Prove Chebyshev's inequality, which states that if $\displaystyle X $ is a random variable with expectation $\displaystyle \mu $ and finite variance $\displaystyle \sigma^2 $, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2} $, for any $\displaystyle \varepsilon > 0 $