Probabilistic inequality for an antisymmetric function

Let $X$ take the values $-\frac 1 2$ with probability $\frac 1 4$ and $1 $ with probability $\frac 3 4$. You can check that $EX >0$ but $E(X(1-|X|) <0$.

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 $$\mathbb E[X(1-\vert X\vert)] = \mathbb E[X(1-\vert 1\vert)] = 0$$