) by (2.2).Notice that the inequalities of Theorem 3.2 provide a refinement and a reverse of (1.2).

In this section, we will present inequalities of Theorem 2.2 for Hilbert-Schmidt norm.

Theorem 4.1   Let $X\in M_{n}$ and $B\in M^{++}_{n}$ for $0< v, \tau<1, $ then we have

$ \begin{eqnarray} &&\frac{||(1-v)X+v XB||^{2}_{2}-||[(1-v)X^{-1}+v B^{-1}X^{-1}]^{-1}||^{2}_{2}}{v(1-v)}\nonumber\\ &&\leq\frac{||(1-\tau)X+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\tau(1-\tau)}; \end{eqnarray} $

(4.1)

for $(B-I)(\tau-v)\geq0;$and

$ \begin{eqnarray} &&\frac{||(1-v)X+\nu XB||^{2}_{2}-||[(1-v)X^{-1}+v B^{-1}X^{-1}]^{-1}||^{2}_{2}}{v(1-v)}\nonumber\\ &&\geq\frac{||(1-\tau)X+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\tau(1-\tau)}. \end{eqnarray} $

(4.2)

for $(B-I)(\tau-v)\leq0.$

Proof   Since $B$ is positive definite, it follows by spectral theorem that there exist unitary matrices $V\in M_{n}, $ such that $B=V\Lambda V^{\ast}, $ where $\Lambda=diag(\nu_{1}, \nu_{2}, \cdots, \nu_{n})$ and $\nu_{i}$ are eigenvalues of $B, $ so $\nu_{l}>0, \ l=1, 2, \cdots, n.$ Let $Y=V^{\ast}XV=[y_{il}], $ then

$ \begin{array}{ll} & \ \ \ \ (1-v)X+vXB\\ &=V[(1-v)Y+vY\Lambda]V^\ast\\ &=V[(1-v+v\nu_{l})y_{il}]V^\ast, \end{array} $

and

$ \begin{array}{ll} & \ \ \ \ [(1-v)X^{-1}+v B^{-1}X^{-1}]^{-1}\\ &=V[(1-v)Y^{-1}+v\Lambda^{-1}Y^{-1}]^{-1}V^\ast\\ &=V[(1-v+v\nu^{-1}_{l})^{-1}y_{il}]V^\ast. \end{array} $

Now, by $(2.6)$ and the unitarily invariant of the Hilbert-Schmidt norm, we have

$ \begin{array}{ll} & \ \ \ \ ||(1-v)X+v XB||^{2}_{2}-||[(1-v)X^{-1}+v B^{-1}X^{-1}]^{-1}||^{2}_{2}\\ &={\sum\limits_{i, l=1}^n}(1-v+v\nu_{l})^{2}|y_{il}|^{2}-{\sum\limits_{i, l=1}^n}(1-v+v\nu^{-1}_{l})^{-2}|y_{il}|^{2}\\ &={\sum\limits_{i, l=1}^n}[(1-v+v\nu_{l})^{2}-((1-v)+v\nu^{-1}_{l})^{-2}]|y_{il}|^{2}\\ &\leq\frac{v(1-v)}{\tau(1-\tau)}{\sum\limits_{i, l=1}^n}[((1-\tau)+\tau\nu_{l})^{2}-((1-\tau)+\tau\nu^{-1}_{l})^{-2}]|y_{il}|^{2}\\ &=\frac{v(1-v)}{\tau(1-\tau)}[{\sum\limits_{i, l=1}^n}((1-\tau)+\tau\nu_{l})^{2}|y_{il}|^{2}-{\sum\limits_{i, l=1}^n}((1-\tau)+\tau\nu^{-1}_{l})^{-2}|y_{il}|^{2} ]\\ &=\frac{v(1-v)}{\tau(1-\tau)}[||(1-\tau)X+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2} ]. \end{array} $

Here we completed the proof of (4.1). Using the same method in (2.7), we can get (4.2) easily. So we omit it.

It is clearly that Theorem 4.1 provided a refinement of Corollary 4.2 in [7].

Remark 4.2   Theorem 4.1 is not true in general when we exchange $I$ for $A, $ where $A$ is a positive definite matrix. That is: Let $X\in M_{n}$ and $A, B\in M^{++}_{n}$ for $0< \nu, \tau<1, $ then we can not have results as below:

$ \begin{eqnarray} && \frac{||(1-\nu)AX+\nu XB||^{2}_{2}-||[(1-\nu)X^{-1}A^{-1}+\nu B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\nu(1-\nu)}\nonumber\\ &&\leq\frac{||(1-\tau)AX+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}A^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\tau(1-\tau)} \end{eqnarray} $

(4.3)

for $(B-A)(\tau-\nu)\geq0$ and

$ \begin{eqnarray} &&\frac{||(1-\nu)AX+\nu XB||^{2}_{2}-||[(1-\nu)X^{-1}A^{-1}+\nu B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\nu(1-\nu)}\nonumber\\ &&\geq\frac{||(1-\tau)AX+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}A^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\tau(1-\tau)}. \end{eqnarray} $

(4.4)

for $(B-A)(\tau-\nu)\leq0.$

Now we give the following example to state it.

Example 4.3   Let $ B= \left ( {\begin{array}{*{20}c} \frac{1}{2} & 0 \\ 0 & 1\\ \end{array}} \right ) $, $ A= \left ( {\begin{array}{*{20}c} \frac{1}{3} & 0\\ 0 & \frac{1}{2}\\ \end{array}} \right ) $ and $ X= \left ( {\begin{array}{*{20}c} 1 & -1 \\ -1 & 2\\ \end{array}} \right ), $ then (4.3) and (4.4) are not true for $\nu=\frac{1}{2}$ and $\tau=\frac{2}{3}$.

Proof   we can compute that

$ ||(1-\nu)AX+\nu XB||_2^2=\frac{53}{36}+\frac{23}{9}\nu +\frac{53}{36}\nu^2, $

and

$ ||((1-\nu)X^{-1}A^{-1}+\nu B^{-1}X^{-1})^{-1}||_2^2=\frac{1}{(6-6\nu+2\nu^2)^2}[53+9\nu^2-40\nu]. $

A careful calculation shows that

$ \begin{eqnarray} && \frac{||(1-\nu)AX+\nu XB||^{2}_{2}-||[(1-\nu)X^{-1}A^{-1}+\nu B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\nu(1-\nu)}\nonumber\\ &&=\frac{1}{(6-6\nu+2\nu^2)^2}[-\frac{53}{9}\nu^4+\frac{173}{9}\nu^3-\frac{123}{9}\nu^2-\frac{231}{9}\nu+26] \end{eqnarray} $

(4.5)

Let $\nu=\frac{1}{2}$ and $\tau=\frac{2}{3}$, then (4.5) implies

$ \frac{||(1-\nu)AX+\nu XB||^{2}_{2}-||[(1-\nu)X^{-1}A^{-1}+\nu B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\nu(1-\nu)}=0.96\cdots, $

and

$ \frac{||(1-\tau)AX+\tau XB||^{2}_{2}-||[(1-\tau)X^{-1}A^{-1}+\tau B^{-1}X^{-1}]^{-1}||^{2}_{2}}{\tau(1-\tau)}=0.88\cdots, $

which implies that (4.3) is not true clearly. Similarly, we can also prove that (4.4) is not true by exchanging $\nu$ and $\tau$.

In this section, we will present inequalities of Theorem 2.1 and Theorem 2.2 for determinant. Before it, we should recall some basic signs. The singular values of a matrix $A$ are defined by $s_{j}(A), j=1, 2, \cdots, n.$ And we denote the values of $\{s_{j}(A)\}$ as a non-increasing order. Besides, $det(A)$ is the determinant of $A.$ To obtain our results, we need a following lemma.

Lemma 5.1   [10] (Minkowski inequality) Let $a=[a_{i}], \ b=[b_{i}], \ i=1, 2, \cdots, n$ such that $a_{i}, \ b_{i}$ are positive real numbers. Then

$ (\prod\limits_{i=1}^n a_{i})^{\frac{1}{n}}+(\prod\limits_{i=1}^n b_{i})^{\frac{1}{n}}\leq(\prod\limits_{i=1}^n (a_{i}+b_{i}))^{\frac{1}{n}}. $

Equality hold if and only if $a=b.$

Theorem 5.2   Let $X\in M_{n}$ and $A, B\in M^{++}_{n}$ for $0< v, \tau<1, $ then we have for $(B-A)(\tau-v)\leq0, $

$ det(A!_{v}B)^{\frac{1}{n}}+\frac{v(1-\nu)}{\tau(1-\tau)}det(A\nabla_{\tau}B-A!_{\tau}B)^{\frac{1}{n}}\leq det(A\nabla_{v}B)^{\frac{1}{n}}. $

(5.1)

Proof   We may assume $0<v<\tau<1, $ then $0<s_{j}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})\leq1$ for $(B-A)(\tau-v)\leq0, $ so we have $A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\leq I.$ By the inequality (2.2) and we denote the positive definite matrix $T=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}, $ then we have

$ \frac{((1-v)+vs_{j}(T))-((1-v)+vs_{j}(T)^{-1})^{-1}}{((1-\tau)+\tau s_{j}(T))-((1-\tau)+\tau s_{j}(T)^{-1})^{-1}}\geq\frac{v(1-v)}{\tau(1-\tau)} $

for $j=1, 2, \cdots, n.$ It is a fact that the determinant of a positive definite matrix is product of its singular values, with Lemma 5.1, we have

$ \begin{eqnarray*} &&det(I\nabla_{v}T)^{\frac{1}{n}}\nonumber\\ &&=det[(1-v)I+vT]^{\frac{1}{n}}\nonumber\\ &&=[\prod\limits_{i=1}^n (1-v+vs_{i}(T))]^{\frac{1}{n}}\nonumber\\ &&\geq[\prod\limits_{i=1}^n (1-v+v s_{i}(T)-(1-v+vs_{i}(T)^{-1})^{-1})]^{\frac{1}{n}}+[\prod\limits_{i=1}^n (1-v+v s_{i}(T)^{-1})^{-1}]^{\frac{1}{n}}\nonumber\\ &&\geq[\prod\limits_{i=1}^n \frac{v(1-v)}{\tau(1-\tau)}(1-\tau+\tau s_{i}(T)-(1-\tau+\tau s_{i}(T)^{-1})^{-1})]^{\frac{1}{n}}\nonumber\\ &&+[\prod\limits_{i=1}^n (1-v+v s_{i}(T)^{-1})^{-1}]^{\frac{1}{n}}\nonumber\\ &&=\frac{v(1-v)}{\tau(1-\tau)}det[(I\nabla_{\tau}T)-(I!_{\tau}T)]^{\frac{1}{n}}+det(I!_{v}T)^{\frac{1}{n}}\nonumber\\ \end{eqnarray*} $

Multiplying $(detA^{\frac{1}{2}})^{\frac{1}{n}}$ on the both sides of the inequalities above, we can get (5.1).

Theorem 5.3   Let $X\in M_{n}$ and $A, B\in M^{++}_{n}$ for $0< \nu, \tau<1, $ then we have for $(B-A)(\tau-\nu)\leq0:$

$ det(A!_{\nu}B)^{\frac{2}{n}}+\frac{\nu(1-\nu)}{\tau(1-\tau)}det(A\nabla_{\tau}B-A!_{\tau}B)^{\frac{2}{n}}\leq det(A\nabla_{\nu}B)^{\frac{2}{n}}. $

(5.2)

Proof   Using the same technique above to $(2.4), $ we can easily get the proof of Theorem 5.3.