Suppose a random variable $X$ has unit variance i.e. $\sigma^{2} = 1$. Is there a positive constant $c > 0$ such that $$\mathbb{E}[\ | X - \mathbb{E}[X] | \ ] \ge c $$

My attempt of a solution is as follows:

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

Also by the triangle inequality,

$$P(|Y-\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k) - P(|\mathbb{E}[Y]| \geq k) $$

which implies

$$P(|Y-\mathbb{E}[Y]| \geq k) + P(|\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k) $$

or $$1/k^2 + \epsilon \geq P(|Y| \geq k) $$

where $\epsilon \in [0, 1]$.

From this statement, can it be concluded that $0$ is the greatest lower bound of $\mathbb{E}[|Y|]$?