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?