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\begin{document}

\title[New Bellman-type inequalities]
{New Bellman-type inequalities}

\author[R. Pashaei]{Ronak Pashaei}
\address{Department of Mathematics, Faculty of Science, Central Tehran Branch,
Islamic Azad University, P. O. Box 13185/768, Tehran, Iran.}
\email{pashai.ronak@gmail.com}
%    Information for  the second author
\author[M. S. Asgari]{Mohammad Sadegh Asgari}
\address{Department of Mathematics, Faculty of Science, Central Tehran Branch,
Islamic Azad University, P. O. Box 13185/768, Tehran, Iran.}
\email{msasgari@yahoo.com ; moh.asgari@iauctb.ac.ir}
\thanks{The authors' work was partially supported by the Central Tehran Branch
of Islamic Azad University.}

%    Information for the third author
%\author[]{}
%\address{}
%\email{}
%\thanks{Support information for the second author.}
%\author{}
%\address{}
%\email{}
%\date{January 1, 2001 and, in revised form, June 22, 2001.}
%\dedicatory{This paper is dedicated to our advisors.}
%    General info
\subjclass[2000]{Primary 47A63, Secondary 26A51, 26D15, 26B25, 39B62.}
\keywords{Operator order, Jensen's inequality, convex functions, norm inequality, positive operators.}

\begin{abstract}
In this paper, we present a general version of operator Bellman inequality. Also, the refinement of inequality due to J. Aujla
and F. Silva for the convex functions is given as well.
\end{abstract}

\maketitle

\section{Introduction and Preliminaries}

Let $\mathcal{H}$ be a Hilbert space equipped with the inner product $\left\langle \cdot,\cdot \right\rangle $, and let
$\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators acting on $\mathcal{H}$ equipped with
the operator norm
\begin{align*} \left\| A \right\|=\sup\big\{\|Ax\|:\text{ }x\in \mathcal{H},\text{ }\left\langle x, x \right\rangle
=1 \big\}.
\end{align*}
Let the symbol $I$ stand for the identity operator on $\mathcal{H}$. An operator $A$ is said to be positive (denoted by $0\le A$)
if $0\le \left\langle Ax,x \right\rangle $ for all $x\in \mathcal{H}$, and also an operator $A$ is said to be strictly positive
(denoted by $0<A$) if $A$ is positive and invertible. For two self-ajoint operators $A,B\in\mathcal{B}(\mathcal{H})$,
we write $A\le B$ if $0\le B-A$.  A linear map $\Phi:\mathcal{B}(\mathcal{H})\to \mathcal{B}(\mathcal{H})$ is said to
be positive if $0\le \Phi \left( A \right)$ when $0\le A$. If, in addition, $\Phi(I)=I,$ it is said to be unital.\par 
For any strictly positive operator $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ and $v\in \left[ 0,1 \right]$, we write
\begin{align*} &A{{\nabla }_{v}}B=(1-v)A+vB,\\& A{{\sharp}_{v}}B={{A}^{\frac{1}{2}}}{{
\big({{A}^{-\frac{1}{2}}}B{{A}^{-\frac{1}{2}}}\big)}^{v}}{{A}^{\frac{1}{2}}},\\&
A{{!}_{v}}B={{\left( {{A}^{-1}}{{\nabla }_{v}}{{B}^{-1}} \right)}^{-1}}.\end{align*}
For the case $v={1}/{2}\;$, we write $\nabla $, $\sharp$, and $!$, respectively. The weighted operator
arithmetic-geometric-harmonic mean inequality asserts that 
\begin{align*} A{{!}_{v}}B\le A{{\sharp}_{v}}B\le A{{\nabla }_{v}}B,\end{align*}
for any positive operators $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ and any $v\in \left[ 0,1 \right]$.
Various inequalities improving this above inequality have been studied in \cite{ro, moradi2}.\\
Let $\sigma $ be an operator mean in the sense of Kubo and Ando \cite{kubo}.
A mean $\sigma $ is called to be symmetric if $A\sigma B=B\sigma A$. According to the general theory
of operator means \cite{kubo}, $\nabla $ is the biggest and $!$ is the smallest among symmetric means. \par 
A real valued function $f$ defined on an interval $J$ is said to be operator convex (resp. operator concave) if 
\begin{equation}\label{12}
f\left( \left( 1-v \right)A+vB \right)\le \left( \text{resp}\text{. }\ge  \right)\left( 1-v \right)f\left( A \right)+vf\left( B \right),
\end{equation}
 for all self-adjoint operators $A,B$ with spectra in $J$ and all $v\in \left[ 0,1 \right]$. A continuous real valued function $f$ defined on an interval $J$ is called operator monotone (more precisely, operator monotone increasing) if $A\le B$
implies that $f\left( A \right)\le f\left( B \right)$, and operator monotone decreasing if  $A\le B$ implies $f\left( B \right)\le f\left( A \right)$ for all self-adjoint operators $A,B$ with spectra in $J$.\par
The scalar Bellman inequality \cite{bellman} says that if $p$ is a positive integer and $a,b,{{a}_{i}},{{b}_{i}}\left( 1\le i\le n \right)$
are positive real numbers such that $\sum\nolimits_{i=1}^{n}{a_{i}^{p}}\le {{a}^{p}}$ and $\sum\nolimits_{i=1}^{n}{b_{i}^{p}}
\le {{b}^{p}}$, then
\begin{align*} {{\Big( {{a}^{p}}-\sum\limits_{i=1}^{n}{t_{i}^{p}} \Big)}^{\frac{1}{p}}}+
{{\Big( {{b}^{p}}-\sum\limits_{i=1}^{n}{s_{i}^{p}} \Big)}^{\frac{1}{p}}}\le 
{{\Big( {{\left(a+b \right)}^{p}}-\sum\limits_{k=1}^{n}{{{\left( {{a}_{i}}+
{{b}_{i}} \right)}^{p}}} \Big)}^{\frac{1}{p}}}.\end{align*}
A multiplicative analogue of this inequality is due to Acz\'el \cite{aczel}.  In 1956, he proved that
\begin{align*} \Big( a_{1}^{2}-\sum\limits_{i=2}^{n}{a_{i}^{2}} \Big)
\Big( b_{1}^{2}-\sum\limits_{i=2}^{n}{b_{i}^{2}} \Big)\le {{\Big( {{a}_{1}}{{b}_{1}}-\sum
\limits_{i=2}^{n}{{{a}_{i}}{{b}_{i}}} \Big)^2}},\end{align*}
where ${{a}_{i}},{{b}_{i}}\left( 1\le i\le n \right)$ are positive real numbers such that 
\begin{align*} a_{1}^{2}-\sum\nolimits_{i=2}^{n}{a_{i}^{2}}>0,\hspace{5mm}\text{or}\hspace{5mm} b_{1}^{2}-\sum\nolimits_{i=2}^{n}{b_{i}^{2}}>0. \end{align*} 
The operator theory related to inequalities in Hilbert space is studied in many papers. In \cite[Corollary 2.2]{3}, Morassaei et al.
showed the following non-commutative version of classical Bellman inequality:
\begin{equation}\label{23}
\Phi \big( {{\left( I-A \right)}^{\frac{1}{p}}}{{\nabla }_{v }}{{\left( I-B \right)}^{\frac{1}{p}}} \big)\le {{\left( \Phi
\left( I-A{{\nabla }_{v }}B \right) \right)}^{\frac{1}{p}}},
\end{equation}
where $0\le v \le 1,~p>1$ and $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ are two contractions (i.e., $0<A,B\le I$) and
$\Phi:\mathcal{B}\left( \mathcal{H} \right)\to \mathcal{B}\left( \mathcal{H} \right)$ is a unital positive linear map. We refer
the reader to \cite{moradi} for some fresh discussion of Bellman inequality.\par 
In this paper, we extend inequality \eqref{23} to two arbitrary operator means $\sigma$ and $\tau $. Naturally, this
generalization imposes additional constant. Some related inequalities for operator concave functions are also presented.

\section{Main Results}

In order to prove our desired  inequalities, we need the following lemmas. The first lemma is the celebrated
Choi-Davis-Jensen inequality (see, e.g. \cite[Theorem 1.20]{pecaric}).
\begin{lemma}
Let $A\in \mathcal{B}\left( \mathcal{H} \right)$ be a self-adjoint operator with the spectra on the interval $J$, $f:J\to \mathbb{R}$
be an operator concave, then for any unital positive linear mapping $\Phi$,
\begin{equation}\label{1}
\Phi \left( f\left( A \right) \right)\le f\left( \Phi \left( A \right) \right),
\end{equation}
or equivalently,
\begin{equation}\label{2}
\Phi \left( A\sigma B \right)\le \Phi \left( A \right)\sigma \Phi \left( B \right),
\end{equation}
where $\sigma $ is an arbitrary operator mean in the Kubo-Ando sense.
\end{lemma}
The second lemma contains a multiplicative reverse for the weighted arithmetic-geometric-harmonic mean inequality.
See \cite{fms}, \cite{gumus}, and references therein.
\begin{lemma}\label{4}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $mI\le A,B\le MI$ for some scalars
$0<m<M$, then \begin{align*} \frac{m{{\sharp}_{\lambda}}M}{m{{\nabla }_{\lambda}}M}A{{\nabla }_{v}}B\le
A{{\sharp}_{v}}B\le \frac{m{{\nabla }_{\lambda}}M}{m{{\sharp}_{\lambda}}M}A{{!}_{v}}B,\end{align*}
where $\lambda =\min \left\{ v,1-v \right\}$.
\end{lemma}
The following lemma contains a well known property for operator monotone increasing (resp. decreasing). However,
we give the proof to keep the present paper self-contained. 
\begin{lemma}\label{5}
Let $f:(0,\infty)\to (0,\infty)$ be a given function and let $\alpha \ge 1$. \begin{enumerate} \item[(a)] If $f$ is operator
monotone, then \begin{equation}\label{needed_1_lemma_oper} f\left( \alpha t \right)\le \alpha f\left( t \right).
\end{equation}
\item[(b)] If $g$ is operator monotone decreasing, then \begin{equation*} g\left( \alpha t \right)\ge \frac{1}{\alpha }g
\left( t \right). \end{equation*}\end{enumerate}\end{lemma}
\begin{proof}
(a) Since $f\left( t \right)$ is operator monotone, then $\frac{t}{f\left(t\right)}$ is operator monotone too
\cite[Corollary 1.14]{pecaric}. Hence for $\alpha \ge 1$, \begin{equation*}
\frac{t}{f\left( t \right)}\le \frac{\alpha t}{f\left( \alpha t \right)}%
\text{ }\Rightarrow \text{ }\alpha f\left( t \right) \ge f\left( \alpha t \right).\end{equation*}
(b) If $g$ is operator monotone decreasing, then ${1}/{g}\;$ is operator monotone. Applying inequality
\eqref{needed_1_lemma_oper} for $f={1}/{g}\;$, we infer that \begin{equation*}
g{{\left( \alpha t \right)}^{-1}}\le \alpha g{{\left( t \right)}^{-1}}\text{}\Rightarrow \text{}g
\left( \alpha t \right)\ge \frac{1}{\alpha }g\left(t\right). \end{equation*}
\end{proof}
It should be mentioned here that part (b) of Lemma \ref{5} will not be used in this paper. It has been given for the
sake of completeness. On making use of the above lemmas, we reach the next result.
\begin{theorem}\label{6}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $mI\le A,B\le MI$  for some scalars
$0<m<M$, and $\tau$, $\sigma$ be two arbitrary operator means, and let $f:(0,\infty)\to (0,\infty)$ be an operator
monotone, then for any unital positive linear mapping $\Phi$,
\begin{align*} \Phi \big(f( A)\tau f(B)\big)\le\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2 
f\big(\Phi(A\sigma B)\big),\end{align*}
where $\lambda =\min \left\{ v,1-v \right\}$ and $0\le v\le 1$.
\end{theorem}
\begin{proof}
Since $f$ is operator concave, by \cite[Corollary 1.12]{pecaric}, Lemma \ref{4} and Lemma \ref{5} (a) we have
\begin{align*} f(A){{\nabla }_{v}}f(B)&\le f(A{{\nabla }_{v}}B) 
\le f\Big(\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2A{{!}_{v}}B\Big)\\&
\le\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2f\left( A{{!}_{v}}B \right)
\le\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2f\left( A\sigma B \right).  
\end{align*}
Applying positive linear mapping $\Phi$, we get
\begin{equation}\label{3}
\begin{align*} \Phi\big(f(A)\big)\tau \Phi\big(f(B)\big)&\le \Phi\big(f(A)\big)
{{\nabla }_{v}} \Phi\big(f(B)\big) \\& =\Phi\big(f(A){{\nabla }_{v}}f(B)\big)
\\& \le \Phi\Big(\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
f(A\sigma B) \Big) \\& =\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
 \Phi\big(f(A\sigma B)\big).  \end{align*}
\end{equation}
By \eqref{1}, we know that
\begin{align*} \Phi\big(f(A\sigma B)\big)\le f\big(\Phi(A\sigma B)\big)\end{align*}
so,
\begin{align*}\Phi\big(f(A)\big)\tau \Phi\big(f(B)\big)\le 
\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
f\big(\Phi(A\sigma B)\big). \end{align*}
Now, by the inequality \eqref{2} we obtain
\begin{align*}\Phi\big(f(A)\tau f(B)\big)&\le\Phi\big(f(A)\big)\tau \Phi\big(f(B)\big)
\\& \le\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
f\big(\Phi(A\sigma B)\big).\end{align*}
This completes the proof of the theorem.
\end{proof}
Now  we  are  in  a  position  to  present  our  promised  generalization  of  \eqref{23}.
\begin{corollary}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $mI\le A,B\le MI$  for some scalars
$0<m<M$, and $\tau$, $\sigma$ be two arbitrary operator means, then for any unital positive linear mapping $\Phi$
and $p>1$,
\begin{align*} \Phi\big((I-A)^{\frac{1}{p}}\tau (I-B)^{\frac{1}{p}}\big)\le 
\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2\big(\Phi(I-A\sigma B)\big)^{\frac{1}{p}},
\end{align*}
where $\lambda =\min \left\{ v,1-v \right\}$ and $0\le v\le 1$.
\end{corollary}
\begin{proof} By \cite[Corollary 1.16]{pecaric} the function $f\left( t \right)={{t}^{r}}$ is operator concave on
$\left( 0,\infty  \right)$ if $0\le r\le 1$. Thus, the function $f\left( t \right)={{\left( 1-t \right)}^{r}}$ on $\left( 0,1 \right)$
is operator concave, for $0\le r\le 1$. Now, since $\Phi $ is unital, it follows from Theorem \ref{6} that
\begin{align*} \Phi\big((I-A)^{\frac{1}{p}}\tau (I-B)^{\frac{1}{p}}\big)&\le 
\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2\big(I-\Phi(A\sigma B)
\big)^{\frac{1}{p}}\\&=\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
\big(\Phi(I)-\Phi(A\sigma B)\big)^{\frac{1}{p}}\\& =\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}
\Big)^2\big(\Phi(I-A\sigma B)\big)^{\frac{1}{p}}.  
\end{align*}
Hence the proof is completed.
\end{proof}
\begin{corollary}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $mI\le A,B\le MI$  for some scalars
$0<m<M$, and $\tau$, $\sigma$ be two arbitrary operator means, then for any unital positive linear mapping $\Phi$,
\begin{align*}\Phi\big(\log A\;\tau\; \log B\big)\le \Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2
\log\Phi(A\sigma B),\end{align*}
where $\lambda =\min \left\{ v,1-v \right\}$ and $0\le v\le 1$.
\end{corollary}
\begin{proof}
Since the function $f\left( t \right)=\log t$ is operator concave on $\left( 0,\infty  \right)$ \cite[Example 1.7]{pecaric},
we infer the desired result from Theorem \ref{6}.
\end{proof}
In \cite[Theorem 2.2]{2}, Moslehian noted the following inequalities for non-negative operator decreasing and operator
concave $f$,
\begin{equation}\label{13} f\left( A \right)\tau f\left( B \right)\le f\left( A\tau B \right). \end{equation}
In particular, he obtained if $p,q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then 
\begin{equation}\label{25} f\left( {{A}^{p}} \right){{\sharp}_{\frac{1}{q}}}f\left( {{B}^{q}} \right)\le 
f\big( {{A}^{p}}{{\sharp}_{\frac{1}{q}}}{{B}^{q}} \big), \end{equation}
and
\begin{equation}\label{26}
\big\langle f(A^p)x, x \big\rangle^{\frac{1}{p}}\big\langle f(B^q)x, x \big\rangle^{\frac{1}{q}}
\le\big\langle f(A^p\sharp_{\frac{1}{q}}B^q)x, x \big\rangle, \end{equation}
for all $x\in \mathcal{H}$. As it is mentioned in \cite[Theorem 2.1]{ando}, the condition operator decreasing is equivalent to
operator convexity. Thus, the inequalities \eqref{13}, \eqref{25}, and \eqref{26} are valid just for the trivial case
$f\left( t \right)=t$. We  conclude  this  paper  with  the  following considerable  generalization  of \eqref{25}.
\begin{theorem}\label{9}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $mI\le A,B\le MI$  for some scalars
$0<m<M$, and $\tau$, $\sigma$ be two arbitrary operator means, and let $f:(0,\infty)\to (0,\infty)$ be an operator
concave, then \begin{equation}\label{14}
f(A)\tau f(B)\le\Big(\frac{m\nabla_\lambda M}{m\sharp_\lambda M}\Big)^2f(A\sigma B).
\end{equation}
In particular,
\begin{equation}\label{15}
f(A)\sharp_v f(B)\le \frac{m\nabla_\lambda M}{m\sharp_\lambda M}f(A\sharp_v B), \end{equation}
where $\lambda =\min \left\{ v,1-v \right\}$ and $0\le v\le 1$.
\end{theorem}\begin{proof}
Letting $\Phi \left( T \right)=T$ for any $T\in \mathcal{B}\left( \mathcal{H} \right)$ in Theorem \ref{6}, we get
\eqref{14}. To prove \eqref{15}, since $f$ is concave, by the arithmetic-geometric mean inequality and Lemma \ref{4},
Lemma \ref{5}(a) we have
\begin{align*}f(A)\sharp_v f(B)&\le f(A)\nabla_v f(B)\le f( A\nabla_v B)\\ 
 & \le f\left( \frac{m{{\nabla }_{\lambda}}M}{m{{\sharp}_{\lambda}}M}A{{\sharp}_{v}}B \right)\le 
 \frac{m{{\nabla }_{\lambda}}M}{m{{\sharp}_{\lambda}}M}f\left( A{{\sharp}_{v}}B \right),
\end{align*}
as desired.
\end{proof}
\begin{remark}
A result similar to \eqref{15} can be found in \cite[Theorem 2]{gumus}.
\end{remark}
\begin{corollary}\label{7}
Let $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ be positive operators such that $m\le A^{p},B^{q}\le M$  for
some scalars $0<m<M$, $1/p+1/q=1,\; p, q>1$. Let $\tau$, $\sigma $ be two arbitrary operator means,
and let $f:(0,\infty)\to (0,\infty)$ be an operator decreasing and operator concave, then
\begin{equation}\label{16}
f(A^p)\sharp_{\frac{1}{q}}f(B^q)\le\frac{m\nabla_\gamma M}{m\sharp_\gamma M}
f(A^p\sharp_{\frac{1}{q}}B^q).\end{equation}
In particular, for any vector $x \in \mathcal{H}$
\begin{equation}\label{17}
\big\langle f(A^p)x, x \big\rangle^{\frac{1}{p}}\big\langle f(B^q)x, x \big\rangle^{\frac{1}{q}}
\le \frac{m\nabla_\gamma M}{m\sharp_\gamma M}\big\langle f(A^p\sharp_v B^q)x, x \big\rangle,
\end{equation}
where $\gamma =\min \left\{ {1}/{p}\;,{1}/{q}\; \right\}$.
\end{corollary}\begin{proof}
By replacing $v={1}/{q}\;$, $A={{A}^{p}}$, and $B={{B}^{q}}$ in Theorem \ref{9}, we get the inequality \eqref{16}.
To prove \eqref{17}, we have \begin{align*} \big\langle f(A^p)x, x \big\rangle^{\frac{1}{p}}
\big\langle f(B^q)x, x \big\rangle^{\frac{1}{q}} &\le \frac{1}{p}\big\langle f(A^p)x, x \big\rangle
+\frac{1}{q}\big\langle f(B^q)x, x \big\rangle\\& \le \frac{m\nabla_\gamma M}{m\sharp_\gamma M}
\big\langle f(A^p\sharp_{\frac{1}{q}}B^q)x, x \big\rangle, \end{align*}
where we have used weighted arithmetic-geometric mean inequality.
\end{proof}
\begin{remark}
It follows from McCarthy inequality \cite[Theorem 1.4]{pecaric},
\begin{align*}\big\langle f(A^p)^{\frac{1}{p}}x, x \big\rangle\big\langle f(B^q)^{\frac{1}{q}}x, x \big\rangle
&\le \big\langle f(A^p)x, x \big\rangle^{\frac{1}{p}}\big\langle f(B^q)x, x \big\rangle^{\frac{1}{q}} \\ 
 & \le \frac{m\nabla_\gamma M}{m\sharp_\gamma M}\big\langle f(A^p\sharp_{\frac{1}{q}}B^q)x, x \big\rangle   
\end{align*}
or equivalently,
\begin{align*}\big\|f(A^p)^{\frac{1}{2p}}x \big\|\big\|f(B^q)^{\frac{1}{2q}}x \big\|\le 
\sqrt{\frac{m\nabla_\gamma M}{m\sharp_\gamma M}}\Big\|f(A^p\sharp_{\frac{1}{q}}B^q)^{\frac{1}{2}}x \Big\|.
\end{align*}\end{remark}
To prove \eqref{23},  Morassaei et al. \cite{3} have proved that if $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ are two
positive operators and  $f:(0,\infty)\to (0,\infty)$ is an operator convex  (resp. concave), then for any unital positive linear
mapping $\Phi$ and $0\le v\le 1$
\begin{align*} f\left( \Phi \left( \left( 1-v \right)A+vB \right) \right)\le \left( \text{resp}\text{. }\ge  \right)
\Phi \left( \left( 1-v \right)f\left( A \right)+vf\left( B \right) \right).\end{align*}
Trivially, the above inequality can be considered as an extension of  inequality \eqref{12} (choose $\Phi \left( T \right)=T$
for any $T\in \mathcal{B}\left( \mathcal{H} \right)$). Another generalization of the inequality \eqref{12} for operators in $\mathcal{B}\left( \mathcal{H} \right)$ (see, \cite{5}) asserts that if $A,B\in \mathcal{B}\left( \mathcal{H} \right)$ are
positive, and $f$ is a non-negative convex (not necessary operator convex) function on $\left( 0,\infty  \right)$, then
\begin{equation}\label{11}
\left\| f\left( \left( 1-v \right)A+vB \right) \right\|\le \left\| \left( 1-v \right)f\left( A \right)+vf\left( B \right) \right\|.
\end{equation}
We refer the reader to \cite{moradi1, 12} as a sample of the extensive use of this inequality.\par 
In the next theorem, we improve \eqref{11}.  In order to reach our purpose, we need the following lemma
\cite[Theorem 2.2]{kosem}.
\begin{lemma}\label{10}
Let $A,B \in \mathcal{B}(\mathcal{H})$ be two positive operators, and
let $f$ be a non-negative convex function on $\left[ 0,\infty  \right)$ with $f\left( 0 \right)=0$. Then,
\begin{align*} \left\| f\left( A \right)+f\left( B \right) \right\|\le \left\| f\left( A+B \right) \right\|. \end{align*}
\end{lemma}
Before stating our theorem we recall that if $T\in \mathcal{B}\left( \mathcal{H} \right)$ is a positive operator, and if $f$
is a non-negative increasing function on $\left[ 0,\infty  \right)$, then 
\begin{align*} \left\| f\left( T \right) \right\|=f\left( \left\| T \right\| \right). \end{align*}
\begin{theorem}
Let $A,B \in \mathcal{B}(\mathcal{H})$ be two positive operators, and let $f(t)$ be a non-negative function on
$\left[ 0,\infty  \right)$ such that $g\left( t \right)=f\left( \sqrt{t} \right)$ is convex and $g\left( 0 \right)=0$.
Then for any $0\le v \le 1$,
\begin{footnotesize}\begin{align*}
\bigg\| f\big((1-v)A+vB \big)+f\bigg(\sqrt{2\lambda\Big(\dfrac{A^2+B^2}{2}-\big(\dfrac{A+B}{2}\big)^2\Big)}\;
\bigg)\bigg\|\le \big\|(1-v)f(A)+vf(B)\big\|,
\end{align*}\end{footnotesize}\noindent
where $\lambda=\min \left\{ v,1-v \right\}$.
\end{theorem}
\begin{proof}
First we assume $0\le v\le {1}/{2}\;$. Since $h\left( t \right)={{t}^{2}}$ is a convex function on $\left( 0,\infty  \right)$
\cite[Corollary 1.16]{pecaric}, we have
\begin{align*} (1-v)A^2+vB^2&-2\lambda\Big(\frac{A^2+B^2}{2}-\big(\frac{A+B}{2}\big)^2\Big)\\& 
\ge(1-2v)A^2+2v\big(\frac{A+B}{2}\big)^2 \\& \ge\Big((1-2v)A+2v\big(\frac{A+B}{2}\big) \Big)^2 \\&
=\big((1-v)A+vB\big)^2. \end{align*}
A similar argument for $1/2\le v\le 1$ implies the above inequality. Thus we have
\begin{small}\begin{equation}\label{8}
\big((1-v)A+vB\big)^2+2\lambda\Big(\frac{A^2+B^2}{2}-\big(\frac{A+B}{2}\big)^2\Big)\le (1-v)A^2+vB^2,
\end{equation}\end{small}\noindent
where $\lambda=\min \left\{ v,1-v \right\}$. Obviously, the operator inequality \eqref{8} implies the following norm inequality
\begin{small}\begin{align*}
\Big\|\big((1-v)A+vB\big)^2+2\lambda\Big(\frac{A^2+B^2}{2}-\big(\frac{A+B}{2}\big)^2\Big)\Big\|\le\big\|(1-v)A^2+vB^2
\big\|.\end{align*}\end{small}\noindent
Since $g$ is non-negative and convex on $\left[ 0,\infty  \right)$, it follows that $g$ is increasing. Now, we obtain
\begin{align*}& \bigg\| f\big((1-v)A+vB \big)+f\bigg(\sqrt{2\lambda\Big(\dfrac{A^2+B^2}{2}-\big(\dfrac{A+B}{2}
\big)^2\Big)}\;\bigg)\bigg\| \\=&\bigg\| g\big(\big((1-v)A+vB\big)^2\big)+g\bigg(2\lambda\Big(\dfrac{A^2+B^2}{2}
-\big(\dfrac{A+B}{2}\big)^2\Big)\bigg)\bigg\| \\ \le &\bigg\| g\bigg(\big((1-v)A+vB\big)^2+2\lambda\Big(
\dfrac{A^2+B^2}{2}-\big(\dfrac{A+B}{2}\big)^2\Big)\bigg)\bigg\|\\=&g\bigg(\bigg\|\big((1-v)A+vB\big)^2+
2\lambda\Big(\dfrac{A^2+B^2}{2}-\big(\dfrac{A+B}{2}\big)^2\Big)\bigg\|\bigg)\\\le &g\big(\big\|(1-v)A^2+vB^2
\big\|\big)=\big\|g\big((1-v)A^2+vB^2\big)\big\|\\\le & \big\|(1-v)g(A^2)+vg(B^2)\big\|=\big\|(1-v)f(A)+vf(B)\big\|,  
\end{align*}
where the first and the last equality follows from the assumption $g\left( t \right)=f\left( \sqrt{t} \right)$, the first inequality
obtained from Lemma \ref{10}, and the inequality \eqref{11} implies the last inequality.
\end{proof}
Applying the above theorem to the power function, we get:
\begin{corollary}
Let $A,B \in \mathcal{B}(\mathcal{H})$ be two positive operators. Then for any $0\le v \le 1$, and $p \ge 2$
\begin{footnotesize}\begin{align*}
\bigg\|\big((1-v)A+vB \big)^p+\bigg(\sqrt{2\lambda\Big(\dfrac{A^2+B^2}{2}-\big(\dfrac{A+B}{2}\big)^2\Big)}\;
\bigg)^p\bigg\|\le \big\|(1-v)A^p+vB^p\big\|,
\end{align*}\end{footnotesize}\noindent
where $\lambda=\min \left\{ v,1-v \right\}$.
\end{corollary}

%\section*{Acknowledgements}
%The authors' work was partially supported by the Central Tehran Branch of Islamic Azad University.
%\bibliographystyle{amsplain}
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