\section{Introduction}

We expect from some physical principles that the spacetime will approach to an asymptotically stationary black hole state  at late times \cite{Wald:1999vt}. Considering the initial Cauchy data with a horizon $\sigma$, if the cosmic censorship conjecture is valid \cite{Penrose:1969pc,Hawking:1970zqf,4}, i.e. the future singularity is surrounded by the event horizon and the spacetime tends to be a stationary state, then the black hole no-hair or uniqueness theorem \cite{5} will imply the final state should be the Kerr metric \cite{Chrusciel:2008js}. The surface area of the Kerr horizon is
\be\label{kerr}
A_h=8\pi M_{\rm B} \big(M_{\rm B}+\sqrt{M_{\rm B}^2-a^2}\big)\le 16\pi M_{\rm B}^2\,,
\ee
where $M_{\rm B}$, $J=a M_{\rm B}$ and $A_h$ are the total mass (the asymptotic value of the Bondi mass), total angular momentum and the area of the event horizon respectively. Following these three points: 1.~since the gravitational waves carry positive energy \cite{Bondi:1962px,Sachs:1962wk}, the Bondi mass can never be larger than the Arnowitt-Deser-Misner (ADM) mass $M$; 2.~from the second law of black hole thermodynamics, the area of the event horizon can not decrease \cite{Schon:1981vd,Bardeen:1973gs}; 3.~any apparent horizon must be hidden inside the event horizon \cite{Hawking:1973uf}, and using (\ref{kerr}), one can write the following inequality
\be\label{penroseineq}
2{M}\geq\sqrt{\fft{A[\sigma]}{4\pi}}\,.
\ee
Here $A[\sigma]$ is the minimal area enclosing the apparent horizon $\sigma$. This is the famous ``Penrose inequality'', and it was proposed by Penrose in the early seventies as a thought experiment for the ``cosmic censorship conjecture.''  (See, e.g.~\cite{Bray:2003ns, Mars:2009cj}.) The .Penrose inequality also has a rigid statement: the saturation appears possible only in the Schwarzschild spacetime \cite{Ben-Dov:2004lmn}.

Although the mathematical expression of \eqref{penroseineq} is very ``simple,'' its proof is a very challenging problem in mathematical Relativity and there is hitherto no general proof. There are however two notable special cases where rigourous proofs have been given. The first covers the general static geometries by assuming the weak energy condition (WEC) \cite{14,15}, and the second deals with dynamical spacetimes but with spherical symmetry \cite{Hayward:1994bu} by assuming the dominant energy condition (DEC). This inequality has also been generalized to asymptotically anti-de Sitter spacetimes (AdS) \cite{Itkin:2011ph,Husain:2017cmj,Engelhardt:2019btp,Xiao:2022obq}. Generally, either the DEC or WEC is assumed to be the necessary requirement in the proof of the static or dynamic spacetimes \cite{Mars:2009cj}. However, it has been recently found that null energy condition (NEC) is actually enough to ensure Penrose inequality, at least for the spherically-symmetric and static black holes\cite{Engelhardt:2019btp,Xiao:2022obq,Yang:2022yye}.

The Penrose inequality involves the apparent horizon of the initial data set, and it does not require that the spacetime is or will finally settle into a stationary black hole. It gives a relation between the energy and the size of the space it occupies. If we focus on a stationary black hole, the area of horizon stands for the Bekenstein-Hawking entropy of the system via
\begin{equation}\label{defS}
	S=\fft14 A\,.
\end{equation}
The Penrose inequality then becomes an entropy bound for a system of given total energy.

In thermodynamical systems, in addition to the entropy, the temperature $T$ is also a very important quantity. Furthermore, in Einstein gravity, the mass, entropy and temperature are all geometric quantities solely determined by the metric of the black hole. It is thus interesting to investigate whether we could obtain any new bounds that involve all these quantities. The fact that the product $T S$ has the same dimension of energy suggests that it may modify the mass bound in the inequality. Ref.~\cite{Yang:2022yye} made a first attempt towards this issue and gave a new inequality for spherically-symmetric and static black holes by assuming only NEC:
\begin{equation}\label{EETineq}
	3{M}\geq\sqrt{\frac{S}{\pi}}+2TS\,.
\end{equation}
As was clarified in Ref.~\cite{Yang:2022yye}, this inequality is independent of the Penrose inequality and the saturation appears to be possible also only by the Schwarzschild black hole.

However, there is something unsatisfactory about these two inequalities since both of them are too slack for extremal or near extremal black holes.  For example, the extremal (zero temperature) Reissner-Nordstr\"{o}m (RN) black hole has $M_{\rm ext}= \sqrt{S/\pi}$. It is twice the saturation value of the Penrose inequality and thrice the value of the inequality \eqref{EETineq}. An interesting question arises: could we find a new lower bound for the total energy such that it is tighter than inequalities \eqref{EETineq} and \eqref{penroseineq}? It can be argued that the essence of the Penrose inequality is about how the appropriate energy-momentum tensor of the minimally-coupled matter constrains the spacetime geometry.
It is thus reasonable to expect that we should be able to utilize all the geometric information to obtain a new bound that is more refined and stringent. Considering the fact that the Maxwell field satisfies all the energy conditions, it is thus not entirely unreasonable to expect that the RN black hole will have the lowest mass for given temperature and entropy. Furthermore, since the vacuum itself satisfies all the energy conditions, we also expect that the new bound should be saturated by the vacuum solution, i.e.~the Schwarzschild black hole.

We observe from the RN black hole that
\be\label{rnmts}
M=\frac{4 Q^2+r_+^2}{2 r_+}\,, \qquad TS=\frac{r_+^2-4Q^2}{4 r_+}\,.
\ee
Here $r_+=\sqrt{S/\pi}$ is the radius of the event horizon. It is easy to check $M=r_+-2TS$. Based on our arguments, we propose a new Penrose-like inequality for a static black hole
\be\label{CI}
{M} \ge\sqrt{\frac{S}{\pi}}-2TS\,.
\ee
(As we shall see later, the positivity of the right-hand side is ensured by the WEC.) For the Schwarzschild black hole, its Smarr relation $M=2TS$ implies that the above inequality is also saturated. For the extremal case, we have $T=0$ and the above inequality gives us a tighter bound of the ADM mass compared to the Penrose inequality.

We shall examine a variety of explicit examples to support inequality~\eqref{CI} in section 2. We also give an interesting counterexample. From these we formulate a conjecture of the energy conditions that guarantee the inequality \eqref{CI}. We then give a general proof in section 3 for the spherically-symmetric case. We find that the required energy condition is the combination of NEC and the trace energy condition (TEC), which contains the WEC. This is not a particularly strong requirement for such a tighter inequality. In doing so, we review four different ``quasi-local masses'' for the most-frequently used four different local energy conditions, i.e.~NEC, WEC, the strong energy condition (SEC) and DEC, and illustrate how every one of them will give us a different lower bound on ADM mass and also a different proof of the ``positive mass theorem.'' We then construct a new quasi-local mass that gives us \eqref{CI} under ``NEC$+$TEC.'' In section 4, we generalize our inequality to higher dimensions, or to include rotations or the cosmological constant. We conclude our paper in section 5.

\section{Examples}\label{Ex}

In this section we study a variety of examples in four dimensions to illustrate the validity of our inequality (\ref{CI}). We also provide an interesting counterexample. We focus on the spherically-symmetric and static black holes, of which the metrics have the following general form
\be\label{metric0}
\mathrm{d}s^2 =-f (r)\mathrm{d}t^2 +\frac{\mathrm{d}r^2}{f(r)} + \rho(r)^2 \mathrm{d}\Omega^2\,,
\ee
Here $\mathrm{d}\Omega^2$ is the standard metric of the round unit 2-sphere. We denote $r_+$ to be the location of the event horizon where $f(r_+)=0$. The radius of the horizon is then $\rho(r_+)$. Defining
\be\label{xe}
X=M + 2 T S - \rho(r_+)\,,
\ee
we would like to verify whether we have $X\ge 0$ for a variety of black holes. Since our inequalities involve only the geometric quantities, we shall present only the metrics of all the examples.


\subsection{Born-Infeld}

The spherically-symmetric and asymptotically-flat solution in Einstein-Born-Infeld (EBI) theory was given in \cite{Breton:2002td,Li:2016nll}. The solution has $\rho=r$ and the blackening function $f$ is
\be
f= 1 - \fft{2M}{r} + \ft16 b^2 \Big(r^2 - \sqrt{\fft{Q^2}{b^2} + r^4}\Big)
+ \fft{Q^2}{3r^2}\, {}_2F_1(\ft14,\ft12;\ft54;-\ft{Q^2}{b^2 r^4})\,.
\ee
Here $b>0$ is the Born-Infeld coupling constant, $(M,Q)$ are the mass and electric charge respectively. It is straightforward to obtain
\be
T S=\frac{1}{8} r_+ \left(-b \sqrt{b^2 r_+^4+Q^2}+b^2 r_+^2+2\right).
\ee
Therefore we have
\be
\fft{6X}{b^2 r_+^3}=q^2 \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-q^2\right)-2 \sqrt{q^2+1}+2\ge 0\,,
\ee
where $q\ge 0$ is the dimensionless parameter defined by $Q=q b r_+^2$. The inequality is saturated by $q=0$, which can be achieved either by $Q=0$, corresponding to the Schwarzschild black hole, or by $b\rightarrow \infty$, corresponding to the RN black hole.

\subsection{Quasitopological electromagnetism}

Considering the theory in which the quasitoplogical electromagnetism is minimally coupled to gravity, one can find exact solutions of general spherically-symmetric and static dyonic black holes \cite{Liu:2019rib} that has $\rho=0r$ and
\be
f=1 - \fft{2M}{r} + \fft{p^2}{r^2} +
\fft{q^2}{r^2} {}_2F_1\Big(\ft14,1;\ft54;-\ft{4\alpha p^2}{r^4}\Big),
\ee
where the three integration constants $(M, q, p)$ are associated with the mass, electric and magnetic charges. Also the black hole satisfies at
least the dominant energy condition once the coupling constant $\alpha$ is positive. This solution gives us
\be
T S=\frac{4 \alpha  p^4+r_+^4 \left(p^2+q^2\right)-4 \alpha  p^2 r_+^2-r_+^6}{4 r_+\left(4 \alpha  p^2+r_+^4\right)}.
\ee
Therefore, the following inequality can be obtained
\be
X=\frac{q^2 \left(\left(4 \tilde p^2+1\right) \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};-4 \tilde p^2\right)-1\right)}{2 \left(4 \tilde p^2+1\right) r_+}\ge 0\,,
\ee
where $\tilde p = \alpha p/r_+^2>0$. The saturation happens when $q=0$, or $p=0$ or $\alpha=0$, corresponding to the magnetic, electric or dyonic RN black holes respectively.

\subsection{The STU supergravity model}

The $D = 4$, $N = 2$ STU  supergravity model coupled to three vector multiples admits 4-charge
black hole solutions, given by (e.g.~\cite{Cvetic:1999xp})
\be
f=\fft{f_0}{\sqrt{H_1 H_2 H_3 H_4}}\,,\quad
f_0= 1 - \fft{\mu}{r}\,,\quad H_i = 1 + \fft{q_i}{r}\,,\quad
\rho=r (H_1 H_2 H_3 H_4)^{\fft14}\,,
\ee
where the solution is specified by five integration constants, namely the electric charge parameters $q_i$ and the blackening parameter $\mu=r_+$. We have
\be
M=\ft12\mu + \ft14 (q_1 + q_2 + q_3 + q_4)\,,\qquad T S = \ft1{4} r_+\,.
\ee
Thus we can easily verify
\be
X=\ft14 (\tilde q_1 + \tilde q_2 + \tilde q_3 + \tilde q_4 -
4 (\tilde q_1 \tilde q_2 \tilde q_3 \tilde q_4)^{\fft14}\ge 0\,,
\ee
where $\tilde q_i = q_i + r_+>0$.

\subsection{An Einstein-scalar theory}

Here is a scalar-hairy black hole in Einstein gravity minimally coupled to a scalar with a specific scalar potential as \cite{Zhang:2014sta}. The metric is specified by
\be
f=1 + \alpha^2 \Big(r(q+r) \log \Big(\fft{r+q}{r}\Big) - \fft12
q(q+2r)\Big)\,,\qquad \rho = \sqrt{r(r+q)}\,.
\ee
In this case, we have
\be
M=\ft1{12} \alpha^2 q^3\,,\qquad T S = \ft14\alpha^2
r_+ \left(q+r_+\right) \left(\left(q+2 r_+\right) \log \left(\frac{q+r_+}{r_+}\right)-2 q\right)\,.
\ee
Thus Eq.~(\ref{xe}) leads to  ($x=r_+/q>0$)
\bea
\fft{X}{q}&=&-x-\sqrt{x} \sqrt{x+1}-\frac{2}{3 \left(-2 x+2 (x+1) x \log \left(\frac{1}{x}+1\right)-1\right)}-\frac{1}{2}\nn\\
&=&
\left\{
\begin{array}{ll}
	\fft16 -\sqrt{x}, &\qquad x\rightarrow 0\,; \\
	\fft{1}{40x}, &\qquad  x\rightarrow \infty\,.
\end{array}
\right.
\eea
To be more specific, $X$ is a monotonously-decreasing function from 1/6 to 0.

\subsection{Bardeen: a counterexample}

As our final example here, the Bardeen spacetime metric can be written as \cite{bardeen}
\be
f= 1 - \fft{2M r^2}{(r^2 + q^2)^{\fft32}}\,,\qquad \rho=r\,.
\ee
in which $q$ is the magnetic charge. For fixed $q$, the extremal black hole has
\be
M_{\rm ext} = \ft34\sqrt3 q\,,\qquad r_+^{\rm ext} = \sqrt2 q\,.
\ee
Since $M_{\rm ext}< r_+^{\rm ext}$, we see this example violates the inequality (\ref{CI}). Since  the Bardeen black hole satisfies WEC, but not DEC nor TEC, it shows that the WEC, which is sufficient for the Penrose inequality, is not enough to guarantee our inequality \eqref{CI}.

\subsection{A conjecture}

While our examples are far from exhaustive, they represent different classes of matter fields. Based on these examples, we formulate the following conjecture:

\textit{For a 4-dimensional asymptotically flat and static black hole in Einstein gravity with minimally coupled matter, if (1) the NEC and TEC are both satisfied and (2) the cross-section of event horizon has $S^2$ topology, then the inequality \eqref{CI} is true}.

In the next section, we shall give a proof for the special spherically-symmetric case. Before we do that, we would like first to make a few comments.
\begin{itemize}

\item Our conjecture requires NEC+TEC. This combination implies the WEC, but not the DEC or SEC, which can however be consistently further imposed without upsetting the NEC+TEC.

\item The Penrose inequality turns out to be a corollary of our inequality~\eqref{CI}. To see this, we note that by assuming WEC, Refs.~\cite{Visser:1992qh,Yang:2020ark} have established an inequality
\begin{equation}\label{bdonts2}
	4TS\leq \sqrt{\fft{S}{\pi}}\,.
\end{equation}
This inequality not only ensures that the right-hand side of \eqref{CI} is positive, but also allows us to deduce the Penrose inequality from \eqref{CI}, namely
\be\label{CI2}
2M\ge 2\Big(\sqrt{\fft{S}{\pi}} - 2TS\Big)\geq \sqrt{\fft{S}{\pi}}=\sqrt{\fft{A}{4\pi}}\,.
\ee

\item Thirdly, since the temperature is always nonnegative for a static black hole, we have following inequality for extremal black holes,
\begin{equation}\label{zertot1}
	M_{\rm ext} \geq\sqrt{\fft{S}{4\pi}}=\sqrt{\fft{A}{4\pi}}\,.
\end{equation}
This gives us a tighter bound on the ADM mass for extremal black hole and cannot be improved any more in the sense that it could be saturated by the extremal RN black hole.

\item Our inequality \eqref{CI} is tighter than the Penrose inequality and the recently proposed \eqref{EETineq} is slaker, but it is interesting to observe that the sum of these two inequalities gives precisely the Penrose inequality.

\end{itemize}

\section{The Proof}

\subsection{The setup}

After examining the explicit examples, we now give a proof for a special class of black holes that are {\it static and spherically symmetric.} The most general such metrics take the form
\be\label{metric}
\mathrm{d}s^2 =-f (r) e^{-\chi(r)} \mathrm{d}t^2 +\frac{\mathrm{d}r^2}{f(r)} +r^2 \mathrm{d}\Omega^2.
\ee
The asymptotic flatness requires that the functions $f$ and $\chi$ have the following falloff behaviors near the infinity
\begin{equation}\label{asymfchi}
	f(r)=1-2M/r+\cdots\,,\qquad \lim_{r\rightarrow\infty}r\,\chi(r)=0\,,
\end{equation}
where $M$ is the ADM mass. Denoting $r_+$ as the radius of the event horizon, the temperature $T$ and entropy $S$ are
\be\label{ET}
T=\frac{f'(r_+)e^{-\chi(r_+)/2}}{4\pi}\,,\qquad S=\pi r_+^2\,.
\ee
For the matter energy momentum tensor
$
{T^\mu}_\nu=\text{diag}\{-\rho(r), p_{r}(r), p_{{T}}(r), p_{{T}} (r)\},
$ and the metric (\ref{metric}), the Einstein's equation leads to three ordinary differential equations
\begin{align}\label{Ein}
	f'&=\frac{1- 8 \pi r^2 \rho-f(r)} {r}\,,\qquad \chi'=\frac{-8\pi r (\rho +p_{r})}{f}\, ,\nonumber\\
	p'_r&=\frac{\rho-3p_r+4p_T}{2r} -\frac{(\rho+p_r)(1+8\pi p_r r^2)}{2 f r}\,.
\end{align}
In the next we shall use these equations to prove the inequality (\ref{CI}). In fact, from the second equation above, we already see that the NEC ensures that $\chi'\leq0$. Since the asymptotic flatness requires $\chi(\infty)=0$, we conclude that the NEC forces $\chi\geq0$ outside of the horizon.

\subsection{Quasi-local masses and proof of inequality}

Our main tool is a new quasi-local mass. Before we present it, we first give a brief review on some quasi-local masses widely used in the proofs of inequalities involving black hole mass. For the spherically-symmetric and static configurations, there exists four known quasi-local masses associated with the NEC, WEC, SEC, DEC. These quasi-local masses are nondecreasing under the corresponding energy conditions and will give us different lower bounds on ADM mass, as well as different proofs on the ``positive mass theorem'' of black holes.

The first one is the famous ``Hawking-Geroch mass''~\cite{Hawking2,Geroch}, which plays the crucial role in the proofs of ``Penrose inequality''. For black holes of the metric \eqref{metric}, it has a rather simple form
\begin{align}\label{defm1}
	m_w(r)=\frac{r}2(1-f)\,.
\end{align}
Evaluating the Hawking-Geroch mass on the horizon and at infinity, one obtains
\begin{align}\label{defm1a}
	m_w(r_+)=\fft12 r_+\,,\qquad m_w(\infty)={M}\,.
\end{align}
Using the first equation in \eqref{Ein}, one finds
\begin{equation}\label{defm1b}
	m'_w=4 \pi r^2 \rho\,.
\end{equation}
We see that the WEC ensures the nondecreasing of Hawking-Geroch mass, from which the Penrose inequality arises naturally and hence the positive mass theorem for spherically-symmetric and static black holes is established.

Since here we consider the static black hole, the event horizon is a null Killing horizon and there exists a corresponding time-like Killing vector $\xi^\mu$ outside the black hole. Our the second quasi-local mass is the Komar mass, which in general is defined by the Komar integration over a closed 2-surface $\mathcal{S}$
\begin{equation}\label{defm2} m_s(\mathcal{S})=-\frac1{8\pi}\oint_{\mathcal{S}}(\mathrm{d}\xi)_{\mu\nu}\mathrm{d}S^{\mu\nu}\,.
\end{equation}
Specifying into the spherically-symmetric case under the coordinates gauge~\eqref{metric}, with an equal-$r$ sphere $\mathcal{S}$ , we have
\begin{equation}\label{defm2a}
	m_s(r)=\frac12r^2e^{\chi/2}(fe^{-\chi})'\,.
\end{equation}
Using the Komar mass we in fact can easily obtain a positive mass theorem for general static and asymptotically-flat black holes, not only for the special spherically-symmetric case. Denoting $\Sigma$ to be a 3-dimensional hypersurface outside the event horizon with a boundary $H$ at Killing horizon and a boundary $\mathcal{S}_\infty$ at the spatial infinity. We choose $\Sigma$ to be orthogonal to the Killing vector $\xi^\mu$. We have
\begin{equation}\label{defm2b}
	m_s(\mathcal{S}_\infty)=M\,,\qquad m_s(H)=\fft12 TS\,,
\end{equation}
and
\begin{equation}\label{defm2d} m_s(\mathcal{S}_\infty)-m_s(H)=-\frac1{8\pi}\oint_{\partial\Sigma}
(\mathrm{d}\xi)_{\mu\nu}\mathrm{d}S^{\mu\nu}=2\int_{\Sigma}(T_{\mu\nu}-g_{\mu\nu}T/2)\xi^\mu n^\mu\mathrm{d} V\,.
\end{equation}
Here $n^\mu$ is the unit normal vector of $\Sigma$ and $\xi^\mu\parallel n^\mu$ with $\xi^\mu n_\mu<0$. From this we can easy see that the SEC can ensure the nonnegativity of $m_s(\mathcal{S}_\infty)-m_s(H)$. Thus, we see following inequality under the assumption of SEC
\begin{equation}\label{inequalitydec}
	M\geq2TS\,.
\end{equation}
This gives us a proof of the ``positive mass theorem'' under the assumption of SEC. (It is worth commenting here that our inequality \eqref{CI}, together with \eqref{bdonts2}, imply that
\eqref{inequalitydec} can also be satisfied with NEC+TEC, which is independent of SEC.)


The third quasi-local mass is very similar to Hawking-Georch mass, and it is defined by
\begin{equation}\label{modifhawingm1}
	m_{d}(r) =\frac{r}2(1 - e^{-\chi/2} f)\,.
\end{equation}
This and Hawking-Georch mass have the same values on the event horizon and at infinity. However, their derivatives with respective to $r$ are different. Using both the first and second equations in \eqref{Ein}, we obtain
\begin{equation}\label{modfihawkingm2}
	m_d' =\frac12(1 - e^{-\chi/2}) + 2\pi r^2 e^{-\chi/2}(\rho-p_r)\,.
\end{equation}
We see that the DEC can ensure $m_d'\geq0$ and therefore it gives us a different proof on Penrose inequality, as well as the ``positive mass theorem.''

The fourth quasi-local mass was recently proposed in Refs.~\cite{Xiao:2022obq,Yang:2022yye}. Although it currently applies only to the spherically-symmetric and static black holes, the result is surprising since it establishes the Penrose inequality and the ``positive mass theorem'' by assuming NEC only. Specially, it is
\begin{equation}\label{defmnec}
	m_n(r)=\frac{r^4e^{\chi/2}}6\left(\frac{fe^{-\chi}}{r^2}\right)'+\frac{r}3\,.
\end{equation}
Evaluating this on the horizon and at infinity give the following results
\begin{equation}\label{defmnec2}
	m_n(r_+)=\frac{2TS}{3}+\frac{r_+}3\,,\qquad m_n(\infty)=M\,.
\end{equation}
Using all the three equations in \eqref{Ein}, one finds
\begin{equation}\label{defmnec3}
	m_n'=\frac{8\pi}3 e^{-\chi/2}r^2(\rho+p_T)+\frac13[1-e^{-\chi(r)/2}]\,.
\end{equation}
It follows that NEC can ensure $m_n$ is nondecreasing and so we obtain the inequality \eqref{EETineq}. Interestingly, using this quasi-local mass one can further prove the Penrose inequality \cite{Xiao:2022obq,Yang:2022yye}.


In this paper, we will propose a new quasi-local mass, which is compatible with trace energy condition and will be used to prove our new inequality~\eqref{CI}. To do so, we define a quasi-local mass that is the most general linear combination of $m_n, m_w, m_s$ and $m_d$ such that
\be\label{atih}
m(\infty) = {M}\,,\qquad m(r_+) = r_+ - 2 TS\,.
\ee
We find that the combination has two free parameters $(\alpha, \beta)$:
\be
m(r) = 2m_w(r) - m_s(r) + \alpha (3m_n(r)-m_s(r) - 2m_w(r)) + 2\beta (m_d - m_w)\,.\label{result}
\ee
It's derivative yields
\be\label{pmf}
m'(r) = \gamma (1-e^{-\fft12\chi}) + (1-\gamma) \pi r^2 \rho (1-e^{-\fft12\chi})
+ 4\pi r^2 e^{-\fft12 \chi} \Big((1-\gamma)\rho - (1+\gamma) p_r - 2p_T\Big)\,.
\ee
Here $\gamma=\alpha+\beta$. For $m'(r) \ge 0$, the first and second terms of the above equation suggest that $\gamma\in [0,1]$ and $\rho \ge 0$ which is WEC. Also the null energy condition implies $\chi'\le 0$, so from the asymptotic flatness we have $\chi\ge 0$ and $0<e^{-\chi(r)/2}\le 1$. If we are only allowed to impose any of the five energy conditions, but nothing more, then it appears that we have to set $\gamma=\alpha+\beta=0$. Therefore, from Eq.~(\ref{pmf}) we have
\be
m'_{\tr}(r)\ge 4 \pi r^2 (\rho-p_r-2p_T) = 4 \pi r^2 (-T),
\ee
in which $T$ denotes the trace of the energy momentum tensor. (There should be no confusion with the temperature.) Imposing TEC ($-T\ge0$) to the above relation we have $m'_{\tr}(r) \ge 0$. We therefore obtain a one-parameter family of quasi-local masses \eqref{result} with $\alpha+\beta=0$ that is nondecreasing under NEC$+$TEC.  Finally, from Eq.~(\ref{atih}) we conclude that the inequality (\ref{CI}) is true when the NEC$+$TEC are satisfied. Note that the WEC will be also satisfied under requirement of NEC+TEC.

\section{Generalizations}

\subsection{To higher dimensions}

Our conjecture and its proof are very narrow, focusing on the spherically-symmetric and static black holes in four dimensions.  Nevertheless, its success prompts us to consider its possible generalizations. The simplest is perhaps to higher dimensions. For static black holes, we expect
\be
M + \fft{D-2}{D-3} T S \ge \fft{D-2}{\pi} \Big(\fft{\Omega_{D-2}}{2^D}\Big)^{\fft1{D-2}}
S^{\fft{D-3}{D-2}}\,,
\ee
where $\Omega_k=2\pi^{\fft{k+1}2}/(\fft12 (k-1))!$ is the volume of the round unit $k$-sphere. This inequality is again saturated by Schwarzschild-Tangherlini and RN-Tangherlini black holes.  For spherically-symmetric black holes, the proof is straightforward.

\subsection{To rotations}

We now consider rotating black holes and for simplicity, we consider only four dimensions. It can be easily checked that the Kerr black hole does not satisfy the inequality \eqref{CI}:
\be
(M + 2 T S)^2 - \fft{S}{\pi}=-a^2<0\,.\label{Kerrvio}
\ee
This perhaps is not surprising since as we explained in the introduction, a tight inequality should utilize all the black hole geometric information, which certainly includes the angular momentum $J$ and its conjugate velocity $\Omega$, when it is rotating. According to the dimensions of these quantities, we can have two simple approaches to modify \eqref{CI}. The first is to take advantage of the fact that $\Omega J$ has the same dimension as the mass, and it is natural to propose
\be
M + 2 T S+ \eta\, \Omega J \ge \sqrt{\fft{S}{\pi}}\,,\label{rotineq}
\ee
for certain order-one purely numerical constant $\eta$. Since we expect that $\Omega J$ is positive, the higher positive $\eta$ value makes the inequality more trivial.  Therefore, the key is to find the minimum value of $\eta$. The examples we shall discuss presently suggest that $\eta \ge 1$.

In the second approach, we note that angular momentum $J$ has the same dimension as the entropy, we can thus modify \eqref{CI} by
\be
M + 2 T S \ge \sqrt{\fft{S}{\pi} - |J|}\,.\label{rotineq2}
\ee
In other words, we make the right-hand side of \eqref{CI} a little smaller. The precise ``$-1$'' factor in front of $|J|$ is determined by requiring that the the inequality is saturated by the extremal Kerr black hole. An inequality established in Ref.~\cite{Dain:2011pi}, namely
\be
\fft{S}{\pi} \ge 2 |J|\,,
\ee
ensures that the right-hand side of \eqref{rotineq2} is always a real and positive number. The full conjecture is thus
\be
M + 2 T S \ge \sqrt{\fft{S}{\pi} - |J|}\,\ge |J|\,.\label{rotineq3}
\ee

\subsubsection{Kerr-Newmann black hole}

We first examine the Kerr-Newmann black hole and its angular velocity, temperature and entropy are \cite{4}
\be
\Omega = \fft{a}{r_+^2 + a^2}\,,\qquad T=\fft{r_+-M}{2\pi(r_+^2 + a^2)}\,,\qquad
S=\pi (r_+^2 + a^2)\,,
\ee
in which the parameters $M, Q$ and $a = J/M$ are the mass, charge and the angular momentum (per mass) respectively. The mass $M$ is related to the horizon location by
$r_+^2 + a^2 + Q^2 -2 M r_+=0.$ It is thus easy to verify that for $\eta=0$, we have \eqref{Kerrvio}, independent of the charge. However, we find that \eqref{rotineq} can be satisfied for $\eta\ge 1$. Specifically, we have
\be
(M + 2 T S + \Omega J)^2 - \fft{S}{\pi}=\frac{2 r_+^2 \left(a^6+3 a^4 Q^2\right)+r_+^4 \left(a^4+4 a^2 Q^2\right)+a^4 \left(a^2+Q^2\right)^2}{4 r_+^2 \left(a^2+r_+^2\right)^2}>0\,.
\ee
To test the sequence of inequalities in \eqref{rotineq3}, we find
\bea
&&(M + 2 T S)^2 - \Big(\fft{S}{\pi}-|J|\Big) = \frac{|a| \left(\left(r_+-a\right){}^2+Q^2\right)}{2 r_+}\ge 0\,,\nn\\
&& \fft{S}{\pi} - 2 |J|= 4 \pi  |a| \left(a^2+r_+^2\right) T+\left(r_+-a\right)^2\ge 0\,.
\eea


\subsubsection{Kerr-Sen black hole}

The Kerr-Sen black hole was generated from the Kerr black hole to carry an electric charge in string theory \cite{Sen:1992ua}.  The solution contains three parameters, the mass $M$, angular momentum $J$ and the electric charge $Q$. The temperature, entropy and angular velocity can be expressed as
\bea\label{kerrsents}
T&=&\frac{\sqrt{\left(2 M^2-Q^2\right)^2-4 J^2}}{4 \pi  M \left(\sqrt{\left(2 M^2-Q^2\right)^2-4 J^2}+2 M^2-Q^2\right)}\,,\nn\\
S &=& 2 \pi  M \left(\sqrt{\left(M-\frac{Q^2}{2 M}\right)^2-\frac{J^2}{M^2}}-\frac{Q^2}{2 M}+M\right),\nn\\
\Omega &=&\frac{J}{2 M^2 \left(\sqrt{\left(M-\frac{Q^2}{2 M}\right)^2-\frac{J^2}{M^2}}-\frac{Q^2}{2 M}+M\right)}\,.
\eea
We find that together with the electric potential $\Phi=Q/(2M)$, the first law $dM=TdS + \Omega dJ + \Phi dQ$ is satisfied. It should be clarified here that the theory considered in \cite{Sen:1992ua} was in the string flame, and the dilaton was not minimally coupled. However, one can perform a conformal transformation such that all the matter fields become minimally coupled. Under this conformal transformation, all the thermodynamic variables and the first law remain the same. Although the expressions in \eqref{kerrsents} are complicated, we actually can prove the inequalities analytically. The parameters must satisfy
\be
M\ge M_{\rm ext} =\sqrt{|J| + \ft12 Q^2}\,.
\ee
For $\eta\ge 1$, we find that the inequality \eqref{rotineq} is satisfied. Specifically, we have
\be
(M + 2 T S + \Omega J)^2 - \fft{S}{\pi} = \fft{\Big(2M^2 + Q^2 - 2 \sqrt{(M^2- \ft12 Q^2)^2 - J^2}\Big)^2}{16M^2} \ge 0\,.
\ee
It is also straightforward to examine the sequence of the inequalities in \eqref{rotineq3}.  We find that
\bea
&&(M + 2 T S)^2 - \Big(\fft{S}{\pi}-|J|\Big) = \fft{1}{4M^2}\Bigg( Q^4 + 2 |J|
\Big(Q^2 + 2(M^2-\ft12 Q^2- |J|)\Big) \Bigg)\ge 0\,,\nn\\
&& \fft{S}{\pi} - 2 |J|= 2(M^2- \ft12 Q^2 - |J|) +
2\sqrt{(M^2 - \ft12 Q^2)^2 - J^2}\ge 0\,.
\eea
Note that the second inequality is saturated by the extremal Kerr-Sen solution.

There are further more complicated and involved rotating black holes, such as the four-charge
Cveti\v c-Youm black hole \cite{Cvetic:1996kv,Chong:2004na}. We shall investigate the inequalities for these black holes in some future studies.

\subsection{To include the cosmological constant}

For simplicity, we shall consider the cosmological constant $\Lambda$ minimally coupled to all fields in the theory. We also assume that the black hole is static and spherically symmetric, and the leading metric falloffs at the asymptotic infinity is associated with the mass term. For the negative cosmological constant, the black hole is asymptotic to AdS. In this case, for given $M$, the horizon becomes smaller and temperature become higher, compared to the case without the cosmological constant. Thus the inequality becomes easier to satisfy, without saturation. This suggests that we have not taken into account of all the geometric information in these AdS black holes. It was proposed that cosmological constant can be viewed as the thermodynamic pressure $P_{\rm th}$ and there is a conjugate thermodynamic volume $V_{\rm th}$
\cite{Kastor:2009wy,Cvetic:2010jb}.  Their product $P_{\rm th} V_{\rm th}$ is again an energy-like quantity.  We can therefore propose
\be
M + 2 T S+ \xi\, P_{\rm th} V_{\rm th} \ge \sqrt{\fft{S}{\pi}}\,,\label{cosmoineq0}
\ee
for some order-one purely numerical constant $\xi$. In the simple case of a minimally coupled cosmological constant $\Lambda$, we have
\be
P_{\rm th}= - \fft{\Lambda}{8\pi}\,,\qquad V_{\rm th} = \fft43 \pi r_+^3\,,
\ee
where $r_+$ is the radius of the black hole horizon. We find that the proposed inequality \eqref{cosmoineq0} can be saturated with $\xi = -4$ by both the Schwarzschild-(A)dS and RN-(A)dS black holes, yielding
\be
M + 2 T S -4 \, P_{\rm th} V_{\rm th} \ge \sqrt{\fft{S}{\pi}}\,.\label{cosmoineq1}
\ee
It is straightforward to establish that the asymptotic (A)dS black holes in Einstein-Born-Infeld and Einstein-quasitopological electromagnetism discussed earlier also satisfy the inequality. It is interesting to note that for negative cosmological constant, the $P_{\rm th} V_{\rm th}$ term tightens the inequality, whilst for positive cosmological constant, it increases the gap for the easier satisfaction of the inequality.

When the cosmological constant couples to scalars non-minimally, e.g.~gauge supergravities, the situation becomes more much complicated, because the definition $V_{\rm th}$ can be ambiguous \cite{Feng:2017wvc}. It is a subject worth further studies.

\section{Conclusion}

Motivated by the observation that the Penrose inequality for extremal black holes have a large gap from saturation, we proposed a tighter bound (\ref{CI}) for static black holes. Our inequality contains all the geometric information associated with the energy-like quantities, including the mass and the product of temperature and entropy. It is therefore a more refined inequality.

We first tested our inequality with a variety of concrete examples of the known black holes in literature, and the results are summarized in the Table~\ref{tab:table1}. We then gave a proof for general spherically-symmetric and static black holes. We combined the known four ``quasi-local" masses and obtained a new one, which allowed us to show that inequality (\ref{CI}) was true when both NEC and TEC were satisfied. We also showed that our inequality~\eqref{CI} was always stronger than the Penrose inequality in the sense that the latter was a corollary of \eqref{CI}. Here we emphasize that the requirement ``NEC+TEC'' is a sufficient condition. As we have summarized in the Table~\ref{tab:table1} that the even though TEC can be violated in the Einstein-scalar model, the inequality \eqref{CI} remains nevertheless true.

\begingroup
\begin{table}
	\begin{center}
		\begin{tabular}{c | c | c| c | c | c|c }
			\textbf{BH/EC} & \textbf{NEC} & \textbf{WEC} & \textbf{DEC} & \textbf{SEC} & \textbf{NEC+TEC}&inequality~\eqref{CI} \\
		
		
			\hline
			\multirow{1}{*}{RN} & True & True & True & True & True&True \\
			\hline
			\multirow{1}{*}{Born-Infeld} & True & True & True & True &True&True
			\\
			\hline
			\multirow{1}{*}{QTE ($\alpha \ge 0)$} & True & True & True & True & True& True
			\\
			\hline
			\multirow{1}{*}{STU} & True & True & True & True & True& True
			\\
			\hline
			\multirow{1}{*}{Pure scalar} & True & False  &  False & False & False& True
			\\
			\hline
			\multirow{1}{*}{Bardeen} & True & True & False & False & False& False
		\end{tabular}
\caption{Summary of the concrete examples}
		\label{tab:table1}
	\end{center}
\end{table}
\endgroup

In addition, we consider generalizations in various directions. The generalization of \eqref{CI} to higher dimensions for spherically-symmetric and static black holes is straightforward. The generalization to include rotation is more subtle. In this case, we need augment the inequality with additional information such as the product of angular momentum and its conjugate angular velocity. We find that the inequality \eqref{rotineq} is satisfied by the Kerr-Newmann and Kerr-Sen black holes provided that $\eta\ge 1$. Alternatively, since the dimensions of $J$ and $S$ are the same, it is also natural to propose the inequality \eqref{rotineq3}. We find that they are satisfied by both the Kerr-Newmann and Kerr-Sen solutions. For the second proposal, the bound is tight in that it is saturated by the extremal Kerr black hole. For asymptotic (A)dS spacetimes, we proposed \eqref{cosmoineq1} that is saturated by both the Schwarzschild-(A)dS or RN-(A)dS black holes. However, whether these inequality can stand against more examples or rigorous proof remains to be investigated.

There is one further issue worth addressing. It is well known that the Penrose inequality is saturated only by the Schwarzschild black hole, which becomes special from the Penrose inequality point of view. Our inequality is saturated by the RN black hole. It is thus intriguing to ask whether this is the only non-vacauum solution that saturates the inequality. To narrow down the statement further, is the extremal RN black hole is the only extremal one that saturates the bound \eqref{zertot1}? It is interesting to investigate.

\section*{Acknowledgement}

This work is supported in part by the National Natural Science Foundation of China (NSFC) grants No.12005155, No.~11875200 and No.~11935009.

