I need to show that $\|u\|_{H^1}^2\leq \|u\|_{L_2}\|u\|_{H^2},\forall u\in H^2\cap H^1_0$ using the Helmholtz equation: \begin{align*}-\Delta u+u&=f,\text{ in }\Omega\\u&=0,\text{ on }\partial\Omega\end{align*}
My attempt:
From the Helmholtz equation I got the bilinear form $a(u,v)=(\nabla u,\nabla v)+(u,v)$ where $(\cdot,\cdot)$ means $L_2$ inner product. From another proof, I know that $(u,v)\leq \|u\|_{H^{-1}}\|v\|_{H^1},\forall u\in H^{-1}, v\in H^1$ and so consequntly I said that $(u,u)\leq \|u\|_{L_{2}}\|u\|_{H^2}$ since both the $L_2$ and $H^2$ norms are greater than or equal to $H^{-1}$ and $H^1$ norms respectively.
Then I thought of getting a similar bound for the first term $(\nabla u,\nabla v)$ in the bilinear form and using the fact that this bilinear form is coercive to finally get to the desired inequality. But I cannot figure out how to get that upper bound.
Is this the correct way to prove the result? If so, how to get the upper bound for that first term?
If $u,v\in H^2(\Omega)\cap H_0^1(\Omega)$, then $$ \int_{\Omega}u\Delta v\,dx=-\int_{\Omega}\nabla u\cdot\nabla v\,dx. $$ Hence $$ \int_{\Omega}u\Delta u\,dx=-\int_{\Omega}|\nabla u|^2\,dx $$ and thus $$ \|u\|_{L^2}\|u\|_{H^2} \ge \int_{\Omega}u(u-\Delta u)\,dx=\int_{\Omega}(u^2+|\nabla u|^2)\,dx=\|u\|_{H^1}^2 $$
which implies tha