Inequality concerning a variational integrator

Question

Let $u\in H^1_0((0,\pi))\setminus\{0\}.$ I want to show that $$\frac{\int_0^\pi u'^2(x)dx}{\int_0^\pi u^2(x)dx}\geq 1,$$ where $H_0^1((0,\pi))$ is the Hilbert space of functions $u\in L^2((0,\pi))$, such that $u(0)=u(\pi)=0$, with weak derivative $u'\in L^2((0,\pi))$ and inner product defined by $$\langle u,v\rangle = \int_I uv+u'v'\ dx.$$ What I've done: considering the minimization problem $$\lambda := \left\{\int_Iu'(x)dx: u\in H_0^1(I),\int_Iu^2(x)dx=1\right\}$$ we take a sequence $(u_n)$ in $H_0^1(I)$ such that $$\lVert u_n\rVert^2\rightarrow \lambda\quad \land\quad \int_0^\pi u_n^2(x)dx=1.$$ $(u_n)$ is bounded because we're working on a Hilbert space, therefore (by a theorem) it is possible to extract a subsequence of $(u_n)$ such that it converges weakly. Let that limit be $u_0$. By another theorem I know that $$\int_0^\pi u_0^2(x)dx=1\quad\land\quad\int_0^\pi u_0'(x)dx\leq\lim\inf\int_0^\pi u_n'^2(x)dx.$$ I feel like I'm close to the answer, but I can't seem to conclude. Any help? Thanks in advance.