Chebyshev's inequality. Let $X$ be a random variable. Let $\phi$ be a strictly increasing function on $(0,\infty)$ and $\phi(u)=\phi(-u)$. Suppose $\mathscr{E}[\phi(X)]<\infty$. Then $\forall\,u>0$, $$\mathscr{P}\{|X|\geq u\}\leq\frac{\mathscr{E}[\phi(X)]}{\phi(u)}.$$
See List of Inequalities.
$\bullet$ Proof.
Write the expectation as integral formulae. $$\begin{array}{rl}\mathscr{E}[\phi(X)]
&=\int\phi(x)\,d\mathscr{P}\\
&\geq\int_{|x|\geq u}\phi(x)\,d\mathscr{P} \\
&\geq\int_{|x|\geq u}\phi(u)\,d\mathscr{P}=\phi(u)\mathscr{P}\{|X|\geq u\}. \end{array}$$Thus, $$\mathscr{P}\{|X|\geq u\}\leq\frac{\mathscr{E}[\phi(X)]}{\phi(u)}.$$
$\Box$
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