[proofplan] Use ellipticity and the nonnegative zeroth-order term to control the gradient. Then use Poincare's inequality on $H_0^1(U)$ to control the full $H_0^1$ norm. [/proofplan] [step:Lower bound the bilinear form] For $u\in H_0^1(U)$, the displayed definition of the energy form $B$, the assumptions $b_i=0$, $c\ge0$, and uniform ellipticity give \begin{align*} B[u,u] &=\int_U \sum_{i,j=1}^n a_{ij}\partial_{x_j}u\,\partial_{x_i}u + c u^2\,d\mathcal L^n \\ &\ge \theta\int_U |\nabla u|^2\,d\mathcal L^n. \end{align*} [/step] [step:Control the full Sobolev norm] Since $U$ is bounded and $u\in H_0^1(U)$, the [Poincare Inequality with Zero Trace](/theorems/76) gives a constant $C_U>0$ such that \begin{align*} \|u\|_{L^2(U)}\le C_U\|\nabla u\|_{L^2(U)}. \end{align*} Hence \begin{align*} \|u\|_{H_0^1(U)}^2 =\|u\|_{L^2(U)}^2+\|\nabla u\|_{L^2(U)}^2 \le (C_U^2+1)\|\nabla u\|_{L^2(U)}^2. \end{align*} Combining this with the previous lower bound yields \begin{align*} B[u,u]\ge \frac{\theta}{C_U^2+1}\|u\|_{H_0^1(U)}^2. \end{align*} Thus the claim holds with $\alpha=\theta/(C_U^2+1)>0$. [/step]