"Obvious" integral inequality for radially decreasing function I would like to show the following inequality in $\mathbb{R}^2$:
$$
\iint_{B\left( x_0, R \right)} \frac{1}{\lVert x \rVert} \,\mathrm{d}A
\leq \iint_{B\left( 0,R \right)} \frac{1}{\lVert x \rVert} \,\mathrm{d}A.
$$
Here, $\mathrm{d}A$ denotes the usual Lebesgue measure on $\mathbb{R}^2$. Intuitively, this should be true since points closer to the origin should on average give larger values, but I'm having a lot of trouble proving it rigorously. Any guidance would be appreciated.

Let $A=B(x_0,R)\cap B(0,R)$ with $A_1=B(x_0,R)-A$ and $A_0=B(0,R)-A$.  Note that $A_1$ and $A_0$ have the same area.
$\int_{A_1} \frac{1}{||x||}dx \le \int_{A_0} \frac{1}{||x||}dx$ since $||x||\ge R$ for all points in $A_1$ while $||x||\le R$ for all points in $A_0$.
Add $\int_A \frac{1}{||x||}dx $ to both and get the desired inequality.
