Prove $\int_{\Sigma_r} |\nabla\varphi|^2 d\sigma_r \ge \frac{2}{u(r)^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r $ Reference: this paper

Given the deSitter-Schwarzschild metric with mass $m > 0$ and scalar
  curvature equal to $2$ is the metric
$$\bigg(  1 -\frac{r^2}{3}-\frac{2m}{r} \bigg)^{-1} dr^2 + r^2
 dg_{\mathbb S^2} \tag{1}$$
defined on $(a_0 ,b_0) \times \mathbb S^2$, where $(a_0,b_0)=\{r>0:1
 -\frac{r^2}{3}-\frac{2m}{r}>0 \}$ and $g_{\mathbb S^2}$ is the standard metric on $\mathbb S^2$ with constant Gauss curvature equal
   to $1$.

In order to deal with the metric in $(1)$, we use the warped product metric $g=dr^2 + u(r)^2 dg_{\mathbb S^2}$ on $\mathbb R \times \mathbb S^2$, where $u(r)$ is a positive real function. If we assume that $g$ has constant scalar curvature equal to $2$, then $u$ solves the following second-order differential equation
$$u''(r)=\frac{1}{2}\bigg( \frac{1-u'(r)^2}{u(r)} - \frac{u(r)}{2}  \bigg) \tag{2}$$
Considering only positive solutions $u(r)$ to $(2)$ which are defined for all $r \in \mathbb R$, we get a one-parameter family of periodic rotationally symmetric metrics $g_a = dr^2 + u_a(r)^2 g_{\mathbb S^2}$ with constant scalar curvature equal to $2$, where $a \in (0,1)$ and $u_a(r)$ satisfies $u_a(0)=a= \text{min} ~u$ and $u'_a(0) =0$. These metrics are precisely the deSitter-Schwarzschild metrics on $\mathbb R \times \mathbb S^2$ defined in $(1)$.

Next, let $(M,g)$ be a three-manifold and consider a two-sided compact surface $\Sigma \subset M$. 
The mass of $\Sigma \subset M$ is defined by
$$m(\Sigma) =\ \bigg( \frac{|\Sigma|}{16\pi}  \bigg)^{1/2}   \bigg( 1
 - \frac{1}{ 16\pi } \int_{\Sigma} H^2 d\sigma 
  - \frac{ \Lambda}{24\pi } |\Sigma|  \bigg) \tag{3}$$
where $\Lambda = \text{inf}_M ~ R$, $R$ is the scalar curvature of
  $M$, $H$ is the mean curvature of $\Sigma$, $K_\Sigma$ is the Gauss
  curvature of $\Sigma$, and $|\Sigma| = \int_\Sigma d\sigma$.

The first variation of $m$:
$$\frac{d}{dt}m(\Sigma(t))\bigg|_{t=0} = -
 \frac{2|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \varphi
 \Delta_\Sigma H d\sigma \\
 + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg[ 2K_\Sigma - \frac{8\pi}{|\Sigma|  }+ \bigg( \frac{1}{2|\Sigma|}\int_\Sigma H^2 d\sigma - |A|^2 \bigg)   \bigg] H \phi d\sigma \\
 + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma (\Lambda - R) H\varphi d\sigma \tag{4}$$

Remark $1$. It follows from $(4)$  that if a two-sphere $\Sigma \subset M$ is umbilic where $|A|^2 = \frac{H^2}{2}$ and has constant Gauss curvature and $M$ has constant scalar curvature equal to $2$ along $\Sigma$, then $\Sigma$ is a critical point of the mass in $(3)$. 

Denote by $\Sigma_r$ the slice $\{r\} \times \mathbb S^2$. By Remark $1$, $\Sigma_r$ is a critical point for the mass in $(\mathbb R \times \mathbb S^2 , g_a)$, for all $r \in \mathbb R$ and $a \in (0,1)$. Moreover the mass of $\Sigma_r \subset (\mathbb R \times \mathbb S^2 , g_a)$ is constant for all $r \in \mathbb R$. It follows by a straightforward computation:
$$\frac{d}{dr} m(\Sigma_r) = \frac{1}{2} u'(r)(1-u'(r)-u(r)^2-2u(r)u''(r)) \tag{5}$$
which is zero once $u(r)$ solves $(2)$, we obtain therefore that $m(\Sigma_r)$ is constant equal to $m(\Sigma_0)$.

Now, since $g_{\Sigma_r} = u(r)^2 g_{\mathbb S^2}$, by the Poincare inequality we have
  $$ \int_{\Sigma_r} |\nabla\varphi|^2  d\sigma_r \ge \frac{2}{u(r)^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r   = \frac{8\pi}{|\Sigma_r|}\int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r \tag{6}$$ 
where $\varphi \in C^\infty(\Sigma_r)$.

Question:
Where does $(6)$ come from?
Thank you.
 A: Although I am not really sure with my calculation, but here is my progress. 
Any suggestions or corrections are very welcome.
Given the conformal metric:
\begin{align}
 g|\Sigma_r = u(r)^2 g_{\Bbb S^2} ~,
 \tag{7}
\end{align}
where $u(r)>0$ is some positive function and solves the second-order differential equation below
\begin{align}
 - 6 \Delta u +  R u = \tilde{R}_{g|\Sigma_r} u^3 \\
\int_{\Sigma_r} |\nabla u|^2 d\sigma_r +  \int_{\Sigma_r} \frac{1}{6}2 u^2 d\sigma_r = \int_{\Sigma_r}\frac{1}{6}\frac{8\pi}{|\Sigma|} u^4 d\sigma_r
\tag{8}
\end{align}
where $R =2 >0$ is the scalar curvature of the metric $g_{S^n}$ and $\tilde{R}_{g|\Sigma_r} = 2K_{\Sigma_r} = 2 . \frac{4\pi}{|\Sigma|} = \frac{8\pi}{|\Sigma|} \in \mathbb R$ is the scalar curvature of $g|\Sigma_r$. 
The Yamabe invariant of a compact manifold $(M^3,g)$ :
\begin{align}
 Y(g) = \text{inf }\frac{\int_{\Sigma_r} \left( |\nabla u|^2 + \frac{1}{6}R u^2 \right) d\sigma_r}
 { \bigg( \int_{\Sigma_r} |u|^4 d\sigma_r \bigg)^{1/2}  } ~ ,
 \tag{9}
\end{align}
where $Y(g)$ is the Yamabe constant.
From $(9)$ and $(8)$, it shows that
\begin{align}
 Y(g) \bigg( \int_{\Sigma_r} |u|^4 d\sigma_r \bigg)^{1/2} = \frac{1}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^4 d\sigma_r ~ .
 \tag{10}
\end{align} 
Substitute (10) to (9) we get
\begin{align}
\int_{\Sigma_r} |\nabla u|^2 d\sigma_r  \ge\ & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^2 d\sigma_r 
\nonumber\\
%
%
& - \frac{u^2}{6} 2{u^2} \int_{\Sigma_r}  u^2  d\sigma_r
~ .
\tag{11}
\end{align}
$\int_{\Sigma_r}  u^2  d\sigma_r$ could also be written as
\begin{align}
 \int_{\Sigma_r}  u^2  d\sigma_r =\ & u \int_{\Sigma_r}  u\  d\sigma_r
 \nonumber\\
 %%
 %
 =\ & \frac{1}{|\Sigma|} u |\Sigma| \int_{\Sigma_r}  u\  d\sigma_r
 \nonumber\\
 %
 %
 =\ & \frac{1}{|\Sigma|} u \int_{\Sigma_r}  d\sigma_r \int_{\Sigma_r}  u\  d\sigma_r
 \nonumber\\
 %
 %
 =\ & \frac{1}{|\Sigma|} \int_{\Sigma_r}  u\ d\sigma_r \int_{\Sigma_r}  u\  d\sigma_r
 \nonumber\\
 %
 %
 =\ & \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r}  u\ d\sigma_r \bigg)^2 ~ ,
\tag{12}
\end{align}
Plugging $(12)$ to the second term on RHS of the inequality $(11)$ we get
\begin{align}
 \int_{\Sigma_r} |\nabla u|^2 d\sigma_r  \ge\ & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^2 d\sigma_r 
 \nonumber\\
 %
 %
 & - \frac{u^{2}}{6} \frac{2}{u^2} \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r}  u\ d\sigma_r \bigg)^2 ~ .
 \tag{13}
\end{align}
Since $\frac{8\pi}{|\Sigma|} = \frac{2}{u^2}$ and $\int_{\Sigma_r} u^2 d\sigma_r - \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r}  u\ d\sigma_r \bigg)^2 = \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r$, we could write $(13)$ as
\begin{align}
\int_{\Sigma_r} |\nabla u|^2 d\sigma_r  \ge\ & \frac{u^2}{6} \frac{2}{u^2} \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r \\
= & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r
\tag{14} ~ ,
\end{align}
which is kind of similar to $(6)$. 
My questions now are:


*

*Where does $\frac{u^2}{6}$ go?

*Despite of the vanishing $\frac{u^2}{6}$ in $(6)$, if we use notation $u = \varphi$, why isn't it just written as
\begin{align}
\int_{\Sigma_r} |\nabla \varphi|^2 d\sigma_r  \ge\ & \frac{2}{\varphi^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r \\
= &  \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r 
\tag{15}
\end{align}
Why $\frac{2}{u^2}$ and not $\frac{2}{\varphi^2}$ ?


*Is it legal if I take $\bar{\varphi}=0$ and why?


Thank you.
A: As mentioned by the @user99914, we just need to know the best constant in the Poincare inequality, which I will simply call here as the "Poincare constant". To be precise, here it means the largest constant $C>0$ such that 
\begin{align}
C\int_M(f-\bar{f})d\mu\leq\int_M|\nabla f|^2d\mu
\end{align}
It turns out that the Poincare constant $C$ is related to the first eigenvalue $\lambda_1$ of the Laplacian on the manifold (in this case, the $2$-sphere); by this I mean $-g^{ij}\nabla_i\nabla_j f=\lambda_1f$; that is, there is a minus sign in front. In fact, we have $C=\lambda_1$. This fact probably can be found in many standard Riemannian geometry or geometric analysis texts, and I'll just quote one which I encountered: Jurgen Jost's book. In short, it follows from the following:
\begin{align}
\lambda_1=\inf\left\{\frac{\int_{M}|\nabla f|^2d\mu}{\int_{M}f^2d\mu}:f\in W^{1,2}(M), \int_{M}fd\mu=0\right\}
\end{align}
(For simplicity, I only consider the case where $M$ is compact.)
Now, it is sort of regarded as a standard fact in geometric analysis that the first eigenvalue of an $n$-sphere with radius $1/\sqrt{K}$ is equal to 
\begin{align}
\lambda_1=nK
\end{align}
For a more general result which includes this as a special case, see Lichnerowicz and Obata's work. Now, the radius of our $\Sigma_r$ is $u(r)$, while its dimension is $2$. Hence
\begin{align}
C=\lambda_1=2\cdot\left(\frac{1}{u(r)}\right)^2
\end{align}
which gives you the desired result. 
P/S: I haven't checked out your new questions in your answer. Maybe I'll also try to answer them when I'm free. 
