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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.

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You are assuming that $X \geq 0$ with probability $1$ or $X \leq 0$ with probability $1$. Your argument fails if $X$ takes both positive and negative values with non-zero probability.

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