It is well known that $|E[X]|\leq E[|X|]$, where $X$ is a random variable taking values in $\mathbb{R}$ and $E[X]$ is its expected value.
Now, suppose that $|E[X]|<a$ for some $a\in \mathbb{R}$.
This is equivalent to $-a<E[X]<a$ and multiplying by -1 we get $a>-E[X]=E[-X]>-a$.
If $X\geq 0$ then $|X|=X$ so that $E[|X|]=E[X]<a$. If $X < 0$ then $|X|=-X$ so that $E[|X|]=E[-X]<a$.
In conclusion, this would show that $|E[X]|<a$ implies $E[|X|]<a$. But since this works for any $a\in \mathbb{R}$ then $|E[X]|< E[|X|]$ would be impossible, violating the well-known inequality.
There must be some beginner's mistake in my elaboration and I can't seem to find it.