1 Introduction
Let \(M_{n} \) be the space of \(n\times n\) complex matrices. Let \(\Vert \cdot \Vert \) denote any unitarily invariant norm on \(M_{n} \). So, \(\Vert {UAV} \Vert =\Vert A \Vert \) for all \(A\in M_{n} \) and for all unitary matrices \(U,V\in M_{n} \). For \(A= [ {a_{ij} } ]\in M_{n} \), the Hilbert-Schmidt norm and the trace norm of A are defined by \(\Vert A \Vert _{2} =\sqrt{\sum_{j=1}^{n} {s_{j}^{2} ( A )} } \), \(\Vert A \Vert _{1} =\sum_{j=1}^{n} {s_{j} ( A )} \), respectively, where \(s_{i} ( A ) \) (\(i=1,\ldots,n\)) are the singular values of A with \(s_{1} ( A )\ge\cdots\ge s_{n} (A )\), which are the eigenvalues of the positive semidefinite matrix \(\vert A \vert = ( {AA^{\ast}} )^{\frac{1}{2}}\), arranged in decreasing order and repeated according to multiplicity.
The classical Young inequality says that if \(a,b\ge0\) and \(0\le v\le1\), then
with equality if and only if \(a=b\).
$$ a^{v}b^{1-v}\le va+ ( {1-v} )b $$
(1)
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Kittaneh and Manasrah [1] obtained an improvement of inequality (1) which can be stated as follows:
where \(r_{0} =\min \{ {v,1-v} \}\).
$$ a^{v}b^{1-v}+r_{0} ( {\sqrt{a} -\sqrt{b} } )^{2}\le va+ ( {1-v} )b, $$
(2)
Recently, Burqan and Khandaqji [2] gave the following reverses of the scalar Young type inequalities:
and
$$ v^{2}a^{2}+ ( {1-v} )^{2}b^{2} \le ( {1-v} )^{2} ( {a-b} )^{2}+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v}, \quad 0\le v\le\frac{1}{2}, $$
(3)
$$ v^{2}a^{2}+ ( {1-v} )^{2}b^{2} \le v^{2} ( {a-b} )^{2}+ ( {va} )^{2v}b^{2-2v},\quad \frac{1}{2}\le v\le1. $$
(4)
A matrix Young inequality, proved in [3], says that if \(A,B\in M_{n} \) are positive semidefinite, then
for \(j=1,\ldots,n\).
$$s_{j} \bigl( {A^{v}B^{1-v}} \bigr)\le s_{j} \bigl( {vA+ ( {1-v} )B} \bigr) $$
Based on the reverses of the scalar Young type inequalities (3) and (4), Burqan and Khandaqji proved the following in [2] if \(A,B,X\in M_{n}\) such that A and B are positive semidefinite. If \(0\le v\le\frac{1}{2}\), then
If \(\frac{1}{2}\le v\le1\), then
$$\begin{aligned} &\bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} \\ &\quad\le ( {1-v} )^{2}\Vert {AX-XB} \Vert _{2}^{2} +2v ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{\frac{1}{2}}} \bigr\Vert _{2}^{2} + ( {1-v} )^{2 ( {1-v} )} \bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} . \end{aligned}$$
(5)
$$ \bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} \le v^{2}\Vert {AX-XB} \Vert _{2}^{2} +2v ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{{\frac{1}{2}}}} \bigr\Vert _{2}^{2} +v^{2v} \bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} . $$
(6)
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At the same time, Burqan and Khandaqji proved the following in [2] if \(A,B\in M_{n}\) such that A and B are positive semidefinite. If \(0\le v\le\frac{1}{2}\), then
If \(\frac{1}{2}\le v\le1\), then
$$\begin{aligned} &( {1-v} )^{1-v} \bigl\Vert {A^{v}} \bigr\Vert _{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2} \\ &\quad\ge \sqrt{v^{2}\Vert A \Vert _{2}^{2} + ( {1-v} )^{2}\Vert B \Vert _{2}^{2} - ( {1-v} )^{2} \bigl( {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr)} . \end{aligned}$$
(7)
$$ v^{v} \bigl\Vert {A^{v}} \bigr\Vert _{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2} \ge \sqrt {v^{2}\Vert A \Vert _{2}^{2} + ( {1-v} )^{2}\Vert B \Vert _{2}^{2} -v^{2} \bigl( {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr)} . $$
(8)
2 Reverses of Young type inequalities for scalars
We begin this section with the reverses of Young type inequalities for scalars.
Theorem 1
Let
\(a,b\ge0\). If
\(0\le v\le\frac{1}{2}\), then
where
\(r_{0} =\min \{ {2v,1-2v} \}\).
$$ v^{2}a^{2}+ ( {1-v} )^{2}b^{2}+r_{0} a \bigl( {\sqrt{ ( {1-v} )b} -\sqrt{a} } \bigr)^{2}\le ( {1-v} )^{2} ( {a-b} )^{2}+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v}, $$
(9)
If
\(\frac{1}{2}\le v\le1\), then
where
\(r_{0} =\min \{ {2v-1,2-2v} \}\).
$$ v^{2}a^{2}+ ( {1-v} )^{2}b^{2}+r_{0} b ( {\sqrt{b} -\sqrt{va} } )^{2}\le v^{2} ( {a-b} )^{2}+ ( {va} )^{2v}b^{2-2v}, $$
(10)
Proof
If \(0\le v\le\frac{1}{2}\), then, by inequality (2), we have
and so
If \(\frac{1}{2}\le v\le1\), then
and so
This completes the proof. □
$$\begin{aligned} & ( {1-v} )^{2} ( {a-b} )^{2}-v^{2}a^{2}- ( {1-v} )^{2}b^{2}-r_{0} a \bigl( {\sqrt{ ( {1-v} )b} -\sqrt{a} } \bigr)^{2}+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v} \\ &\quad=a \bigl[ { ( {1-2v} )a+2v ( {1-v} )b} \bigr]-r_{0} a \bigl( { \sqrt{ ( {1-v} )b} -\sqrt{a} } \bigr)^{2}-2 ( {1-v} )ab+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v} \\ &\quad\ge a \bigl\{ {a^{1-2v} \bigl[ { ( {1-v} )b} \bigr]^{2v}} \bigr\} -2 ( {1-v} )ab+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v} \\ &\quad=a^{2-2v} \bigl[ { ( {1-v} )b} \bigr]^{2v}+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v}-2 ( {1-v} )ab \\ &\quad= \bigl[ {a^{1-v} ( {1-v} )^{v}b^{v}-a^{v} ( {1-v} )^{1-v}b^{1-v}} \bigr]^{2}\ge0, \end{aligned}$$
$$v^{2}a^{2}+ ( {1-v} )^{2}b^{2}+r_{0} a \bigl( {\sqrt{ ( {1-v} )b} -\sqrt{a} } \bigr)^{2}\le ( {1-v} )^{2} ( {a-b} )^{2}+a^{2v} \bigl[ { ( {1-v} )b} \bigr]^{2-2v}. $$
$$\begin{aligned} &v^{2} ( {a-b} )^{2}-v^{2}a^{2}- ( {1-v} )^{2}b^{2}-r_{0} b ( {\sqrt{b} -\sqrt{va} } )^{2}+ ( {va} )^{2v}b^{2-2v} \\ &\quad= ( {2v-1} )b^{2}+ ( {2-2v} )vab-r_{0} b ( {\sqrt{b} - \sqrt{va} } )^{2}-2vab+ ( {va} )^{2v}b^{2-2v} \\ &\quad=b \bigl[ { ( {2v-1} )b+ ( {2-2v} )va-r_{0} ( {\sqrt{b} - \sqrt{va} } )^{2}} \bigr]-2vab+ ( {va} )^{2v}b^{2-2v} \\ &\quad\ge b \bigl[ {b^{2v-1} ( {va} )^{2-2v}} \bigr]-2vab+ ( {va} )^{2v}b^{2-2v} \\ &\quad= \bigl[ {b^{v} ( {va} )^{1-v}- ( {va} )^{v}b^{1-v}} \bigr]^{2}\ge0, \end{aligned}$$
$$v^{2}a^{2}+ ( {1-v} )^{2}b^{2}+r_{0} b ( {\sqrt{b} -\sqrt{va} } )^{2}\le v^{2} ( {a-b} )^{2}+ ( {va} )^{2v}b^{2-2v}. $$
3 Reverses of Young type inequalities for matrices
Based on the reverses of the scalar Young type inequalities (9) and (10), we obtain matrix versions of these inequalities.
Theorem 2
Let
\(A,B,X\in M_{n} \)
such that
A
and
B
are positive semidefinite. If
\(0\le v\le\frac{1}{2}\), then
where
\(r_{0} =\min \{ {2v,1-2v} \}\).
$$\begin{aligned} & \bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} +r_{0} \bigl[ { ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{{\frac{1}{2}}}} \bigr\Vert _{2}^{2} + \Vert {AX} \Vert _{2}^{2} -2\sqrt{1-v} \bigl\Vert {A^{\frac{3}{4}}XB^{{\frac{1}{4}}}} \bigr\Vert _{2}^{2} } \bigr] \\ &\quad\le ( {1-v} )^{2}\Vert {AX-XB} \Vert _{2}^{2} +2v ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{\frac{1}{2}}} \bigr\Vert _{2}^{2} + ( {1-v} )^{2 ( {1-v} )} \bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} , \end{aligned}$$
(11)
If
\(\frac{1}{2}\le v\le1\), then
where
\(r_{0} =\min \{ {2v-1,2-2v} \}\).
$$ \begin{aligned}[b] & \bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} +r_{0} \bigl[ {v \bigl\Vert {A^{\frac{1}{2}}XB^{{\frac{1}{2}}}} \bigr\Vert _{2}^{2} + \Vert {XB} \Vert _{2}^{2} -2\sqrt{v} \bigl\Vert {A^{\frac{1}{4}}XB^{{\frac{3}{4}}}} \bigr\Vert _{2}^{2} } \bigr]\\ &\quad \le v^{2}\Vert {AX-XB} \Vert _{2}^{2} +2v ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{{\frac{1}{2}}}} \bigr\Vert _{2}^{2} +v^{2v} \bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} , \end{aligned} $$
(12)
Proof
Since every positive semidefinite matrix is unitarily diagonalizable, it follows that there are unitary matrices \(U,V\in M_{n} \) such that \(A=UDU^{\ast}\) and \(B=VEV^{\ast}\), where
Let \(Y=U^{\ast}XV= [ {y_{ij} } ]\). Then
and
If \(0\le v\le\frac{1}{2}\), by inequality (9), we have
and so
If \(\frac{1}{2}\le v\le1\), then by inequality (10) and the same method above, we have inequality (12). This completes the proof. □
$$D=\operatorname{diag} ( {\lambda_{1} ,\ldots,\lambda_{n} } ),\qquad E=\operatorname{diag} ( {\mu_{1} ,\ldots,\mu_{n} } ),\quad \mbox{and } \lambda_{i} ,\mu _{i} \ge 0, i=1, \ldots,n. $$
$$\begin{aligned}& vAX+ ( {1-v} )XB=U \bigl( {vDY+ ( {1-v} )YE} \bigr)V^{\ast}=U \bigl[{ \bigl( {v\lambda_{i} + ( {1-v} )\mu_{j} } \bigr)y_{ij} } \bigr]V^{\ast}, \\& AX-XB=U \bigl[ { ( {\lambda_{i} -\mu_{j} } )y_{ij} } \bigr]V^{\ast},\qquad A^{\frac{1}{2}}XB^{\frac{1}{2}}=U \bigl[ {\lambda_{i}^{\frac{1}{2}} \mu _{j}^{\frac{1}{2}} y_{ij} } \bigr]V^{\ast}, \end{aligned}$$
$$A^{v}XB^{1-v}=U \bigl[ {\lambda_{i}^{v} \mu_{j}^{1-v} y_{ij} } \bigr]V^{\ast}. $$
$$\begin{aligned} &\bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} \\ &\quad=\sum_{i,j=1}^{n} { \bigl( {v \lambda_{i} + ( {1-v} )\mu _{j} } \bigr)} ^{2} \vert {y_{ij} } \vert ^{2} \\ &\quad\le ( {1-v} )^{2}\sum_{i,j=1}^{n} { ( {\lambda_{i} -\mu _{j} } )^{2}} \vert {y_{ij} } \vert ^{2}+ ( {1-v} )^{2(1-v)}\sum _{i,j=1}^{n} { \bigl( {\lambda_{i}^{v} \mu _{j}^{1-v} } \bigr)^{2}} \vert {y_{ij} } \vert ^{2} \\ &\qquad{}-r_{0} \sum_{i,j=1}^{n} \lambda_{i} \bigl( {\sqrt{ ( {1-v} )\mu_{j} } -\sqrt{ \lambda_{i} } } \bigr) ^{2}\vert {y_{ij} } \vert ^{2}+2v ( {1-v} )\sum_{i,j=1}^{n} { \bigl( {\lambda _{i}^{\frac{1}{2}} \mu_{j}^{\frac{1}{2}} } \bigr)^{2}} \vert {y_{ij} } \vert ^{2} \\ &\quad= ( {1-v} )^{2}\sum_{i,j=1}^{n} { ( {\lambda_{i} -\mu _{j} } )^{2}} \vert {y_{ij} } \vert ^{2}+ ( {1-v} )^{2(1-v)}\sum _{i,j=1}^{n} { \bigl( {\lambda_{i}^{v} \mu _{j}^{1-v} } \bigr)^{2}} \vert {y_{ij} } \vert ^{2} \\ &\qquad{}+ ( {2v-r_{0} } ) ( {1-v} )\sum _{i,j=1}^{n} { \bigl( \lambda_{i}^{\frac{1}{2}} \mu_{j}^{\frac{1}{2}} \bigr)^{2}} \vert y_{ij} \vert ^{2}\\ &\qquad{}-r_{0} \sum _{i,j=1}^{n} {\lambda_{i}^{2} \vert {y_{ij} } \vert ^{2}+2r_{0} \sqrt{ ( {1-v} )} } \sum_{i,j=1}^{n} { \bigl( { \lambda_{i}^{\frac{3}{4}} \mu_{j}^{\frac{1}{4}} } \bigr)^{2}} \vert {y_{ij} } \vert ^{2} \\ &\quad= ( {1-v} )^{2}\Vert {AX-XB} \Vert _{2}^{2} + ( {1-v} )^{2 ( {1-v} )}\bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} \\ &\qquad{}+ ( {2v-r_{0} } ) ( {1-v} )\bigl\Vert {A^{\frac{1}{2}}XB^{\frac{1}{2}}} \bigr\Vert _{2}^{2} -r_{0} \Vert {AX} \Vert _{2}^{2} +2r_{0} \sqrt{1-v} \bigl\Vert {A^{\frac{3}{4}}XB^{\frac{1}{4}}} \bigr\Vert _{2}^{2} , \end{aligned}$$
$$\begin{aligned} & \bigl\Vert {vAX+ ( {1-v} )XB} \bigr\Vert _{2}^{2} +r_{0} \bigl[ { ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{{\frac{1}{2}}}} \bigr\Vert _{2}^{2} +\Vert {AX} \Vert _{2}^{2} -2\sqrt{1-v} \bigl\Vert {A^{\frac{3}{4}}XB^{{\frac{1}{4}}}} \bigr\Vert _{2}^{2} } \bigr] \\ &\quad\le ( {1-v} )^{2}\Vert {AX-XB} \Vert _{2}^{2} +2v ( {1-v} ) \bigl\Vert {A^{\frac{1}{2}}XB^{\frac{1}{2}}} \bigr\Vert _{2}^{2} + ( {1-v} )^{2 ( {1-v} )} \bigl\Vert {A^{v}XB^{1-v}} \bigr\Vert _{2}^{2} . \end{aligned}$$
Remark 2
In the end, we present two new inequalities, by means of inequalities (9) and (10). To do this, we need the following lemmas.
Lemma 1
(Cauchy-Schwarz inequality) [10]
Let
\(a_{i} \ge0\), \(b_{i} \ge0\), for
\(i=1,2,\ldots,n\), then
$$\sum_{i=1}^{n} {a_{i} b_{i} } \le \Biggl( {\sum_{i=1}^{n} {a_{i}^{2} } } \Biggr)^{\frac{1}{2}} \Biggl( {\sum _{i=1}^{n} {b_{i}^{2} } } \Biggr)^{\frac{1}{2}}. $$
Lemma 2
[10]
Let
\(A,B\in M_{n} \), then
$$\sum_{j=1}^{n} {s_{j} ( {AB} )} \le\sum_{j=1}^{n} {s_{j} ( A )} s_{j} ( B ). $$
Theorem 3
Let
\(A,B\in M_{n} \)
such that
A
and
B
are positive semidefinite. If
\(0\le v\le\frac{1}{2}\), then
where
\(r_{0} =\min \{ {2v,1-2v} \}\), \(M_{1} =r_{0} [ { ( {1-v} )\Vert {AB} \Vert _{1} +\Vert A \Vert _{2}^{2} -2\sqrt{1-v} \Vert {A^{\frac{3}{2}}} \Vert _{1} \Vert {B^{\frac{1}{2}}} \Vert _{1} } ]\).
$$\begin{aligned} &( {1-v} )^{1-v} \bigl\Vert {A^{v}} \bigr\Vert _{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2} \\ &\quad\ge\sqrt{v^{2}\Vert A \Vert _{2}^{2} + ( {1-v} )^{2}\Vert B \Vert _{2}^{2} - ( {1-v} )^{2} \bigl( {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr)+M_{1} } , \end{aligned}$$
(13)
If
\(\frac{1}{2}\le v\le1\), then
where
\(r_{0} =\min \{ {2v-1,2-2v} \}\), \(M_{2} =r_{0} [ {v\Vert {AB} \Vert _{1} +\Vert B \Vert _{2}^{2} -2\sqrt{v} \Vert {A^{\frac{1}{2}}} \Vert _{1} \Vert {B^{\frac{3}{2}}} \Vert _{1} } ]\).
$$ v^{v} \bigl\Vert {A^{v}} \bigr\Vert _{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2} \ge \sqrt {v^{2}\Vert A \Vert _{2}^{2} + ( {1-v} )^{2}\Vert B \Vert _{2}^{2} -v^{2} \bigl( {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr)+M_{2} } , $$
(14)
Proof
If \(0\le v\le\frac{1}{2}\), then using Lemma 1, Lemma 2, and inequality (9), we have
Thus
If \(\frac{1}{2}\le v\le1\), then by inequality (10) and the same method above, we have inequality (14). This completes the proof. □
$$\begin{aligned} &\operatorname{tr} \bigl( {v^{2}A^{2}+ ( {1-v} )^{2}B^{2}} \bigr) \\ &\quad=v^{2}\operatorname{tr} A^{2}+ ( {1-v} )^{2} \operatorname{tr}B^{2} \\ &\quad=\sum_{j=1}^{n} { \bigl( {v^{2}s_{j}^{2} ( A )+ ( {1-v} )^{2}s_{j}^{2} ( B )} \bigr)} \\ &\quad\le ( {1-v} )^{2} \Biggl[ {\sum_{j=1}^{n} {s_{j}^{2} ( A )+\sum_{j=1}^{n} {s_{j}^{2} ( B )-2\sum_{j=1}^{n} {s_{j} ( A )s_{j} ( B )} } } } \Biggr] \\ &\qquad{}+\sum_{j=1}^{n} { ( {1-v} )^{2 ( {1-v} )} \bigl[ {s_{j} \bigl( {A^{v}} \bigr)s_{j} \bigl( {B^{1-v}} \bigr)} \bigr]^{2}-r_{0} } \sum_{j=1}^{n} {s_{j} ( A ) \bigl[ {\sqrt{ ( {1-v} )s_{j} ( B )} -\sqrt{s_{j} ( A )} } \bigr]^{2}} \\ &\quad\le ( {1-v} )^{2} \Biggl[ {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2 \sum_{j=1}^{n} {s_{j} ( {AB} )} } \Biggr]+ ( {1-v} )^{2 ( {1-v} )} \Biggl[ {\sum_{j=1}^{n} {s_{j} \bigl( {A^{v}} \bigr)s_{j} \bigl( {B^{1-v}} \bigr)} } \Biggr]^{2} \\ &\qquad{}-r_{0} \Biggl[ {(1-v)\sum_{j=1}^{n} {s_{j} ( A )s_{j} ( B )+\sum_{j=1}^{n} {s_{j}^{2} ( A )-2\sqrt{1-v} \Biggl( {\sum _{j=1}^{n} {s_{j}^{\frac{3}{2}} ( A )s_{j}^{\frac{1}{2}} ( B )} } \Biggr)} } } \Biggr] \\ &\quad\le ( {1-v} )^{2} \bigl[ {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr]+ ( {1-v} )^{2 ( {1-v} )} \Biggl[ {\sum _{j=1}^{n} {s_{j}^{2} \bigl( {A^{v}} \bigr)\sum_{j=1}^{n} {s_{j}^{2} \bigl( {B^{1-v}} \bigr)} } } \Biggr] \\ &\qquad{}-r_{0} \Biggl[ {(1-v)\sum_{j=1}^{n} {s_{j} ( {AB} )+\Vert A \Vert _{2}^{2} -2 \sqrt{1-v} \Biggl( {\sum_{j=1}^{n} {s_{j}^{\frac{3}{4}} ( A )s_{j}^{\frac{1}{4}} ( B )} } \Biggr)^{2}} } \Biggr] \\ &\quad\le ( {1-v} )^{2} \bigl[ {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr]+ ( {1-v} )^{2 ( {1-v} )}\bigl\Vert {A^{v}} \bigr\Vert _{2}^{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2}^{2} \\ &\qquad{}-r_{0} \Biggl[ {(1-v)\Vert {AB} \Vert _{1} + \Vert A \Vert _{2}^{2} -2\sqrt {1-v} \Biggl( {\sum _{j=1}^{n} {s_{j}^{\frac{3}{2}} ( A )} \sum_{j=1}^{n} {s_{j}^{\frac{1}{2}} ( B )} } \Biggr)}\Biggr]. \end{aligned}$$
$$\begin{aligned} v^{2}\Vert A \Vert _{2}^{2} + ( {1-v} )^{2}\Vert B \Vert _{2}^{2} \le{}& ( {1-v} )^{2} \bigl( {\Vert A \Vert _{2}^{2} +\Vert B \Vert _{2}^{2} -2\Vert {AB} \Vert _{1} } \bigr)+ ( {1-v} )^{2 ( {1-v} )} \bigl\Vert {A^{v}} \bigr\Vert _{2}^{2} \bigl\Vert {B^{1-v}} \bigr\Vert _{2}^{2} \\ &{}-r_{0} \bigl[ { ( {1-v} )\Vert {AB} \Vert _{1} + \Vert A \Vert _{2}^{2} -2\sqrt{1-v} \bigl\Vert {A^{\frac{3}{2}}} \bigr\Vert _{1} \bigl\Vert {B^{\frac{1}{2}}} \bigr\Vert _{1} } \bigr]. \end{aligned}$$
Acknowledgements
The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.