$\newcommand{\Ric}{\text{Ric}}$ Let $M$ be a smooth closed oriented Riemannian surface.
I am searching for a reference (or a sketch of proof) for the following inequality:
$$ \int_M | \nabla V|^2 \ge \int_{M} \Ric(V,V)=\int_{M} K|V|^2, \tag{1}$$
for every vector field $V \in \Gamma(TM)$, where $ \nabla$ is the Levi-Civita connection, and the integration is against the Riemannian volume form. ($K$ is the Gauss curvature).
I guess some kind of Bochner identity is needed. I am also interested to know if this inequality holds for manifolds of higher dimensions.
BTW, specializing for the case of the round $2$-sphere, we get
$$ \int_{\mathbb{S}^2} | \nabla V|^2 \ge \int_{\mathbb{S}^2} |V|^2. \tag{2}$$
A proof of this specific case can be found here.
You probably figured it out by now but since it's such a nice inequality I'll just point out that it follows from the identity in OP's other question :
How to prove $\int_M-\text{Ric}(V,V)+ |\nabla V|^2 =\int_M \frac{1}{2}|L_Vg|^2-\big(\text{div}V\big)^2$?
Since $(\mathcal{L}_Vg)_{ij} = \nabla_iV_j + \nabla_jV_i$, on a manifold of dimension $n$ we find $$\frac12|\mathcal{L}_Vg|^2 \geq \frac{1}{2n}|\text{trace}(\mathcal{L}_Vg)|^2 = \frac{2}{n}(\text{div}V)^2.$$ Thus the RHS of that identity is bounded below by $(\frac{2}{n} - 1)(\text{div}V)^2$ which is non-negative when $n \leq 2$.