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I have a question which seems related in some way to the isoperimetric inequality. I want to calculate $$\sup_{A \subset \mathbb R^n, \lvert A \rvert = 1} \int_{A} \frac{1}{\lvert y \rvert^{n-1}} dy$$ where $\lvert A \rvert$ denotes the Lebesgue measure of $A$.

Of course, I know the supremum should be achieved when $A$ is taken to be a ball centered at the origin (with radius $r_1$ chosen so that the volume is 1). This would give $$\int_{B_{r_1}(0)} \frac{1}{\lvert y\rvert^{n-1}} dy = \lvert \mathbb S^{n-1} \rvert r_1$$ where $\lvert \mathbb S^{n-1} \rvert$ is the surface area of $\mathbb S^{n-1} = \{x \in \mathbb R^n : \lvert x \rvert= 1 \}.$

Fixing $A \subset \mathbb R^n$ with $\lvert A \rvert = 1$, I tried covering $A$ with balls $B_k$ so that $$\sum^\infty_{k=1} \lvert B_k \rvert \le 1+\epsilon$$ where $\epsilon > 0$ is arbitrary. Then we would have $$\int_A \frac{1}{\lvert y\rvert^{n-1}} dy \le \sum^\infty_{k=1} \int_{B_k}\frac{1}{\lvert y\rvert^{n-1}} dy$$ but this didn't seem to lead anywhere.

Any help proving this would be appreciated!

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