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

Question

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 $$ |x -y|^{-2}= |x|^{-2}\left(1-2\frac{x \cdot y}{|x|^{2}}+ \frac{|y|^2}{|x|^2}\right)^{-1} $$

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

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Majda, Andrew J.; Bertozzi, Andrea L., Vorticity and incompressible flow, Cambridge Texts in Applied Mathematics. Cambridge: Cambridge University Press (ISBN 0-521-63057-6/hbk; 0-521-63948-4/pbk). xii, 545 p. (2002). ZBL0983.76001.

Answer 1

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 $\frac 3{4R^2} \not \in O(x^{-3})$

Answer 2

This follows because

1. $\frac1{1+t} = 1 - t + O(t^2)$ by Taylor expansion,
2. $\left| \frac{x\cdot y}{|x|^2} \right| < \frac R{|x|} = O(|x|^{-1}) $ by Cauchy-Schwarz and the given estimates, and
3. $|y|^2/|x|^2 = O(|x|^{-2})$ by the given estimates.

This gives

$$ \frac1{1-2\frac{x \cdot y}{|x|^{2}} + \frac{|y|^2}{|x|^2} } = \frac1{1+O(|x|^{-1})} = 1 + O(|x|^{-1})$$ which proves the result once you sub into the given identity for $|x-y|^{-2}$.

Remark on Ross's answer / interpretation - the asymptotic is as $x\to\infty$ under the conditions $|x|>2R$ and $|y|<R$ for a fixed constant $R$, so fixing $|x|=2R$ isn't allowed (but $|y|=cR$ is). The above is actually dimension independent.