I tried to prove on my own theorem 2 of chapter 6 of Evans partial differential equations second edition, but my proof of the coercive estimate doesn't use the Poincare inequality whereas Evan's does.(The next is for reference)
Here we consider linear uniformly elliptic equations in divergence form $$-(a^{ij}(x)u_{x_i})_{x_j} + b^i(x)u_{x_i}+c(x)u $$
in a bounded domain $U$, we assume that the coefficients are in $L^\infty, a^{ij}=a^{ji}$ and the ellipticty condition, that is there is a constant $\theta>0$ such that for a.e $x \in U$ and all $\xi \in \mathbb{R}^n$ we have $$a^{ij}(x)\xi_i\xi_j \geq \theta |\xi|^2 $$
the bilinear form is $$B[u,v] = \int_U a^{ij} u_{x_i}v_{x_j} + b^i u_{x_i}v+cuv$$ where $u,v \in H_0^1$ and the estimate I want to show is $$\beta \| u\|_{H_0^1}^2 \leq B[u,u] + \gamma \|u\|_{L^2}^2 $$ where $\beta > 0$ and $\gamma \geq 0$.
My proof: By the ellipticity condition, choosing $\xi = Du(x)$ we have $$\theta \int_U |Du|^2 \leq \int_U a^{ij}u_{x_i}v_{x_j} = B[u,u] - \int_U b^iu_{x_i}+cu^2 \leq B[u,u] + C \int_U |Du| |u| + C \|u\|_{L_2}^2 $$
Then using the epsilon Cauchy inequality $ab \leq \epsilon a^2 + \frac{b^2}{4\epsilon}$ we have $$\int_U |Du| |u| \leq \epsilon \| Du\|_{L^2}^2 + \frac{1}{4\epsilon} \| u\|_{L^2}^2 $$
Choosing $\epsilon$ such that $C\epsilon < \theta/2$ we obtain $$\theta/2 \|Du \|_{L^2}^2 \leq B[u,u] + \gamma \| u \|_{L^2}^2 $$ and adding $\theta/2 \|u\|_{L^2}^2 $ to both sides we obtain $$\theta/2 \|u\|_{H_0^1}^2 \leq B[u,u] + \gamma \|u\|_{L^2}^2 $$ which is what we wanted.
In Evan's book, the proof is similar except that after he gets the estiamte $$\theta/2 \int_U |Du|^2 \leq B[u,u] + C \int_U u^2 $$ he uses the Poincare inequality $$ \|u\|_{L^2} \leq C \| Du\|_{L^2} $$ and then says that it easily follows that $$\beta \|u\|_{H_0^1}^2 \leq B[u,u] + \gamma \|u\|_{L^2}^2 $$
I don't understand why is necessary to bring the Poincare inequality instead of doing what I did and I don't understand how exactly he is using the inequality, what I think he does is $$\theta/2 \|u\|_{L^2}^2 + \theta/2 \|Du\|_{L^2}^2 \leq \theta C/2 \|Du\|_{L^2}^2 \leq CB[u,u] + C\gamma \|u\|_{L^2}^2 $$ but this wont work unless $C \leq 1$