Reference: this paper page 7.
The B$\ddot{\text{o}}$chner-Weitzenb$\ddot{\text{o}}$ck identity applied on $\Sigma_r$
\begin{align} \frac{1}{2} \Delta |\nabla\phi|^2 =\ & |\nabla^2 \phi|^2 + \langle \nabla \Delta \phi , \nabla \phi \rangle + \text{Ric}(\nabla\phi,\nabla\phi) \tag{1} \\ % % \ge\ & \frac{1}{2} (\Delta\phi)^2 + \langle \nabla \Delta \phi , \nabla\phi \rangle + K_{\Sigma_r}|\nabla\phi|^2 \tag{2} \\ % % =\ & \frac{1}{2} (\Delta\phi)^2 + \langle \nabla \Delta \phi , \nabla\phi \rangle + \frac{4\pi}{|\Sigma_r|}|\nabla\phi|^2 \end{align}
Integrate over $\Sigma_r$ we get
\begin{align} - \frac{1}{2} \int_{\Sigma_r} (\Delta\phi)^2 d\sigma_r \le\ & - \frac{4\pi}{|\Sigma_r|} \int_{\Sigma} |\nabla\phi|^2 d\sigma_r \tag{3} \end{align}
My attempt:
I want to prove (3).
\begin{align} \frac{1}{2} \int_{\Sigma_r} \Delta |\nabla\phi|^2 d\sigma_r \ge\ & \frac{1}{2} \int_{\Sigma_r} (\Delta\phi)^2 d\sigma_r + \int_{\Sigma_r} \langle \nabla \Delta \phi , \nabla\phi \rangle d\sigma_r + \frac{4\pi}{|\Sigma_r|} \int_{\Sigma} |\nabla\phi|^2 d\sigma_r \\ % % - \frac{1}{2} \int_{\Sigma_r} (\Delta\phi)^2 d\sigma_r \ge\ & - \frac{1}{2} \int_{\Sigma_r} \Delta |\nabla\phi|^2 d\sigma_r + \int_{\Sigma_r} \langle \nabla \Delta \phi , \nabla\phi \rangle d\sigma_r + \frac{4\pi}{|\Sigma_r|} \int_{\Sigma} |\nabla\phi|^2 d\sigma_r \\ % % \frac{1}{2} \int_{\Sigma_r} (\Delta\phi)^2 d\sigma_r \le\ & \frac{1}{2} \int_{\Sigma_r} \Delta |\nabla\phi|^2 d\sigma_r - \int_{\Sigma_r} \langle \nabla \Delta \phi , \nabla\phi \rangle d\sigma_r - \frac{4\pi}{|\Sigma_r|} \int_{\Sigma} |\nabla\phi|^2 d\sigma_r \tag{4} \end{align}
It seems like
\begin{align} \int_{\Sigma_r} (\Delta\phi)^2 d\sigma_r = \frac{1}{2} \int_{\Sigma_r} \Delta |\nabla\phi|^2 d\sigma_r - \int_{\Sigma_r} \langle \nabla \Delta \phi , \nabla\phi \rangle d\sigma_r \tag{5} \end{align}
But I do not know how to achieve (5).
Please help.