Let $s<0$ and $D$ be a bounded subset of $\mathbb R^d$ and assume that $B_{D}\subset \mathbb R^d$ is the ball centered at the origin with $|D|=|B_{D}|$ (here $|S|$ denotes the Lebesgue measure of the set $S\subset \mathbb R^d$)
For example: Let $D=[3,5]\subset \mathbb R$ and $B_D=[-1,1]$ Then $|D|=|B_{D}|=2.$
Question: How to show that $\int_{3}^5 (1+|\xi|^2)^{s} d\xi \leq \int_{-1}^1 (1+|\xi|^2)^s d\xi$ for $s<0$?
Note. If we the draw the graph of the integrand then it is clear from the picture. The integrand function is symmetric and decreasing. But I do not know how to justify rigorously.
Put $A_{\lambda}=\{x \in \mathbb R^d: (1+|x|^2)^{s/2} >\lambda \} \ (\lambda >0)$
Can we say that $|A_{\lambda} \cap D| \leq |A_{\lambda} \cap B_{D}|$ for $\lambda >0$ ?
Can we say that $\int_{D} (1+ |\xi|^2)^s d\xi \leq \int_{B_{D}} (1+|\xi|^2)^{s} d\xi$?
The integral over $D$ can be split up into the part contained in $B = B_D$ and the part not contained in $B$, and the may be done of the integral over $B$. That is, if $f(\xi) = (1 + |\xi|^2)^s$ then
$$ \int_D f(\xi)d\xi = \int_{D \cap B} f(\xi)d\xi + \int_{D \cap B^c} f(\xi)d\xi $$ and $$ \int_B f(\xi)d\xi = \int_{B \cap D} f(\xi)d\xi + \int_{B \cap D^c} f(\xi)d\xi. $$
To compare $\int_{D \cap B^c}$ to $\int_{B \cap D^c}$ we note on $D \cap B^c$ that $f(\xi) \leq 1/(1 + r^2)^s = c_r$ where $r = r(|D|)$ is the radius such that $|B_r(0)| = |D|$, and on $B \cap D^c$ that $f(\xi) \geq c_r$. It follows $$\int_{B \cap D^c} f(\xi)d\xi \geq |B \cap D^c| c_r = |D \cap B^c| c_r \geq \int_{D \cap B^c} f(\xi)d\xi.$$