If a RV \(X\) is non-negative with mean \(\mu\), then
\[ \text{P}(X \geq a) \leq \frac{\mu}{a}, \;\; \text{for all $a > 0$.} \]
Proof (For the discrete case):
\[ \begin{aligned} \mu&=\sum_x xp_x(x) \\ &=\sum_{0 \leq x < a} xp_x(x) + \sum_{x \geq a} xp_x(x) \\ &\geq 0 + \sum_{x \geq a} ap_x(x) \;\; \color{gray}{\leftarrow\text{Since $x$ is non-negative}} \\ &= a\sum_{x \geq a} p_x(x) \\ &= a\text{P}(X \geq a) \\ \end{aligned} \]
Similarly, for the continuous case:
\[ \begin{aligned} \mu &= \int_0^{+\infty} xf_x(x)dx \\ &=\int_0^a xf_x(x)dx + \int_a^{+\infty} xf_x(x)dx \\ &\geq 0 + \int_a^{+\infty} af_x(x)dx \\ &= a\int_a^{+\infty} f_x(x)dx \\ &= a\text{P}(X \geq a) \\ \end{aligned} \]
\[ \text{P}(X \geq a) \leq \frac{\mu}{a}, \;\; \text{for all $a > 0$.} \]
An alternative form: letting \(a=k\mu\), where \(k > 0\), we have
\[ \text{P}(X \geq k\mu) \leq \frac{1}{k}, \;\; \text{for all $k > 0$.} \]
If \(X\) is a RV with mean \(\mu\) and variance \(\sigma^2\), then
\[ \text{P}\big(|X - \mu| \geq c \big) \leq \frac{\sigma^2}{c^2}, \;\;\;\;\;\; \text{for all $c > 0$.} \]
Proof
Let \(Y=(X-\mu)^2\). We know \(Y\) is non-negative.
Apply the Markov inequality with \(a=c^2\), we have
\[ \text{P}\big((X-\mu)^2 \geq c^2\big) \leq \frac{\text{E}[(X-\mu)^2]}{c^2}=\frac{\sigma^2}{c^2} \]
Event \((X-\mu)^2 \geq c^2\) is identical to \(|X-\mu| \geq c\).
Thus, we have
\[ \text{P}\big(|X - \mu| \geq c\big) = \text{P}\big((X-\mu)^2 \geq c^2\big) \leq \frac{\sigma^2}{c^2} \]
\[ \text{P}(|X - \mu| \geq c) \leq \frac{\sigma^2}{c^2}, \;\; \text{for all $c > 0$.} \]
An alternative form: letting \(c=k\sigma\), where \(k > 0\). We have
\[ \text{P}\big(|X - \mu| \geq k\sigma \big) \leq \frac{\sigma^2}{k^2\sigma^2} = \frac{1}{k^2}, \;\; \text{for all $k > 0$.} \]