= Solution
For a nonnegative <random variable> $Y$ and $a>0$, <Markov inequality> states
$$
P(Y\geq a)\leq\frac{\mathbb EY}{a}.
$$
For a <random variable> $X$ of finite mean $\mu$ and <variance> $\sigma^2$, <Chebyshev inequality> states that for every $t>0$,
$$
\boxed{P(|X-\mu|\geq t)\leq\frac{\sigma^2}{t^2}.}
$$
To prove it, apply <Markov inequality> to the nonnegative variable $(X-\mu)^2$ at threshold $t^2$:
$$
P(|X-\mu|\geq t)=P((X-\mu)^2\geq t^2)
\leq\frac{\mathbb E[(X-\mu)^2]}{t^2}.
$$
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