mean curvature $H+\<\bar\nabla u,\nu\>>0$, then the following inequality holds: $$n\di\int_{\O}e^{\tau u}d\O\le\di (n-1)\int_{\S}\frac{e^{\tau u}}{H+\<\bar\nabla u,\nu\>} dA,
             \label{Heintze-Karcher coro 1}$$ where $\tau=\frac{n+1}{n-1}$. Moreover, if equality in holds, then $\S$ is umbilical.* ]{}

    We remark that the weight in is $e^{\tau u}$ instead of $e^u$. The new weight volume form $e^{\tau u}d\O$ has a property that it is parallel under the affine connection $D^{\a,\g}$. In the special case of $1$-Bakry-Émery Ricci curvature, this parallel property has been observed by Wylie-Yeroshkin [@WY] recently.

In particular, for a Riemannian triple $(M, g, V)$ whose static Ricci tensor has a positive lower bound, we get the first eigenvalue estimate for the operator $V\bar \De - \bd V\cdot$.

\[coro\] Let $(M^n, \bar g, V)$ be an $n$-dimensional Riemannian triple. Let $\O$ be a bounded domain in $M$ with smooth boundary $\S$. Assume the static Ricci tensor satisfies $$V Ric- \mathrm{\bar \nabla^2}  V+ \bar \De V \bar g \geq (n-1) \bar g.$$ Then we have\
[**a)** ]{} if $\S=\emptyset$, then $\lambda_1(V\bar \De - \bd V\cdot)\ge n;$\
[**b)** ]{} if $\S \neq\emptyset$ and $\S$ satisfies $H^D\ge 0$, then $\lambda_1^{Dir}(V\bar \De - \bd V\cdot)\ge n;$\
[**c)** ]{} if $\S\neq\emptyset$ and $h-\g u_\nu g\ge 0$, then $\lambda_1^{Neu}(V\bar \De - \bd V\cdot)\ge 