Integral estimate on a two dimensional Riemannian manifold

For my Master's thesis, I'd like to prove the following (but I'm not sure it's true):

On a two-dimensional Riemannian manifold (oriented and closed), for any smooth function $f$, it holds that $$\int_M \left( 2 |\nabla^2 f |^2 + \text{Scal} |\nabla f|^2 \right) \text{d} V \geq 0,$$ where Scal denotes the scalar curvature. It feel like I have to use some divergence/integration by parts theorem and use that $M$ is Einstein but I just can't make it. Thanks in advance!

• Why are you trying to prove this? Is there some motivation? – Igor Rivin Jul 12 '13 at 20:33
• Look up Weitzenböck formulae and Bochner identities: these introduce a curvature term that will reduce to $\mathrm{Scal}$ in yr $2$-dimensional setting. – Fran Burstall Jul 14 '13 at 11:30

You have the following idenity on a general closed Riemannian manifold: $$\int_M (\Delta_g f)^2 \, dv_g = \int_M \vert \nabla^2_g f\vert^2 + Ricc(\nabla f,\nabla f) \, dv_g$$ You can porve it by integrating by part the first term and commuting the covariant derivatives which make appear the curvature. In your special case $Ricc= \frac{Scal}{2} g$, then your are done.