Exists $C$ where $\int_{\mathbb{R}^n} {{u^2}\over{|x|^2}}\,dx \le C \int_{\mathbb{R}^n} |Du|^2\,dx$, $u \in H^1(\mathbb{R}^n)$? For each $n \ge 3$, does necessarily exist a constant $C$ so that$$\int_{\mathbb{R}^n} {{u^2}\over{|x|^2}}\,dx \le C \int_{\mathbb{R}^n} |Du|^2\,dx$$for all $u \in H^1(\mathbb{R}^n)$?

Ideas. I probably want to use the fact that$$\left|Du + \lambda {x\over{|x|^2}}u\right|^2 \ge 0$$for each $\lambda \in \mathbb{R}$ somehow, but I do not what do from here on out. Could anybody help?
Let $u\in C_0^\infty(\mathbb{R}^n)$. Note that
$$0\leq \left|Du + \lambda {x\over{|x|^2}}u\right|^2=|Du|^2+\lambda^2\frac{|u|^2}{|x|^2}+2\lambda\sum_{i=1}^nu_{x_i}\frac{x_i}{|x|^2}u$$
Integrating over $\mathbb{R}^n$ (here we use the compact support of $u$), we get
$$0\leq \int_{\mathbb{R}^n} |Du|^2\;dx+\lambda^2\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx-\lambda n\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx+2\lambda\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx$$

Taking $\lambda$ such that $\lambda(n-2-\lambda)&gt;0$ (here we use $n\geq 3$) we get 
$$\int_{\mathbb{R}^n}\frac{|u|^2}{|x|^2}\;dx\leq C \int_{\mathbb{R}^n}|Du|^2\;dx$$
with $C=\frac{1}{\lambda(n-2-\lambda)}$.

As we can see here, a density argument extends the result for $u\in H^1(\mathbb{R}^n)$.
You can take $C = 4/(n-2)^2$. This is Hardy inequality.