Show that $\int_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^n}$ is finite.
Here $|x|^2 := x_1^2+\dots+x_n^2$ is the norm of $x$. The statement is clearly true for $n=1$. For $n=2$ we have $$ \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(1+x_1^2+x_2^2)^2}\le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{(x_1^2+x_2^2)^2} \le \int_{\mathbb{R}^2} \frac{dx_1 dx_2}{x_1^2+x_2^2} \le\,\, ?$$ How do I proceed with this. I noticed that the integrandum is even, and I feel like I need to find an upper bound for this integral that is the $n$-th power of an integral that is finite and that can be easily computed (for example the $n=1$-integral); something like $$ \int_{\mathbb{R}} \frac{dx_1}{1+x_1^2}\int_{\mathbb{R}} \frac{dx_2}{1+x_2^2}\cdots\int_{\mathbb{R}} \frac{dx_n}{1+x_n^2}.$$ Thanks in advance.
$\frac 1 {1+|x|^{2}} = \frac 1 {1+x_1^{2}+x_2^{2}+...+x_n^{2}} \leq \frac 1 {1+x_i^{2}}$ for each $i$. Multiply these $n$ inequalities and integrate.