STATEMENT: let $\Omega \in \mathbb{R^d}$ be bounded and regular set, then for all $u \in H^1(\Omega)$ there exists $c > 0$ such that: $$||u||^2_{H^1(\Omega)} \le c \left(\int_{\Omega}|u|^2 +\int_{\Omega}|\nabla u+\nabla u^t|^2\right)$$
PROOF: we argue by contradiction,assume that there exists a sequence of functions $(u_n)_n$ in $H^1(\Omega)$ such that $$||u_n||^2_{H^1(\Omega)} \ge n \left(\int_{\Omega}|u_n|^2 +\int_{\Omega}|\nabla u_n+\nabla u^t_n|^2\right) $$
By homogeneity, we assume that $||u_n||^2_ {H^1(\Omega)} =1$ then we have the inequality: $$\int_{\Omega}|\nabla u_n+\nabla u_n^t|^2 \le \frac{1}{n} .$$
This means that $ (\nabla u_n+\nabla u^t_n)_n$ is bounded in $L^2(\Omega)$ but $(u_n)_n$ is bounded in $H^1(\Omega)$ so $\nabla u_n$ is bounded in $L^2(\Omega)$. Hence $\nabla u^t_n$ is bounded in $L^2(\Omega)$.
Now, $ (\nabla u_n+\nabla u^t_n)_n$ converges to $ (\nabla u+\nabla u^t)$ in the sense of distributions and to zero in $L^2(\Omega)$.
I deduce that in the sense of distribution I have: $$ (\nabla u+\nabla u^t) =0.$$
Is the above argument true? And how can I complete it to come up with a contradiction? Any help please