This is an old prelim problem. Let $\Omega \subset \mathbb{R}^n$ open and bounded with smooth boundary. If $u\in C^3(\bar\Omega)$ solves $$ -\sum_{i,j=1}^n a_{ij}(x)\,u_{x_ix_j}(x)=f(x)\quad \mbox{in }\Omega $$ with a uniformly elliptic operator and $a_{ij},f\in C^1(\bar\Omega)$. Use the function $v=|\nabla u|^2+\lambda u^2$ with a constant $\lambda >0$ to be chosen, to show there exists a constant $C$ depends on $\Omega$ and $a_{ij}$ such that $$\sup_{\bar\Omega}|\nabla u|^2\leq C(\sup_{\Omega}u^2+\sup_{\Omega}f^2+\sup_{\Omega}|\nabla f|^2)+\sup_{\partial\Omega}|\nabla u|^2.$$
My attempt: I was trying to show that $Lv\leq 0$ for a certain $\lambda$ where $L= -\sum_{i,j=1}^n a_{ij} \frac{\partial^2}{\partial x_i \partial x_j}$, then apply maximum principle to have $\sup_{\bar\Omega}v\leq \sup_{\partial \Omega}v$. But if suppose we proved $Lv\leq 0$, then $\sup_{\bar\Omega}|\nabla u|^2\leq \sup_{\bar\Omega}v\leq \sup_{\partial\Omega}|\nabla u|^2+\lambda \sup_{\partial\Omega}u^2$ which for me doesn't look like it will help. Anyway, $$L(\lambda u^2)= -2\lambda \sum_{i,j=1}^n a_{ij} u_{x_i}u_{x_j}+2\lambda u f\leq -2\lambda \alpha |\nabla u|^2+2\lambda u f $$ where $\alpha>0$ is the ellipticity constant. \begin{align} L((u_{x_k})^2)&=-2 \sum_{i,j=1}^n a_{ij}u_{x_kx_i}u_{x_kx_j}-2u_{x_k}\sum_{i,j=1}^n a_{ij}u_{x_ix_jx_k}\\\\ &\leq -2\alpha |\nabla(u_{x_k})|^2+2u_{x_k}f_{x_k}+2u_{x_k}\sum_{i,j=1}^n (a_{ij})_{x_k}u_{x_ix_j}. \end{align} Thus, \begin{align} L(|\nabla u|^2)&\leq -2\alpha |D^2u|^2+2\sum_{k=1}^n u_{x_k}f_{x_k}+2C_0\sum_{k=1}^n u_{x_k}\sum_{i,j=1}^n u_{x_ix_j}\\\\ &\leq -2\alpha |D^2u|^2+2|\nabla u|\,|\nabla f|+2C_0 |\nabla u||D^2u| \end{align} where $D^2u$ is Hessian matrix and $C_0=\sup_{\bar\Omega}|\nabla(a_{ij})|$. Since $ab\leq \epsilon a^2+\frac{1}{4\epsilon}b^2$ for all $\epsilon>0$, with $a=|D^2u|$, $b=|\nabla u|$, and $\epsilon=\alpha/C_0$, we obtain \begin{align} L(|\nabla u|^2)&\leq 2|\nabla u|\,|\nabla f|+\frac{C_0^2}{2\alpha} |\nabla u|^2. \end{align} Hence, \begin{align} L(v)&\leq -2\lambda \alpha |\nabla u|^2+2\lambda u f+2|\nabla u|\,|\nabla f|+\frac{C_0^2}{2\alpha} |\nabla u|^2. \end{align} Then I don't know what to do. Any help is appreciated.