I am reading through a proof that a Brownian motion trail almost surely has dimension $1 \frac{1}{2}$. While attempting to derive a lower bound for the dimension they use this inequality:
$(x^2 + h^2)^{-s/2} \leq x^{-s} + h^{-s}$
Where it is assumed that $1 < s < 2$.
It doesn't seem that complicated an inequality, but I'm not sure how to prove it, or even make a heuristic argument to justify it.
Edit: It seems I misread. The actual inequality they used is as follows:
$\int_0^\infty (x^2 + h^2)^{-s/2} dx \leq \int_0^h h^{-s} dx + \int_h^\infty x^{-s} dx$
For $s>0$ have both $$ \frac{1}{(x^2+h^2)^{s/2}} \le \frac{1}{x^s}$$ and $$ \frac{1}{(x^2+h^2)^{s/2}} \le \frac{1}{h^s}$$
so we can replace the integrand by either of these in the inequality.
The reason they split it up like and choose a different inequality for each that is so that both the integrals converge.