Heisenberg uncertainty principle in $d$ dimensions.

Suppose $f(x)$ is a $d$-dimensional real function and $\int_{R^{d}}|f(x)|^2dx=1$. Show that

$$ (\int_{R^{d}}|x|^2|f(x)|^2dx)(\int_{R^{d}}|\xi|^2|\hat f(\xi)|^2d\xi)\geq\frac{d^2}{16\pi^2}$$

I derived that $$1=\int_{R^{d}}x(\frac{d}{dx})|f(x)|^2dx$$
but I lost my way. I need your help.

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Consider the equation
$$
\sum_{i=1}^n\frac12x_i\frac{\mathrm{d}}{\mathrm{d}x_i}|f|^2=\mathrm{Re}\left(\nabla f\cdot\overline{xf}\right)\tag{1}
$$
Integrating $(1)$ over $\mathbb{R}^n$ and then integrating by parts on the left side:
$$
\begin{align}
\frac n2\|f\|_2^2
&=\mathrm{Re}\left(\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right)\\
&\le\left|\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right|\\[6pt]
&\le\|\nabla f\|_2\|xf\|_2\\[9pt]
&=2\pi\|\xi\hat{f}\|_2\|xf\|_2\tag{2}
\end{align}
$$
Thus,
$$
\|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{3}
$$
The last inequality says that the $L^2$ support radius for $f$ and $\hat{f}$ cannot have a product less than $\frac{n}{4\pi}$. This inequality is sharp as can be seen using the function $f(x) = e^{-\pi x\cdot x}$, whose Fourier Transform is itself, and whose $L^2$ support radius is $\sqrt{\frac{n}{4\pi}}$.