Bounding $P(|X-\mu|\geq \alpha)$ using $E[|X|]$ Given $ X $ is a random variable with $E[X] = \mu < \infty$, show that for every $\alpha \in \mathbb{R} $ $$ P(|X-\mu|\geq \alpha) \leq \frac{2E[|X|]}{\alpha} $$ I tried using Chebyshev's inequality, but I don't know that $\text{Var}(X)$ exists, and even if it does exist, it only gives me $$ P(|X-\mu|\geq\alpha) \leq \frac{\text{Var}(X)}{\alpha^{2}}=\frac{E[X^2]-\mu^2}{\alpha^{2}} $$Which doesn't seem to help me very much.

Hint:
$$
E\big(|X-\mu|\big)\le E\big(|X|+|\mu|\big)
$$
by the triangle inequality.
(Note that you need to specify $\ \alpha>0\ $ for the inequality to be true, not merely $\ \alpha\in\mathbb{R}\ $, as you currently have)
