Let $X$ be a random variable with mean $\mu$ and let $E\left[(X-\mu)^{2 k}\right]$ exist. Show, with $d>0$, that $P(|X-\mu| \geq d) \leq E\left[(X-\mu)^{2 k}\right] / d^{2 k}$. This is essentially Chebyshev's inequality when $k=1$. The fact that this holds for all $k=1,2,3, \ldots$, when those $(2 k)$ th moments exist, usually provides a much smaller upper bound for $P(|X-\mu| \geq d)$ than does Chebyshev's result.