Inequality involving the weak second moment I want to ask the following probability inequality:
Is it true that for any random variable $X\ge 0$, we have 
$$
\sup_{t>0}(t\mathbb E(X\mathbf 1_{X\ge t})) 
\le 
 2\sup_{t>0}(t^2 \mathbb P(X \ge t))?
$$
 A: This is an interesting inequality.
Let $f$ be the density of $|X|$.
Let 
$$g(t)=t\int_t^\infty xf(x)dx.$$
Then our inequality is
$$\sup_t g(t)\leq \sup_t t^2\int_t^\infty f(x)dx.$$
Let us assume that $\sup$ of the LHS is attained at some point $t=a$,
so that $g(a)$ be the maximal value of $g$.
(It will be easy to get rid of this assumption, as well as of the assumption
about existence of the density in the end). Our $g$ is bounded.
Then the RHS evaluated at the same point $a$ is
$$-2a^2\int_a^\infty\left(\frac{g(t)}{t}\right)^\prime\frac{dt}{t}.$$
I integrate this by parts and obtain
$$2a^2\left( g(a)a^{-2}-\int_a^\infty\frac{g(t)}{t^3} dt\right).$$
Now, using that $a$ is the maximum of $g$, I estimate this from below as
$$2g(a)-2g(a)a^2\int_a^\infty\frac{dt}{t^3}\geq g(a),$$
which completes the proof.
Getting rid of the assumptions made before is routine: just approximate
your distribution by a distribution satisfying those assumptions.
A: Notice that for a non negative random variable $Y$, we have $$
\mathbb E(Y)=\int_0^{\infty}\mu(Y\geqslant s)\mathrm ds.$$
Fix $t\geqslant 0$. We have for $s\leqslant t$ that $\{X\mathbf 1_{X\geqslant t}\geqslant s\}=\{X\geqslant t\}$ and if $s\gt t$ then $\{X\mathbf 1_{X\geqslant t}\geqslant s\}=\{X\geqslant s\}$. Consequently, we have 
$$\mathbb E(X\mathbf 1_{X\geqslant t})=t\mu(X\geqslant t)+\int_t^\infty\mu(X\geqslant s)\mathrm ds\leqslant t\mu(X\geqslant t)+\sup_{s\in\mathbb R}s^2\mu(X\geqslant s)\cdot \int_t^\infty s^{-2}\mathrm ds.$$
Multiplying by $t$ on both sides yields the wanted result. 
