If $-\Delta + a$ is coercive, then $\int_\Omega |\nabla u|^2 + a u^2 \ge C \int_\Omega|\nabla u|^2$. In my functional analysis course, we are studying the first eigen value of the laplacian on $H^1_0 (\Omega)$,
$$-\Delta u + a u = \lambda_1^a(\Omega)u$$
for $a \in C^{0, \eta}(\Omega)$ ($0 < \eta < 1$) and $\lambda_1^a(\Omega)$ obtained from the minimization of the functional
$$I_a(u) = \int_\Omega |\nabla u|^2 + a u^2$$
with the constraint $\|u\|_{L^2(\Omega)}^2 = 1$. Then, we said that the operator $-\Delta + a$ on $H^1_0(\Omega)$ is coercive iff $\lambda_1^a(\Omega) > 0$.
Thereafter, my teacher said that we clearly have that, in the case where $-\Delta + a$ is coercive,
$$I_a(u) \ge C \int_\Omega|\nabla u|^2 = C\|u\|^2_{H^1_0(\Omega)}.$$
for some $C > 0$. Nevertheless, it doesn't seem so obvious to me, because we clearly have
$$I_a(u) \ge \lambda_1^a(\Omega) \int_\Omega u^2,$$
by the definition of $\lambda_1^a(\Omega)$, but how can we deduce the other inequality from this one ? Any help ?
 A: An integration by parts result
\begin{align}
b(u,v) := \int_{\Omega} (-\Delta u v+auv) &= -\int_{\partial\Omega} (\nabla u \cdot n)v+ \int_{\Omega} (\nabla u \cdot \nabla v+auv) \\
&= \int_{\Omega} (\nabla u \cdot \nabla v+auv)  \quad \forall v \in H^1_0(\Omega).
\end{align}
Then combining the $b: H^1_0(\Omega) \times H^1_0(\Omega) \to \mathbb{R} $ coercivity with an eigenvector $u \in H^1_0(\Omega)$ results.
$$\int_{\Omega} \lambda^a(\Omega) u^2 =  b(u,u)\geq C\|u\|_{H^1_0(\Omega)}^2$$
The equation above implies that  $\lambda^a(\Omega)\geq 0$. Note if $\lambda^a(\Omega)=0$ implies that $u=0$ but that is a contradiction because $\lambda^a(\Omega)$ is an eigenvalue. Conversely,
suposse that $\lambda^a(\Omega)>0$, let $u_\lambda$ an eigenvector in $H^1(\Omega)$ minimizing $I_a$ under the constrain $\|u\|_{L^2(\Omega)}=1$ then
\begin{align}
b(u,u)=I_a(u) \geq I_a(u_\lambda) = \lambda^a(\Omega) \|u_\lambda\|^2_{L^2(\Omega)}= \lambda^a(\Omega) \quad \forall u \in H^1_0(\Omega).
\end{align}
Now taking $u : = \frac{v}{\|v\|_{H^1_0(\Omega)}} \quad \forall v \in H^1_0(\Omega)$ the coercivity follows.
