I am working on a problem and I have the weak formulation of Poisson's problem in $2$ spatial dimensions i.e. $u = u(x,y)$: $$a(u,v) = \ell(v) $$ where $$a(u,v)=\int_{\Omega}\nabla u\nabla v\quad\text{ and }\quad \ell(v) = \int_\Omega fv.$$ I have the following Sobolev space
$$\mathcal{H}_0^1(\Omega) = \left\{f:\Omega\subset\mathbb{R}^2\to\mathbb{R}\,\bigg| \int_\Omega |f|^2<\infty,\, \int_\Omega \left|\frac{\partial f}{\partial x}\right|^2<\infty\,, \int_\Omega \left|\frac{\partial f}{\partial y}\right|^2<\infty\,, f(\partial\Omega) = 0 \right\} $$ which I have equipped with the following norm $$\|u\|_{\mathcal{H}_0^1} = \left\|\frac{\partial u}{\partial x}\right\|_2+\left\|\frac{\partial u}{\partial y}\right\|_2 $$ where $\|\cdot\|_2$ is the usual $\mathcal{L}^2(\Omega)$-norm.
Now for my problem: I wanted to use Lax-Milgram theorem to show existence of solutions to the weak formulation, however I am unable to show that $a(\cdot,\cdot)$ is coercive, i.e.
$$a(u,u)\geq \alpha\|u\|_{\mathcal{H}_0^1}^2 $$ and I even suspect that it in fact isn't coercive.
Question: Is $a(\cdot,\cdot)$ coercive (with respect to the norm I have chosen)? If so, how do I show it?
Attempt: We have that $$a(u,u) = \int_\Omega |\nabla u|^2 = \int_\Omega \left[\left|\frac{\partial u}{\partial x}\right|^2 + \left|\frac{\partial u}{\partial y}\right|^2\right] = \left\|\frac{\partial u}{\partial x}\right\|_2^2+\left\|\frac{\partial u}{\partial y}\right\|_2^2$$ and $$\|u\|_{\mathcal{H}_0^1}^2 = \left(\left\|\frac{\partial u}{\partial x}\right\|_2+\left\|\frac{\partial u}{\partial y}\right\|_2\right)^2 >\left\|\frac{\partial u}{\partial x}\right\|_2^2+\left\|\frac{\partial u}{\partial y}\right\|_2^2 = a(u,u)$$ hence I believe that $a(u,u)< \|u\|_{\mathcal{H}_0^1}^2$, so the only chance seems to find $\alpha>0$ little enough, but I don't know how.
Since for each non-negative $a$ and $b$, $(a+b)^2\leqslant 2a^2+2b^2$, we have $$\|u\|_{\mathcal{H}_0^1}^2 = \left(\left\|\frac{\partial u}{\partial x}\right\|_2+\left\|\frac{\partial u}{\partial y}\right\|_2 \right)^2\leqslant 2\left\|\frac{\partial u}{\partial x}\right\|_2^2+2\left\|\frac{\partial u}{\partial y}\right\|_2^2=2\cdot a(u,u),$$ hence $\alpha=1/2$ does the job.