# Elementary estimate for $|x-y|^{-2}$ When $x$ is large

I want to prove that $$|x -y|^{-2} \leq \frac{1}{|x|^2} + O(|x|^{-3})$$ when $$| y| \leq R$$, and $$|x| \geq 2R$$, $$R>0$$. The hint is to use that $$|x -y|^{-2}= |x|^{-2}(1-2\frac{x \cdot y}{|x|^{2}}+ \frac{|y|^2}{|x|^2})^{-1}$$

I tired different attempts, but I am not able to show it. I would really appreciate any hint. Thanks!

EDIT: I found this estimate in the book (Vorticity and incompressible flow ) by Majda and Bertozzi. Here is a screen shot • The dot product can be written as $|x||y|\cos\theta$ and you can make further approximations from there. – Everiana Jul 30 at 1:27
• Are you trying to prove the hint, or to use it to prove the problem? – Ross Millikan Jul 30 at 2:40
• Are $x$ and $y$ real numbers here or are they vectors? If they're vectors, the problem needs to be restated with $|x|^2$ and so on. – Ted Shifrin Jul 30 at 2:41
• They are vectors. Yes you are right! – Demha Jul 30 at 14:24

The statement is not correct, even if $$x,y$$ are reals. If the are vectors, let them both be along the same axis. Then if $$|x|=2R, |y|=R$$ and they are in the same direction $$|x-y|^{-2}=\frac 1{R^2}$$ while $$\frac 1{|x|^2}=\frac 1{4R^2}$$ and the difference is $$\frac 3{4R^2} \not \in O(x^{-3})$$
• The statement is about vectors. There is a typo in what I wrote. There should absolute value signs on the right-hand side. The statement should be $|x-y|^{-2} \leq \frac{1}{|x|^2} +O(\frac{1}{|x|^3})$. – Demha Jul 30 at 14:15
• The statement is true if $y$ is limited to a fixed size, but if it is allowed to grow with $x$ it is not. You can just count powers of $R$ in the term $\frac {x\cdot y}{|x|^2}$. They all cancel, so the term is the same size as the $\frac 1{|x|^2}$ out front. – Ross Millikan Jul 30 at 14:36