How to show bilinear form is coercive? For $\sigma\in \mathbb R$ , let us now define the operator $L_{\sigma}$ by $L_{\sigma}u=Lu-\sigma u$. We have to show that associated bilinear form is coercive if either $\sigma $ is sufficiently large or $|\Omega|$ is sufficiently small.
\begin{equation}\label{eq:81}
    Lu=D_i(a^{ij}(x)D_ju+b^i(x)u)+c^i(x)D_iu+d(x)u
\end{equation}.
In weak sense $u$ is said to satisfy $Lu=0$ in $\Omega$ as
\begin{equation}\label{eq:82}
    \mathfrak L(u,v)=\int \{(a^{ij}D_ju+b^iu)D_iv-(c^iD_iu+du)v)\}dx=0
\end{equation}
I understand following Lemma.
Lemma:Let $L$ satisfy conditions
\begin{align*}
a^{ij}(x)\xi_i\xi_j &\geq \lambda |\xi|^2\\
\sum|a^{ij}(x)|^2&\leq \Lambda^2\\
\lambda^{-2}\sum (|b^i(x)|^2+|c^i(x)|^2)+\lambda^{-1}|d(x)|&\leq \nu^2.
    \end{align*}
Then
\begin{equation}\label{eq:811}
    \mathfrak L(u,u)\geq \frac{\lambda}{2}\int_{\Omega}|Du|^2dx-\lambda\nu^2\int_{\Omega}u^2dx
\end{equation}
Proof:
\begin{align*}
    \mathfrak L(u,u)&=\int \{(a^{ij}D_ju+b^iu)D_iu-(c^iD_iu+du)u)\}dx\\
    &=\int_{\Omega}\left((a^{ij}D_juD_iu+(b^i-c^i)uD_iu-du^2\right)\\
    &\geq \int_{\Omega}\lambda|Du|^2-\frac{\lambda}{2}|Du|^2-\lambda\nu^2u^2\qquad \text{By Schwartz inequality }\\
    &=\frac{\lambda}{2}\int_{\Omega}|Du|^2dx-\lambda\nu^2\int_{\Omega}u^2dx.
\end{align*}
My attempt:
\begin{align*}
    \mathfrak L_{\sigma}(u,u)&=\int \{(a^{ij}D_ju+b^iu)D_iu-(c^iD_iu+du)u-\sigma u^2)\}dx\\
    &=\int_{\Omega}\left((a^{ij}D_juD_iu+(b^i-c^i)uD_iu-du^2-\sigma u^2\right)\\
    &\geq \int_{\Omega}\lambda|Du|^2-\frac{\lambda}{2}|Du|^2-\lambda\nu^2u^2-\sigma u^2 \\
    &=\frac{\lambda}{2}\int_{\Omega}|Du|^2dx-(\lambda\nu^2+\sigma)\int_{\Omega}u^2dx\\
    &\geq \left(\frac{\lambda}{2}-\lambda\nu^2-\sigma\right)||u||_2^2\qquad \text{ By Poincare's inequality}
\end{align*}
I am stuck here as I think RHS is negative so don't make sense . I don't know how to proceed further. Also for coercive norm on u should be of $W^{1,2}$ but I get of $L^2$
Any help will be appreciated
 A: You have not carried the change in sign over to the term involving the $\sigma$. According to your working what you should have gotten is,
\begin{align}
\int_{\Omega}(a^{ij}D_{j}u+b^{i}u)D_{i}u-(c^{i}D_{i}u+du)u+\sigma u^{2}dx.
\end{align}
Now, when you apply the lemma you have above you should end up at,
\begin{align}
\mathfrak{L}_{\sigma}(u,u)\geq \int_{\Omega}\frac{\lambda}{2}|Du|^{2}dx+\int_{\Omega}(\sigma-\lambda\nu^{2})u^{2}dx,
\end{align}
So fix $\sigma>\lambda\nu^{2}$ and you achieve the desired result.
With regards to the norm you are ending up with, notice that if $\sigma$ is chosen to satisfy the above condition, set $M:=\min\{\frac{\lambda}{2},\sigma-\lambda\nu^{2}\}$ then,
\begin{align}
\mathfrak{L}_{\sigma}(u,u)\geq M\bigg(\int_{\Omega}|Du|^{2}dx+\int_{\Omega}u^{2}dx\bigg)=M\|u\|_{W^{1,\,2}}^{2}.
\end{align}
As an aside I will note that I am unsure of how you achieved the final coefficients in the lemma you have. When I worked through it I ended up with $-\frac{3\lambda\nu^{2}}{2}$ as the coefficient of $u^{2}$. At some point you need to also use Young's inequality to split the middle term involving $|u||Du|$.
