Tenemos $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}=\underbrace{\sum_{i,j=1}^{d}\int_{U}a^{ij}\Delta u\mathrm{d}x}_{A}-\underbrace{\int_{U}\left(b\cdot\nabla u\right)\Delta u\mathrm{d}x}_{B}-\underbrace{\int_{U}cu\Delta u\mathrm{d}x}_{C}.$$ Obtenemos fácilmente por la desigualdad de Young que $B\ge-\left\Vert b\right\Vert _{\infty}\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}+\frac{1}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}\right]$ y que $C\ge-\left\Vert c\right\Vert _{\infty}\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}+\frac{1}{4\varepsilon}\left\Vert u\right\Vert _{L^{2}}^{2}\right]$ . Observemos ahora que basta con demostrar nuestra desigualdad para la función en $C_{c}^{\infty}(U)$ desde $H_{0}^{1}(U)$ es el cierre de $C_{c}^{\infty}(U)$ en $H^{1}(U)$ y tenemos que tratar con funciones continuas. Así, para $u\in C_{c}^{\infty}(U)$ podemos integrar por partes sin obtener partes límite y por lo tanto obtenemos \begin{alignat*}{1} A= & \sum_{i,j,k}\int_{U}a^{ij}u_{x_{i}x_{j}}u_{x_{k}x_{k}}\mathrm{d}x=-\sum_{i,j,k}\int_{U}u_{x_{k}}\left(a^{ij}u_{x_{i}x_{j}}\right)_{x_{k}}\mathrm{d}x\\ = & -\sum_{i,j,k}\int_{U}u_{x_{k}}a_{x_{k}}^{ij}u_{x_{i}x_{j}}\mathrm{d}x-\sum_{i,j,k}\int_{U}u_{x_{k}}a^{ij}u_{x_{i}x_{j}x_{k}}\mathrm{d}x\\ = & -\sum_{i,j,k}\int_{U}a_{x_{k}}^{ij}u_{x_{i}x_{j}}u_{x_{k}}\mathrm{d}x+\sum_{i,j,k}\int_{U}\left(u_{x_{k}}a^{ij}\right)_{x_{i}}u_{x_{j}x_{k}}\mathrm{d}x\\ = & \underbrace{-\sum_{i,j,k}\int_{U}a_{x_{k}}^{ij}u_{x_{i}x_{j}}u_{x_{k}}\mathrm{d}x}_{A_{1}}+\underbrace{\sum_{i,j,k}\int_{U}a_{x_{i}}^{ij}u_{x_{j}x_{k}}u_{x_{k}}\mathrm{d}x}_{A_{2}}+\underbrace{\sum_{i,j,k}\int_{U}a^{ij}u_{x_{i}x_{k}}u_{x_{j}x_{k}}\mathrm{d}x}_{A_{3}}. \end{alignat*} Si consideramos $M:=\sup\left\{ \left\Vert a_{x_{k}}^{ij}\right\Vert :1\le i,j,k\le d\right\} $ (que es finito ya que $a^{ij}$ son suaves) obtenemos inmediatamente $A_{1}+A_{2}\ge2M\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}+\frac{1}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}\right]$ y también $A_{3}\ge\theta\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}$ . En definitiva, obtenemos $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}\ge\left(\theta-\left(2M+\left\Vert b\right\Vert _{\infty}+\left\Vert c\right\Vert _{\infty}\right)\varepsilon\right)\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}-\frac{\left\Vert b\right\Vert _{\infty}+2M}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}-\frac{\left\Vert c\right\Vert _{\infty}}{4\varepsilon}\left\Vert u\right\Vert _{L^{2}}^{2}$$ Ahora usando la desigualdad $\left\Vert Du\right\Vert _{L^{2}}^{2}\le\eta\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}+C\left(\eta\right)\left\Vert u\right\Vert _{L^{2}}^{2}$ para $\eta=\frac{4\varepsilon^{2}}{\left\Vert b\right\Vert _{\infty}+2M}$ obtenemos que $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}\ge\underbrace{\left(\theta-\left(2M+\left\Vert b\right\Vert _{\infty}+\left\Vert c\right\Vert _{\infty}+1\right)\varepsilon\right)}_{\beta\left(\varepsilon\right)}\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}-\underbrace{\left(\frac{\left\Vert c\right\Vert _{\infty}}{4\varepsilon}+C\left(\varepsilon\right)\right)}_{\gamma\left(\varepsilon\right)}\left\Vert u\right\Vert _{L^{2}}^{2} $$ por lo que basta con elegir un $\varepsilon$ lo suficientemente pequeño como para que $\beta(\varepsilon)>0$ y $\gamma(\varepsilon)\ge 0$ y aplicar la desigualdad de Poincare y ya está.