For any real random variable $X$, define
$$\|X\|\_{2,1}=\int\_0^\infty \sqrt{\Pr(|X|>t)}dt.$$
This quantity (it is not a norm) appears in various problems, e.g. the multiplier central limit theorem (see, e.g., Section 2.9 in [this book](https://link.springer.com/book/10.1007/978-1-4757-2545-2)) or in L-statistics (see, e.g., [this paper](https://arxiv.org/abs/1910.07572)). Problem 2.9.1 of the book cited above mentions the inequality $\|X\|\_{2,1}^2\ge E(X^2)/4$. I think we have actually better. For all $x\ge 0$,
$$\|X\|\_{2,1} \ge \int\_0^x \sqrt{\Pr(|X|>t)}dt\ge x \sqrt{\Pr(|X|>x)},$$
which implies that
$$\begin{array}{rcl}
\|X\|\_{2,1}^2 & = & \int\_0^\infty \|X\|\_{2,1} \sqrt{\Pr(|X|>x)}dx \\
& \ge & \int\_0^\infty x \Pr(|X|>x)dx \\
& =& E[X^2]/2.
\end{array}$$
My question is: is this bound sharp (I don't think it is)? If not, what is the best constant in the inequality?



Comparison between $\|X\|_2$ and $\|X\|_{2,1}$
$$\begin{aligned}
\|X\|\_{2,1}^2&=\int\_0^\infty\int\_0^\infty ds\,dt\,\sqrt{P(|X|>s)}\sqrt{P(|X|>t)} \\ 
&\ge\int\_0^\infty\int\_0^\infty ds\,dt\,P(|X|>\max(s,t)) \\ 
&=E\int\_0^\infty\int\_0^\infty ds\,dt\,1(|X|>\max(s,t)) \\ 
&=E|X|^2=EX^2.
\end{aligned}$$
So, we have an improvement of your bound. Moreover, the lower bound $EX^2$ on $\|X\|\_{2,1}^2$ is exact: it is attained when $P(|X|=c)=1$ for some real $c$.


