How to show $\frac{n}{4}\ln [1+\int |\nabla \phi|^2 dV]\le \frac{1}{2\pi}\int |\nabla \phi|^2 dV + C(g,n)$? [closed]
There is nothing special about $2\pi$. You can choose this constant as long as you make $C(g, n)$ large.
Since $\ln(1+x)\le \sqrt{x}$, together with $2ab \le \epsilon a^2 + \epsilon ^{-1} b^2$, we have
$$ \frac n4 \ln (1+x) \le \frac n4 \sqrt x \le \frac n4 ( \epsilon x + \epsilon^{-1}) = \frac{n\epsilon}{4} x + \frac{n}{4\epsilon}. $$
Now choose $\epsilon$ so that $\frac{n\epsilon}{4} = \frac{1}{2\pi}$ and $x = \int _M |\nabla \phi|^2 dV$, one obtain your inequality (with a different $C(g, n)$, as suggested in the text).