Show that $\mathbb{E}[ | X - \mathbb{E}[X] | ] \ge c > 0$ if $\mathbb{V}(X) = 1$? I have a random variable $X$ with unit variance, i.e. $\mathbb{V}(X) = 1$. Is there a positive constant $c > 0$ such that
$$\mathbb{E}[\ | X - \mathbb{E}[X] | \ ] \ge c > 0  ?$$
How do we prove it? If it is false, can we find a counter-example?

Let $Y = X-\mathbb{E}(X)$. Then by Chebyshev's inequality, for any $k > 0$, $P(|Y-\mathbb{E}[Y] | \geq k) \leq 1/k^{2} \geq P(|Y| \geq k)$. By Chebyshev's inequality, think you can reach a contradiction.
