Show that if $||g||_2 = 1$, $\int_\mathbb{R} x^2 g^2(x) dx \cdot \int_\mathbb{R} \xi^2 |\mathcal{F}g|^2(\xi) dx \geq \pi / 2$

Question

I want to show that, given $g \in C^1(\mathbb{R})$ such that $g, xg, g' \in L^2(\mathbb{R})$ and $\int_\mathbb{R} g^2 d\lambda = 1$, we have: \begin{equation} \int_\mathbb{R} x^2 g^2(x) dx \cdot \int_\mathbb{R} \xi^2 |\mathcal{F}g|^2(\xi) dx \geq \pi / 2 \end{equation} First, I thought that it looked a little bit like the Hölder inequality, with: \begin{equation} \int_\mathbb{R} x^2 g^2(x) dx \cdot \int_\mathbb{R} \xi^2 |\mathcal{F}g|^2(\xi) dx \geq (\int_\mathbb{R} |x g(x) \cdot x (\mathcal{F}g)(x)| dx)^2) \end{equation} However, I have no idea how I could bound the second part. Could someone please help me?