I would like if someone could help me out with this inequality, Let $S \subset \mathbb{R}^d$ be a Legesgue measurable set ( with non-zero measure) then, $$ \int_{S} |x| \ dx \geq C |S|^{\frac{d+1}{d}}, $$ where C is a constant independent of $S$, depending only on dimension.
One can see that this is true for balls, using polar coordinates, but I just don't know how to generalize this to any Lebesgue measurable set.
Thank you for your help.
Let $|A|$ denote the measure of $A\subset\mathbb R^d$.
For given $s>0$ consider the problem of minimizing $\int_S|x|dx$ subject to $|S|=s$. By Lyapunov's theorem about continuous vector measures, this minimum is attained by some $S$ for which $\nu(S)=(s,t)$, where $\nu$ is the vector measure $\nu: A\mapsto \int_A (1,|x|) dx=(\int_A 1, \int_A |x|)$. Now let $B$ be the $0$-centered ball with $|B|=s$, let $r$ be the radius of $B$. Write $S=S_A\cup S_B$ where $S_A = S\setminus B$ and $S_B=S\cap B$; write $B=(B\setminus S)\cup (B\cap S)$. Since $|B|=|S|$ we have $|B\setminus S|=|S_A|$. Since $|x|\ge r$ on $S_A$ and $|x|\le r$ on $B$, we have $$\int_B|x|=\int_{B\setminus S}|x| + \int_{B\cap S}|x| \le \int_{S_A}|x| +\int_{S_B}|x|=\int_S|x|.$$ But $\int_S|x|\le\int_B|x|$ by assumption, so $\int_S|x|=\int_B|x|$, and thus $|S_B|=|B\setminus S|=0$, so $S$ and $B$ differ by measure $0$.