Let $X$ be a real valued random variable. Does $\mathbb{P}( | X | \ge c) \le \frac{\text{Var}[X]}{c^2}$ hold for $c > 0$?
My counterexample:
Let $X \sim \text{Bin}(n,0)$ for $n \in \mathbb{N}$. Then we have $\text{Var}[X] = 0$ but (since $|X| = X$) \begin{align*} \mathbb{P}[X \ge 2] & = 1 - \mathbb{P}(X \le 1) + \mathbb{P}(X = 1) = 1 - \left(\sum_{k = 0}^{1} \binom{n}{k} 0^k 1^{1 - k}\right) + \binom{n}{k} 0^k 1^{1 - k} \\ & = 1 - 0 + 0 = 1 > 0. \end{align*}
Is this correct? Is there an easier counterexample?
Edit 1: Since the above counterexample is wrong, here's a second idea.
Let $X \in \text{Ber}(1)$. Then we have $\textsf{V}[X] = 1 \cdot (1 - 1) = 0$ but since $\mathbb{P}(X = k) \ne 0$ only for $k = 1$, we have $$ \mathbb{P}(X \ge 1) = \mathbb{P}(X = 1) + \mathbb{P}(X = 0) = 1 + 0 > 0. $$