Provide a weak formulation for the following boundary value problem $$ (*) \quad L u=: \sum_{i, j=1}^{n} \partial_{j}\left(a_{i j} \partial_{i} u\right)+c u=f, \quad \text { in } \Omega, \quad \text { and }\left.\quad u\right|_{\partial \Omega}=0 $$ Prove that for any $f \in L^{2}(\Omega)$ there exists a unique weak solution of $(*)$ if $L$ is uniformly elliptic and $c(x) \geq 0$ for $x \in \Omega$.
My attempt:
Proof. Define $$ B[u, v]=\int_{U} \sum_{i, j=1}^{n} a^{i j} u_{x_{i}} v_{x_{j}}+c u v d x, u, v \in H_{0}^{1}(U) $$ Clearly, $$ |B[u, v]| \leq\left\|a^{i j}\right\|_{\infty}\|D u\|_{L^{2}(U)}\|D v\|_{L^{2}(U)}+\|c\|_{\infty}\|u\|_{L^{2}(U)}\|v\|_{L^{2}(U)} \leq \alpha\|u\|_{H_{0}^{1}(U)}\|v\|_{H_{0}^{1}(U)} $$ for some positive constant $\alpha$. For $c(x) \geq-\mu$, where $\mu$ is a fixed constant to be assigned later. Using the uniform ellipticity, we have $$ \begin{aligned} \theta \int_{U}|D u|^{2} d x & \leq \int_{U} \sum_{i, j=1}^{n} a^{i j} u_{x_{i}} u_{x_{j}} \\ &=B[u, u]-\int_{U} c u^{2} d x \\ & \leq B[u, u]+\mu \int_{U} u^{2} d x \\ & \leq B[u, u]+C \mu \int_{U}|D u|^{2} d x \end{aligned} $$ where the Poincaré inequality is applied in the last inequality. Now we set $\mu=\frac{\theta}{2 C}$ and see that if $c(x) \geq \frac{\theta}{2 C}$, then $$ \frac{\theta}{2} \int_{U}|D u|^{2} d x \leq B[u, u] . $$
Therefore, $$ \begin{aligned} \|u\|_{H_{0}^{1}(U)}^{2} &=\|u\|_{L^{2}(U)}^{2}+\|D u\|_{L^{2}(U)}^{2} \\ & \leq(1+C)\|D u\|_{L^{2}(U)}^{2} \\ & \leq \frac{2}{\theta}(1+C) B[u, u] \end{aligned} $$
You have the correct bilinear form. You should then say $u \in H^1_0(\Omega)$ is a weak solution if $B[u,v]=(u,v)_{L^2(\Omega)}$ for all $v\in H^1_0(\Omega)$ which I expect is what you had in mind.
Your calculation shows that if $c \geqslant -\mu$ for your chosen $\mu>0$ then there is a unique weak solution by Lax-Milgram. But you have $c\geqslant 0 >-\mu$ regardless of what $\mu$ is which proves your claim. (Maybe your issue is that you have a small typo - after you set $\mu = \frac \theta {2C}$ you go on to state 'if $c \geqslant \frac \theta {2C}$ ...' when you mean 'if $c \geqslant -\frac \theta {2C}$ ...')
However, you don't need to introduce the $\mu$ at all since you are assuming $c\geqslant 0$. Indeed,
$$ \theta \int_\Omega \vert Du \vert^2 \, dx \leqslant \sum_{i,j}a^{ij}D_{i}u D_ju = B[u,u]-\int_\Omega c u^2 \, dx \leqslant B[u,u]. $$ Therefore, $$\| u \|_{H^1(\Omega)}^2 \leqslant C \| Du \|_{L^2(\Omega)}^2\leqslant CB[u,u]. $$ ($C>0$ changes each line).