Prove $\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2$ if $-\Delta u(x)+f(x)u(x)=0.$ Let $\Omega $ a smooth domain of $\mathbb R^d$ ($d\geq 2$), $f\in \mathcal C(\overline{\Omega })$. Let $u\in \mathcal C^2(\overline{\Omega })$ solution of $$-\Delta u(x)+f(x)u(x)=0\ \ in\ \ \Omega .$$
Assume that $f(x)\geq 0$ for $x\in \Omega $. Prove that $$\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2,$$
for all $x\in \Omega $ and $r>0$, $B(x,2r)\subset \subset \Omega $ for somme $C\geq 0$ independant of $u,f,x$ and $r$.
My attempts
Using divergence theorem and that $\Delta u=fu$ in $\Omega $, I have that $$\int_{B(x,r)}|\nabla u|^2=\int_{B(x,r)}\text{div}(u\nabla u)-\int_{B(x,r)}u\Delta u=\int_{\partial B(x,r)}u\nabla u\cdot \nu-\int_{B(x,r)}fu^2.$$
But I can't do better. Any help would be welcome.

Denote $B(x,r)$ by $B_r$ and choose a smooth function $\xi$ to have the following property
$$ \xi=\left\{\begin{array}{lcr}1,&\text{if }x\in B_r\\
\ge 0,&\text{if }x\in B_{2r}\setminus B_r\\
0,&\text{if }x\in\Omega\setminus B_{2r},
\end{array}
\right. $$
satisfying $|\nabla \xi|\le \frac{C}{r}$. Using $u\xi^2$ as a test function in the equation, one has
$$ \int_{\Omega}\nabla u\nabla(u\xi^2)dx=-\int_{\Omega}fu^2\xi^2dx\le0. $$
Thus
$$ \int_{\Omega}\xi^2|\nabla u|^2dx\le-2\int_{\Omega}(\xi\nabla u)(u\nabla\xi) dx. $$
By using $2ab\le \frac{1}{2}a^2+2b^2$, one has
$$ \int_{\Omega}\xi^2|\nabla u|^2dx\le \frac12\int_{\Omega}\xi^2|\nabla u|^2dx+2\int_{\Omega}|u\nabla\xi|^2 dx $$
and hence
$$ \int_{\Omega}\xi^2|\nabla u|^2dx\le 4\int_{\Omega}|u\nabla\xi|^2 dx. $$
So using the property of $\xi$, one has
$$ \int_{B_r}|\nabla u|^2dx\le \frac{C}{r^2}\int_{B_{2r}}|u|^2 dx. $$
