How to show that $\frac{1}{V}\int_MR \, dV\ge\mathcal{F(g_{ij},f)}$?

Question

When I read the proof of Corollary 1.5.5 of this paper (204th page),I get stuck in the red box in picture below.How to show it ?

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Answer 1

I don't know whether I am right.

Let $f=k$,$k$ is constant. Because $\int_Me^{-f}dV=1$ , then $1=\int_Me^{-k}dV=e^{-k}\int_MdV=e^{-k}V$. So,$e^{-k}=\frac{1}{V}$.

So, $\int_M(R+|\nabla f|^2)e^{-k}dV=\frac{1}{V}\int_MRdV$.

So,$\frac{1}{V}\int_MRdV\ge\lambda(g_{ij}(t))$

Because the $f$ is continuous about $t$ , So ,when $t\rightarrow t_0$, we still have the red box.Because the $\lambda(g_{ij}(t_0))\le0$,then $-\int_M(R+|\Delta f|^2)e^{-f}dv\ge 0$. So we can get the last inequality.

For proving $\lambda(g_{ij}(t_0)) \ge \int_M (R + |\nabla f|^2)e^{-f} dV$.Under coupled flow of 201 page,we have
$$ \frac{d}{dt}\int_M(R+|\nabla f |^2)e^{-f}dV=2\int_M|R_{ij}+\nabla_i\nabla_jf|^2e^{-f}dV\ge 0 $$ So, we have $\lambda(g_{ij}(t_0)) \ge \int_M (R(t) + |\nabla f(t)|^2)e^{-f(t)} dV_t$ ,when $t\le t_0$.