1 Introduction
For two positive numbers a, b and \(v\in[0,1]\), the quantity
is called Heron mean. And the inequality
is called Young’s inequality. Even though these inequalities look very simple, they have attracted many researchers in this field, where adding a positive term to refine the inequalities is possible.
$$ F_{v}(a,b)=(1-v)\sqrt{ab}+v\frac{a+b}{2} $$
(1.1)
$$ a^{v}b^{1-v} \leq va+(1-v)b,\quad a,b > 0\text{ and }0 \leq v \leq1 $$
(1.2)
Heron mean is the interpolation between arithmetic and geometric means for \(a,b\geq0\) and \(v\in[0,1]\). We can see papers [1, 2], and [3] for some new results about Heron mean and arithmetic–geometric mean.
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The first refinements of Young’s inequality is the squared version proved in [4]
Later, the authors in [5] obtained the other interesting refinement
A common fact about refinements (1.2) and (1.3) is having one refining term.
$$ \bigl(a^{v}b^{1-v}\bigr)^{2}+\min\{ v,1-v \} ^{2} (a-b)^{2} \leq \bigl(va+(1-v)b\bigr)^{2}. $$
(1.3)
$$ a^{v}b^{1-v}+\min\{ v,1-v \} (\sqrt{a}- \sqrt {b})^{2} \leq va+(1-b). $$
(1.4)
In the recent paper [6], some reverses and refinements of Young’s inequality were presented. It was proved that
where \(r_{1}=\min\{ 2r,1-2r\}\) for \(r=\min\{ v,1-v\}\). In the same paper, the following reversed versions were proved:
where \(r_{1}=\min\{2r,1-2r\}\) for \(r=\min\{v,1-v\}\).
$$ \textstyle\begin{cases} a^{v}b^{1-v}+v(\sqrt{a}-\sqrt{b})^{2}+r_{1}(\sqrt[4]{ab}-\sqrt {b})^{2}\leq va+(1-v)b,& 0\leq v\leq\frac{1}{2},\\ a^{v}b^{1-v}+(1-v)(\sqrt{a}-\sqrt{b})^{2}+r_{1}(\sqrt[4]{ab}-\sqrt {a})^{2}\leq va+(1-v)b,&\frac{1}{2}\leq v\leq1, \end{cases} $$
(1.5)
$$ \textstyle\begin{cases}va+(1-v)b+r_{1}(\sqrt[4]{ab}-\sqrt{a})^{2}\leq a^{v}b^{1-v}+(1-v)(\sqrt{a}-\sqrt{b})^{2},& 0\leq v\leq\frac {1}{2},\\ va+(1-v)b+r_{1}(\sqrt[4]{ab}-\sqrt{b})^{2}\leq a^{v}b^{1-v}+v(\sqrt {a}-\sqrt{b})^{2},&\frac{1}{2}\leq v\leq1, \end{cases} $$
(1.6)
In this paper, our main results are to give refinements of Heron mean for scalars and matrices in Sect. 2; and in Sect. 3, in a different way, to get an operator version of (1.5), which is the refinement of (1.4). Besides, in the same section, the refinements of Young’s inequalities for the Hilbert–Schmidt norm will be presented using the same technology as in Sect. 2.
For our convenience, we firstly give some denotations.
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Throughout the paper, H is a Hilbert space and \(B(H)\) denotes the set of all bounded linear operators on a complex Hilbert space H. An operator \(A\in B^{+}(H)\) is positive invertible if \((Ax,x)>0\) for every vector \(x\in H\setminus\{0\}\). \(M_{n}\) denotes the space of all \(n\times n\) complex matrices. The Hilbert–Schmidt norm of \(A=[a_{ij}]\in M_{n}\) is defined by
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 is more, we define
denoted by \(A \nabla B\) and \(A\sharp B\), respectively, when \(v=\frac{1}{2}\).
$$\Vert A \Vert _{2}=\sqrt{{\sum _{i,j=1}^{n}} \vert a_{ij} \vert ^{2}}. $$
$$\begin{gathered} A\nabla_{v} B= (1-v)A+vB,\quad v\in[0,1], \\ A\sharp_{v} B=A^{\frac{1}{2}}\bigl(A^{-\frac{1}{2}}BA^{-\frac {1}{2}} \bigr)^{v}A^{\frac{1}{2}},\quad v\in R\end{gathered} $$
2 Main results
2.1 Refinements of Heron mean
Heron mean is defined by
$$ F_{v}(a,b)=(1-v)\sqrt{ab}+v\frac{a+b}{2}. $$
(2.1)
It is easy to see that \(F_{v}(a,b)\) is an increasing function in v on \([0,1]\) and
$$ \sqrt{ab}=F_{0}(a,b)\leq F_{v}(a,b)\leq F_{1}(a,b)= \frac{a+b}{2}. $$
(2.2)
Our purpose of this section is to give refinements of Heron mean for a scalar and some other auxiliary results.
Theorem 2.1
For
\(a,b\geq0\), and
\(v\in[0,1]\), we have
where
\(F_{v}(a,b)=(1-v)\sqrt{ab}+v\frac{a+b}{2}\).
$$\begin{aligned}& \frac{1}{2}v(1-v) (\sqrt{a}-\sqrt{b})^{2}+\sqrt {ab}\leq F_{v}(a,b), \end{aligned}$$
(2.3)
$$\begin{aligned}& F_{v}(a,b)\leq\frac{a+b}{2}-\frac{1}{2}v(1-v) (\sqrt {a}- \sqrt{b})^{2}, \end{aligned}$$
(2.4)
Proof
Firstly,
$$\begin{gathered} F_{v}(a,b)-\frac{1}{2}v(1-v) ( \sqrt{a}-\sqrt{b})^{2} \\ \quad=\frac{1}{2}(va+vb+2\sqrt{ab}-2v\sqrt{ab})-\frac {1}{2} \bigl(va+vb-2v\sqrt{ab}-v^{2}a-v^{2}b+2v^{2} \sqrt{ab}\bigr) \\ \quad=v^{2}\biggl(\frac{a+b}{2}-\sqrt{ab}\biggr)+\sqrt{ab} \\ \quad\geq\sqrt{ab}. \end{gathered} $$
Next, we also have
□
$$\begin{gathered} \frac{a+b}{2}-\frac{1}{2}v(1-v) (\sqrt{a}- \sqrt{b})^{2}-F_{v}(a,b) \\ \quad=\frac{1}{2}\bigl(a+b-va-vb+2v\sqrt{ab}+v^{2}a+v^{2}b-2v^{2} \sqrt {ab}-va-vb-2\sqrt{ab}+2v\sqrt{ab}\bigr) \\ \quad=\frac{1}{2}\bigl[(1-2v)a+(1-2v)b-2(1-2v)\sqrt{ab}+v^{2}(a+b-2 \sqrt {ab})\bigr] \\ \quad=\frac{1}{2}(1-v)^{2}(\sqrt{a}-\sqrt{b})^{2} \\ \quad\geq0. \end{gathered} $$
It is clear that \(\frac{1}{2}v(1-v)(\sqrt{a}-\sqrt{b})^{2}\geq0\), so (2.3) and (2.4) are refinements of (2.2).
With Theorem 2.1 in hand, we will give refinements of Heron mean for operators by the monotonicity property of operator functions.
Lemma 2.2
Let
\(X\in B(H)\)
be self-adjoint, and let
f
and
g
be continuous real functions such that
\(f(t)\geq g(t)\)
for all
\(t\in \operatorname{Sp}(X)\) (the spectrum of
X). Then
\(f(X)\geq g(X)\).
For more details about this property, the reader is referred to [7].
Theorem 2.3
Let
\(A,B\in B^{+}(H)\)
be positive invertible operators, I
be the identity operator, and
\(v\in[0,1]\), then we have
and
where
\(F_{v}(A,B)=vA\bigtriangledown B+(1-v)A\sharp B\).
$$ v(1-v) (A\bigtriangledown B-A\sharp B)+A\sharp B\leq F_{v}(A,B) $$
(2.5)
$$ F_{v}(A,B)\leq A\bigtriangledown B-v(1-v) (A\bigtriangledown B-A\sharp B), $$
(2.6)
Proof
Let \(b=1\) in (2.3) and expand the summand to get
Note that the operator \(X=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\) has a positive spectrum, and by Lemma 2.2 and (2.7) we have
Finally, multiplying inequality (2.8) by \(A^{\frac{1}{2}}\) on the left- and right-hand sides, we can get
which is equivalent to (2.5).
$$ \frac{1}{2}v(1-v) (a+1-2\sqrt{a})+\sqrt{a}\leq\frac {1}{2}v(a+1)+(1-v) \sqrt{a}. $$
(2.7)
$$ \frac{1}{2}v(1-v) \bigl(X+I-2X^{\frac{1}{2}}\bigr)+X^{\frac {1}{2}}\leq \frac{1}{2}v(X+I)+(1-v)X^{\frac{1}{2}}. $$
(2.8)
$$\frac{1}{2}v(1-v) (A+B-2A\sharp B)+A\sharp B\leq\frac {1}{2}v(A+B)+(1-v)A \sharp B, $$
Next, we will present the refinements of Heron mean for the Hilbert–Schmidt norm.
Theorem 2.4
Suppose
\(A,B,X\in M_{n}\)
such that
A, B
are two positive definite matrices for
\(0\leq v\leq1\), then we have
$$ \begin{aligned}[b] &\frac{1}{2}v(1-v)|\!|\!|AX-XB |\!|\!|^{2}_{2}+\big|\!\big|\!\big|A^{\frac {1}{2}}XB^{\frac{1}{2}} \big|\!\big|\!\big|^{2}_{2} \\ &\quad\leq\frac{1}{2}v|\!|\!|AX+XB|\!|\!|^{2}_{2}+(1-2v) \big|\!\big|\!\big|A^{\frac{1}{2}}XB^{\frac{1}{2}}\big|\!\big|\!\big|^{2}_{2} \\ &\quad\leq\frac{1}{2}|\!|\!|AX+XB|\!|\!|^{2}_{2}- \frac {1}{2}v(1-v)|\!|\!|AX-XB|\!|\!|^{2}_{2}-\big|\!\big|\!\big|A^{\frac {1}{2}}XB^{\frac{1}{2}}\big|\!\big|\!\big|^{2}_{2}. \end{aligned} $$
(2.9)
Proof
Replace a, b by \(a^{2}\), \(b^{2}\) in (2.3) and (2.4), then we have
Since A and B are positive definite, it follows by the spectral theorem that there exist unitary matrices \(U,V\in M_{n}\) such that
where \(\Lambda_{1}=\operatorname{diag}(\lambda_{1},\lambda_{2},\ldots,\lambda _{n})\), \(\Lambda_{2}=\operatorname{diag}(\nu_{1},\nu_{2},\ldots,\nu_{n})\), \(\lambda _{i},\nu_{i}>0\), \(i=1,2,\ldots,n\).
$$ \begin{aligned}[b] &\frac{1}{2}v(1-v) (a-b)^{2}+ \bigl(a^{\frac{1}{2}}b^{\frac{1}{2}}\bigr)^{2} \\ &\quad\leq\frac{1}{2}v(a+b)^{2}+(1-2v) \bigl(a^{\frac{1}{2}}b^{\frac {1}{2}} \bigr)^{2} \\ &\quad\leq\frac{1}{2}(a+b)^{2}-\frac{1}{2}v(1-v) (a-b)^{2}-\bigl(a^{\frac {1}{2}}b^{\frac{1}{2}}\bigr)^{2}. \end{aligned} $$
(2.10)
$$A=U\Lambda_{1}U^{\ast},\qquad B=V\Lambda_{2}V^{\ast}, $$
Let \(Y=U^{\ast}XV=[y_{il}]\), then
and
Now, by (2.10) and the unitary invariance of the Hilbert–Schmidt norm, we have
So we finished the proof. □
$$\begin{gathered} AX-XB=U\bigl[(\lambda_{i}-\nu_{l})y_{il} \bigr]V^{\ast}; \\ A^{\frac{1}{2}}XB^{\frac{1}{2}}=U\bigl[\bigl(\lambda_{i}^{\frac {1}{2}} \nu_{l}^{\frac{1}{2}}\bigr)y_{il}\bigr]V^{\ast};\end{gathered} $$
$$AX+XB=U\bigl[(\lambda_{i}+\nu_{l})y_{il} \bigr]V^{\ast}. $$
$$\begin{gathered} \frac{1}{2}v(1-v)|\!|\!|AX-XB |\!|\!|^{2}_{2}+\big|\!\big|\!\big|A^{\frac {1}{2}}XB^{\frac{1}{2}} \big|\!\big|\!\big|^{2}_{2} \\ \quad={\sum_{i,l=1}^{n}}\biggl\{ { \frac{1}{2}v(1-v) (\lambda_{i}-\nu _{l})^{2}+ \bigl(\lambda_{i}^{\frac{1}{2}}\nu_{l}^{\frac{1}{2}} \bigr)^{2}}\biggr\} \vert y_{il} \vert ^{2} \\ \quad\leq{\sum_{i,l=1}^{n}}\biggl\{ \frac{1}{2}v(\lambda_{i}+\nu _{l})^{2}+(1-2v) \bigl(\lambda_{i}^{\frac{1}{2}}\nu_{l}^{\frac {1}{2}} \bigr)^{2}\biggr\} \vert y_{il} \vert ^{2} \\ \quad=\frac{1}{2}v|\!|\!|AX+XB|\!|\!|^{2}_{2}+(1-2v) \big|\!\big|\!\big|A^{\frac {1}{2}}XB^{\frac{1}{2}}\big|\!\big|\!\big|^{2}_{2} \\ \quad\leq{\sum_{i,l=1}^{n}}\biggl\{ \frac{1}{2}(\lambda_{i}+\nu _{l})^{2}- \frac{1}{2}v(1-v) (\lambda_{i}-\nu_{l})^{2}- \bigl(\lambda _{i}^{\frac{1}{2}}\nu_{l}^{\frac{1}{2}} \bigr)^{2}\biggr\} \vert y_{il} \vert ^{2} \\ \quad=\frac{1}{2}|\!|\!|AX+XB|\!|\!|^{2}_{2}- \frac{1}{2}v(1-v)|\!|\!|AX-XB|\!|\!|^{2}_{2}-\big|\!\big|\!\big|A^{\frac{1}{2}}XB^{\frac{1}{2}}\big|\!\big|\!\big|^{2}_{2}. \end{gathered} $$
2.2 Refinements of Young’s inequalities
It is well known that
with equality if and only if \(a=b\) is called Young’s inequality.
$$ a^{1-v}b^{v} \leq(1-v)a+vb,\quad a,b > 0\text{ and }v\in[0,1] $$
(2.11)
An operator version of (2.11) in [7] says that
for \(A,B\in B^{+}(H)\) and \(v\in[0,1]\). Kittaneh and Manasrah [5] gave a different type of improvement of Young’s matrix inequalities:
for \(A, B\in B^{+}(H)\), \(v\in[0,1]\), \(r=\min\{v, 1-v\}\), and \(s=\max \{v, 1-v\}\).
$$ A\sharp_{v}B\leq A\nabla_{v} B $$
(2.12)
$$ 2r(A\nabla B-A\sharp B)\leq A\nabla_{v} B-A\sharp _{v}B \leq2s(A\nabla B-A\sharp B) $$
(2.13)
Here, we give the first inequalities’ refinements of (2.13). Before that, we need a lemma.
Lemma 2.5
([8])
$$A\nabla_{\mu}(A\sharp_{\nu}B)=A\nabla_{\mu\nu}B-\mu(A \nabla _{\nu}B-A\sharp_{\nu}B) $$
Proof
$$\begin{gathered} A\nabla_{\mu}(A\sharp_{\nu}B) \\ \quad=(1-\mu)A+\mu A\sharp_{\nu}B \\ \quad=A+\mu\nu B-\mu\nu A-\mu\bigl[(1-\nu)A+\nu B-A\sharp_{\nu}B\bigr] \\ \quad=A\nabla_{\mu\nu}B-\mu(A\nabla_{\nu}B-A\sharp_{\nu}B). \end{gathered} $$
Theorem 2.6
(1)
If
\(0\leq\nu\leq\frac{1}{2}\), then
$$ 2r_{1}(A\nabla_{\frac{1}{4}}B-A\sharp_{\frac {1}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B)\leq A\nabla_{v} B-A\sharp _{v}B. $$
(2.14)
(2)
If
\(\frac{1}{2}\leq\nu\leq1\), then
where
\(r=\min\{\nu, 1-\nu\}\)
and
\(r_{1}=\min\{2r, 1-2r\}\).
$$ 2r_{1}(A\nabla_{\frac{3}{4}}B-A\sharp_{\frac {3}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B)\leq A\nabla_{v} B-A\sharp _{v}B, $$
(2.15)
Proof
For \(0\leq\nu\leq\frac{1}{2}\), then we have \(0\leq2\nu\leq1\). Substituting B by \(A\sharp B\) and ν by 2ν in the first inequality (2.13), we have
By computing directly with Lemma 2.5, then we have
Exchanging A for B and ν for \(1-\nu\) in (2.17) for \(\frac {1}{2}\leq\nu\leq1\), we get
So we completed the proof. □
$$ 2\min\{2r, 1-2r\}\bigl(A\nabla(A\sharp B)-A\sharp(A\sharp B)\bigr)\leq A \nabla_{2v} (A\sharp B)-A\sharp_{2v}(A\sharp B). $$
(2.16)
$$ 2r_{1}(A\nabla_{\frac{1}{4}}B-A\sharp_{\frac {1}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B)\leq A\nabla_{v} B-A\sharp _{v}B. $$
(2.17)
$$ 2r_{1}(A\nabla_{\frac{3}{4}}B-A\sharp_{\frac {3}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B)\leq A\nabla_{v} B-A\sharp _{v}B. $$
(2.18)
Remark 2.7
Our inequalities (2.14) and (2.15) are stronger than the first inequality (2.13), that is,
(1)
for \(0\leq\nu\leq\frac{1}{2}\),
$$ 2r(A\nabla B-A\sharp B)\leq2r_{1}(A\nabla_{\frac {1}{4}}B-A \sharp_{\frac{1}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B); $$
(2.19)
(2)
for \(\frac{1}{2}\leq\nu\leq1\),
where \(\nu\in[0, 1]\), \(r=\min\{\nu, 1-\nu\}\) and \(r_{1}=\min\{2r, 1-2r\}\).
$$ 2r(A\nabla B-A\sharp B)\leq2r_{1}(A\nabla_{\frac {3}{4}}B-A \sharp_{\frac{3}{4}}B)+(2r-r_{1}) (A\nabla B-A\sharp B), $$
(2.20)
Proof
For \(0\leq\nu\leq\frac{1}{4}\), then \(r=\nu\), \(r_{1}=2\nu\). So (2.19) is equivalent to
that is,
So we only need to prove
which is clearly true for A, B are positive definite operators.
$$ 4\nu(A\nabla_{\frac{1}{4}}B-A\sharp_{\frac {1}{4}}B)\geq2\nu(A\nabla B-A\sharp B), $$
(2.21)
$$ \frac{3}{2}A+\frac{1}{2}B-2A^{\frac{1}{2}}\bigl(A^{-\frac {1}{2}}BA^{-\frac{1}{2}} \bigr)^{\frac{1}{4}}A^{\frac{1}{2}}\geq\frac {A+B}{2}-A^{\frac{1}{2}} \bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac {1}{2}}A^{\frac{1}{2}}. $$
(2.22)
$$ I+\bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac {1}{2}}\geq2 \bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac{1}{4}}, $$
(2.23)
For \(\frac{1}{4}\leq\nu\leq\frac{1}{2}\), then \(r=\nu\), \(r_{1}=1-2\nu\). Equation (2.19) is equivalent to
that is,
By (2.21), we can prove (2.25) directly.
$$ 2(1-2v) (A\nabla_{\frac{1}{4}}B-A\sharp_{\frac {1}{4}})+(2v-1+2v) (A\nabla B-A \sharp B)\geq2\nu(A\nabla B-A\sharp B), $$
(2.24)
$$ 2(A\nabla_{\frac{1}{4}}B-A\sharp_{\frac {1}{4}}B)\geq(A\nabla B-A\sharp B). $$
(2.25)
Similarly, we can prove (2.20).
For \(\frac{3}{4}\leq\nu\leq1\), then \(r=1-\nu\), \(r_{1}=2-2\nu\). So (2.20) is equivalent to
that is,
Multiplying by \(A^{-\frac{1}{2}}\) on both sides and dividing by \((A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}\), we get
which is clearly true for A, B are positive definite operators.
$$ (4-4v) (A\nabla_{\frac{3}{4}}B-A\sharp_{\frac {3}{4}}B)\geq2(1-v) (A\nabla B-A \sharp B), $$
(2.26)
$$ B+A^{\frac{1}{2}}\bigl(A^{-\frac{1}{2}}BA^{-\frac {1}{2}}\bigr)^{\frac{1}{2}}A^{\frac{1}{2}} \geq2A^{\frac{1}{2}}\bigl(A^{-\frac {1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac{3}{4}}A^{\frac{1}{2}}. $$
(2.27)
$$ \bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac {1}{2}}+I\geq2 \bigl(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\bigr)^{\frac{1}{4}}, $$
(2.28)
For \(\frac{1}{2}\leq\nu\leq\frac{3}{4}\), then \(r=1-\nu\), \(r_{1}=2\nu-1\). So (2.20) is equivalent to
that is,
which can be got directly from (2.26). □
$$ (4v-2) (A\nabla_{\frac{3}{4}}B-A\sharp_{\frac {3}{4}}B)\geq(2\nu-1) (A\nabla B-A \sharp B), $$
(2.29)
$$ 2(A\nabla_{\frac{3}{4}}B-A\sharp_{\frac {3}{4}}B)\geq(A\nabla B-A\sharp B), $$
(2.30)
Here, we should remind the readers that the reverse of Theorem 2.6 is stronger than (2.13) and only holds for \(0\leq\nu\leq\frac{1}{4}\) and \(\frac{3}{4}\leq\nu\leq1\) in Zhao and Li [8]. We also can see Zhao and Wu [6] for a different method to get Theorem 2.6.
Next, we will present refinements of Young’s inequalities (2.14) and (2.15) for the Hilbert–Schmidt norm. Firstly, we give their scalar type inequalities. That is to say, for \(0\leq\nu\leq\frac{1}{2}\), we have
where \(\nu\in[0, 1]\), \(r=\min\{\nu, 1-\nu\}\), and \(r_{1}=\min\{2r, 1-2r\}\).
$$ a^{1-v}b^{v}+v(\sqrt{a}-\sqrt{b})^{2}+r_{1} \bigl(\sqrt [4]{ab}-\sqrt{a}\bigr)^{2} \leq(1-v)a+vb, $$
(2.31)
Substituting a by \(a^{2}\) and b by \(b^{2}\) in (2.31) respectively, we get
that is,
Similarly, for \(\frac{1}{2}\leq\nu\leq1\), we have
$$ \bigl(a^{1-v}b^{v}\bigr)^{2}+v(a-b)^{2}+r_{1}( \sqrt{ab}-a)^{2} \leq(1-v)a^{2}+vb^{2}, $$
(2.32)
$$ \bigl(a^{1-v}b^{v}\bigr)^{2}+v(a-b)^{2}+r_{1}( \sqrt {ab}-a)^{2}+2v(1-v)ab \leq\bigl((1-v)a+vb\bigr)^{2}. $$
(2.33)
$$ \bigl(a^{1-v}b^{v}\bigr)^{2}+(1-v) (a-b)^{2}+r_{1}(\sqrt {ab}-b)^{2}+2v(1-v)ab \leq \bigl((1-v)a+vb\bigr)^{2}. $$
(2.34)
Corollary 2.8
Suppose
\(A,B,X\in M_{n}\)
such that
A, B
are two positive definite matrices, for
\(0\leq\nu\leq\frac{1}{2}\), we have
for
\(\frac{1}{2}\leq\nu\leq1\), we have
$$ \begin{aligned}[b] & \bigl\Vert A^{1-v}XB^{v} \bigr\Vert _{2}^{2}+r \Vert AX-XB \Vert _{2}^{2}+r_{1} \bigl\Vert A^{\frac {1}{2}}XB^{\frac{1}{2}}-AX \bigr\Vert _{2}^{2}+2v(1-v) \bigl\Vert A^{\frac {1}{2}}XB^{\frac{1}{2}} \bigr\Vert _{2}^{2} \\ &\quad\leq \bigl\Vert (1-v)AX-vXB \bigr\Vert _{2}^{2}; \end{aligned} $$
(2.35)
$$ \begin{aligned}[b] & \bigl\Vert A^{1-v}XB^{v} \bigr\Vert _{2}^{2}+r \Vert AX-XB \Vert _{2}^{2}+r_{1} \bigl\Vert A^{\frac {1}{2}}XB^{\frac{1}{2}}-XB \bigr\Vert _{2}^{2}+2v(1-v) \bigl\Vert A^{\frac {1}{2}}XB^{\frac{1}{2}} \bigr\Vert _{2}^{2} \\ &\quad\leq \bigl\Vert (1-v)AX-vXB \bigr\Vert _{2}^{2}. \end{aligned} $$
(2.36)
3 Discussion
In the theory of operators, the operator means and operator inequalities are two key concepts. An effective method to study operators is to find some refinements among some operator means, and these inequalities are usually based on scalars or matrices.
4 Conclusion
In order to better estimate the Heron mean, a refinement inequality about the classical interpolation between arithmetic mean and geometric mean by Heron mean is obtained, which is also applicable to establishing the inequalities for operators and matrices. Next an operator version refinement inequality about Young’s inequality is also established, which is a generalization on the results obtained previously by Kittaneh and Manasrah [5]. It is worth noting that the inequality mentioned can also give the refinement inequality about Young’s inequality, which was presented by Zhao and Wu [6].
Competing interests
The authors declare that they have no competing interest.
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