If $|f(x)|\leq\frac{C}{(1+|x|^2)^N}$, prove its Fourier transform$\hat{f}$ is well-defined, smooth and in $\mathcal{L}^2(\mathbb{R}^n)$

Question

Given a measurable function $f:\mathbb{R}^n\longrightarrow\mathbb{C}$ with the property that for all natural number $N$, there exists a constant $C$ (depending on $N$) such that the inequality $|f(x)|\leq\frac{C}{(1+|x|^2)^N}$ holds for all $x$ in $\mathbb{R}^n$, how can I show that $f$ has a Fourier transform that (a) is well-defined, (b) is smooth, and (c) belongs to $\mathcal{L}^2(\mathbb{R}^n)$? I know that a function has a well-defined Fourier transform if it is integrable. The Fourier transform would belong to $\mathcal{L}^2(\mathbb{R}^n)$ if also the original function belongs to this space. Having the inequality stated above, is there a way of proving that $f$ belongs to both spaces $\mathcal{L}^1(\mathbb{R}^n)$ and $\mathcal{L}^2(\mathbb{R}^n)$? And how can one show that in this case, its Fourier transform is even smooth? The result would follow if $f$ were a Schwartz function. I have not learned about Sobolev spaces yet, so I need a method that does not use them. Thanks for your help!