Let $f$ be measurable, $\epsilon\gt0$, and $p\gt0$. Prove that: $\mu(\{|f|\geq\epsilon\})$ $\leq$ $1/\epsilon^p\int|f|^pd\mu$.
Let $E=\{|f|\geq\epsilon\}.$ Thus $E^c=\{|f|\lt\epsilon\}$ and we have:
$\int|f|^pd\mu=\int_E|f|^pd\mu+\int_{E^c}|f|^pd\mu =\int|f|^p1_Ed\mu +\int|f|^p1_{E^c}d\mu$.
I can deduce that $\int|f|^p1_Ed\mu\geq\int\epsilon^p1_Ed\mu$ which means $1/\epsilon^p\int|f|^pd\mu\geq\int1_Ed\mu=\mu(E).$
After this I am completely stuck. How do I deal with the integral over $E^c$? Is everything I have said valid up to this point? (i.e. can I split up the integral the way I have?)
You have basically already solved the exercise. You have that if $E = \{x\mid |f(x)|\geq \varepsilon\}$ then $0\leq \varepsilon^p 1_{E}\leq |f|^p1_{E}\leq |f|^p$ and therefore
$$\varepsilon^p\mu(E) = \int \varepsilon^p1_{E}d\mu \leq \int|f|^pd\mu.$$
Note that you have shown that
$$\varepsilon^p\mu(E)\leq \int_E |f^p|d\mu$$
notice now that $\int_E |f^p|d\mu \leq \int |f^p|d\mu$ and you are done.