Tchebycheff inequality: consider any probability density $p(x)$ for $(-\infty<x<\infty)$, with mean $\lambda$, and variance $\sigma^2$. Show that the total probability of outcomes that are more than $n \sigma$ away from $\lambda$ is less than $1 / n^2$, that is,
$$
\int_{|x-\lambda| \geqslant n \sigma} d x p(x) \leq \frac{1}{n^2}
$$
Hint. Start with the integral defining $\sigma^2$, and break it up into parts corresponding to $|x-\lambda|>n \sigma$, and $|x-\lambda|<n \sigma$.