$\nabla \frac{1}{|x|}$ is Lipschitz continuous

Question

Let $x, y \in \mathbb{R}^3.$ How to prove $\nabla \frac{1}{|x|}$ is Lipschitz continuous for $|x|\geq1$?

I tried that WLOG suppose $|y|\geq |x|$, then

$$ \left| \frac{x}{|x|^3} - \frac{y}{|y|^3} \right| = \frac{1}{|x|^3} \left| x - \frac{|x|^3}{|y|^3}y \right| \leq C(?)|x-y|$$

I don't know how to construct $C$.

Answer 1

Hint: $$\frac{x|y|^3-y|x|^3}{|x|^3|y|^3}=\frac{(x-y)|y|^3}{|x|^3|y|^3}+\frac{y(|y|^3-|x|^3)}{|x|^3|y|^3}.$$Now factor the numerator in the second fraction on the right and note that $\left||y|-|x|\right|\le|x-y|.$

Answer 2

You can just try to be a bit more explicit and compute the square norm:

$$\left|\dfrac{x}{|x|^3} - \dfrac{y}{|y|^3}\right|^2 = \dfrac{(x^T|y|^3 - y^T|x|^3)(x|y|^3 - y|x|^3)}{|x|^6|y|^6} = \dfrac{|x|^4+|y|^4 - 2|x||y|x^Ty}{|x|^4|y|^4}.$$

Now use Cauchy-Schwarz as: $$x^Ty\leq |x||y|$$ then you are basically done.