Prove the inequality $m(x \in \mathbb{R}^d: \Vert x \Vert >r)\leq \frac{1+r^2}{r^2}\int_{\mathbb{R}^d}\frac{\Vert x\Vert^2}{1+\Vert x \Vert^2}m(dx)$ [closed]

When I read the lecture notes about the Levy process, I get stuck in proving the inequality:

$m(x \in \mathbb{R}^d: \Vert x \Vert >r)\leq \frac{1+r^2}{r^2}\int_{\mathbb{R}^d}\frac{\Vert x\Vert^2}{1+\Vert x \Vert^2}m(dx),$

where m is a Borel measure.

I try to： $m(x \in \mathbb{R}^d: \Vert x \Vert >r)=\int_{\{x \in \mathbb{R}^d: \Vert x \Vert >r\}}m(dx)\leq \int_{\{x \in \mathbb{R}^d: \frac{\Vert x\Vert}{1+\Vert x \Vert}>r\}}m(dx),$ but how to get the inequality?

• Where do you get stuck? – uniquesolution Aug 12 '18 at 18:43
• You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Aug 12 '18 at 18:59
• @uniquesolution I forget how to get the inequlaity. Is it easy to get it? – user469065 Aug 12 '18 at 19:21
• @Shaun Hey! It is not my exercise and it appears in a lecture note about Levy process. Here is the Lecture – user469065 Aug 12 '18 at 19:29
• Your questions should come with context regardless of where they came from. – Shaun Aug 12 '18 at 20:04

Note that $f(u)=\frac{u^2}{1+u^2}$ is a strictly increasing function for positive $u$. Therefore: $$m(x \in \mathbb{R}^d: \| x \|> r)= m\left(x\in\mathbb R^d: \frac{\Vert x\Vert^2}{1+\Vert x \Vert^2}>\frac{r^2}{1+r^2}\right).$$ Apply Markov's inequality to get your result.