Proof utilizing Chebyshev's inequality I'm being asked to show that $P(|X-\mu|\geq t) \leq \beta_4/t^4,$ where $\beta_4=E((X-\mu)^4)$.
I'm familiar with Chebyshev's Inequality, which similarly states that $P(|X-\mu|\geq t) \leq \sigma^2/t^2,$ and I know that $\sigma^2 = E((X-\mu)^2), $ but the furthest I get with this knowledge is that $P(|X-\mu|\geq t) \leq E((X-\mu)^2)/t^2.$
Apparently I somehow need to turn $E((X-\mu)^2)/t^2$ into $E((X-\mu)^4)/t^4.$ It doesn't seem like squaring this would help since $E((X-\mu)^2)^2 \neq E((X-\mu)^4),$ and I'd also have to square the $P(|X-\mu|\geq t)$ term also, which was not done in the final answer I'm asked to show. How can I prove this inequality?

In statistics, it's important to remember what a theorem says, so you can apply it.  But in mathematics it's usually more important to remember why a theorem is true, so that you really understand it and can generalize it.  This is that kind of situation.

Observe that when the power $p \ge 1,$ the gray area, weighted by the probability of $X,$ cannot exceed the area under the curve $y = |(x-\mu)/t|^p$ (yellow plus gray), weighted by the same probability distribution.

Write this inequality in terms of expectations.  The case $p=2$ proves Chebyshev's Inequality.  Pick a suitable value of $p$ to finish your proof.
