<s>[INST] Quick way to see that $\sqrt{1+x^2+y^2}^{-4}$ is $L^2$-integrable Is there a quick trick to determine that a function like $$f(x,y):=\sqrt{1+x^2+y^2}^{-4}$$ (which is essentially a negative power of the norm on $\mathbb R^d$ with the singularity at $0$ removed) is in $L^2(\mathbb R^2)$? $f$ is clearly bounded (it takes valeus in $[0,1]$), but that doesn't help here.
 [/INST] Hint
$$
\sqrt{1+x^2+y^2}^{-4}=\frac{1}{(1+x^2+y^2)^2}=\frac{1}{1+x^2+y^2} \cdot \frac{1}{1+x^2+y^2}\leq \frac{1}{1+x^2} \cdot \frac{1}{1+y^2}
$$
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