Probabilistic inequality for an antisymmetric function Let $X$ be a random variable with $\mathbb P(-1 \leq X \leq 1) = 1$. Does $\mathbb E(X) \geq 0$ imply
$$\mathbb E[X(1-|X|)] \geq 0?$$
The function is antysymmetric around $0$ and has more probability mass on the positive side. Intuitively, it should be correct.
If not, is it true if I assume $\mathbb E(X) > 0$?

The answer is no.
A simple counter-example is letting $X\equiv 1$.
meaning, $X$ is a constant random variable equal to 1. 
We get $$\mathbb E[X(1-\vert X\vert)] = \mathbb E[X(1-\vert 1\vert)] = 0$$
