If $f\in L^2(\mathbb{R})\cap C^1(\mathbb{R})$, establish an inequality involving the Fourier transform. Assume $f\in L^2(\mathbb{R})\cap C^1(\mathbb{R})$. I would like to show that
$$
\left(\int_{\mathbb{R}} x^{2}|f(x)|^{2} d x\right)\left(\int_{\mathbb{R}} \xi^{2}|\hat{f}(\xi)|^{2} d \xi\right) \geq \frac{1}{16 \pi^{2}}\left(\int_{\mathbb{R}}|f(x)|^{2} d x\right)^{2}
$$
where $\hat{f}(\xi)=\int_\mathbb{R} f(x)e^{-2\pi i\xi x}dx$ is the Fourier transform.
Analysis is definitely not my strong suit so here goes, and Fourier analysis is still new to me. Also, I perused the questions like this and that to see what techniques are used.
The right hand side looks like an inner product of $f(x)/4\pi$ with itself in $L^2(\mathbb{R})$ and then squared. Applying the Cauchy-Schwarz inequality does not look reasonable in this case. At the very least, I can take the square root of both sides, I can rewrite this as
$$
\left \langle \frac{f(x)}{4\pi},\frac{f(x)}{4\pi} \right \rangle\leq \|xf(x)\|_2\|\xi\hat{f}(\xi)\|_2.
$$
Naively applying the Cauchy-Schwarz inequality, I get
$$
\left \langle \frac{f(x)}{4\pi},\frac{f(x)}{4\pi} \right \rangle\leq \|f(x)/4\pi\|_2\|f(x)/4\pi\|_2.
$$
I suspect I need to use Fourier inversion in some nontrivial manner. In this case, $f^2\in L^1$ so Fourier inversion says that $\hat{f^2}^\vee=f^2$ where the $\vee$ indicates the Fourier transform inverse given by $g^\vee=\hat{g}(-x)$. Then, I could appeal to the Hausdorff-Young inequality. My only issue here though is that I have no clue where the $x^2,\xi^2$ terms are coming from. Actually, I'm not even sure how I can cancel out the $\frac{1}{4\pi}$ terms from the upper bound I get.
A hint is preferred, but a general suggestion on how to approach similar problems is even better. Thanks.
 A: In case $\int_\mathbb{R} \vert x \vert^2 \vert f(x) \vert^2 dx =\infty$, then the inequality is trivial. Thus, we assume for the rest $\int_\mathbb{R} \vert x \vert^2 \vert f(x) \vert^2 dx <\infty$.
By the fundamental theorem of calculus we get
$$ \int_\mathbb{R} \vert f (x) \vert^2 dx = \int_\mathbb{R} \overline{f}(x) x \frac{1}{x} \left( f(x_0) + \int_{x_0}^x f'(y) dy \right)dx. $$
First we consider the second term. We have
$$ \int_\mathbb{R} \overline{f}(x) x \frac{1}{x}  \int_{x_0}^x f'(y) dy dx
\leq \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)^{1/2}\left( \int_\mathbb{R} \frac{1}{\vert x \vert} \int_{x_0}^x \vert f'(y) \vert^2  \right)^{1/2}
\leq \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)^{1/2}\left( 4\int_\mathbb{R} \vert f'(x) \vert^2 dx \right)^{1/2}, $$
where we used Hardy's inequality to pass to the last line.
For the second term we note that we just need to show that $f\in L^1(\mathbb{R})$, then we just choose $x_0$ such that $f(x_0)$ gets smaller and smaller to eliminate this term. For this we use Cauchy-Schwarz and that $f$ is locally bounded
$$ \int_\mathbb{R} \vert f(x) \vert dx = \int_{[-1, 1]} \vert f(x) \vert dx + \int_{\mathbb{R}} \vert x \vert \vert f(x) \vert \frac{1}{x} dx
\leq 2 \sup_{x\in [-1,1]} \vert f(x) \vert + \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 \right)^{1/2} \left( 2 \int_1^\infty \frac{1}{x^2} dx \right)^{1/2} <\infty. $$
Hence, we have shown so far
$$ \left( \int_\mathbb{R} \vert f (x) \vert^2 dx \right)^2 
\leq 4 \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)\left( \int_\mathbb{R} \vert f'(x) \vert^2 dx \right). $$
Finally we note that by Parseval's identity we have
$$ \int_\mathbb{R} \vert f'(x) \vert^2 dx = \int_{\mathbb{R}} \vert \widehat{f'}(\xi) \vert^2 d\xi. $$
Now use that $\widehat{f'}(\xi)= 2\pi i \xi \widehat{f}(\xi)$ to conclude.
