= Solution
Let $Z$ be a real <random variable> with finite <expectation> $m$ and finite <variance> $\sigma^2$. \b[The <Chebyshev inequality> is]
$$
\boxed{\mathbb P(|Z-m|\geq t)\leq\frac{\sigma^2}{t^2}\quad(t>0).}
$$
The proof is the pointwise comparison
$$
t^2\mathbf1_{\{|Z-m|\geq t\}}\leq(Z-m)^2.
$$
Taking <expectations> and using the definition of <variance> gives
$$
t^2\mathbb P(|Z-m|\geq t)\leq\mathbb E(Z-m)^2=\sigma^2.
$$
Division by the positive number $t^2$ proves the <Chebyshev inequality>, including the case $\sigma^2=0$. Equivalently, this is the <Markov inequality> applied to the nonnegative <random variable> $(Z-m)^2$.
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