Prove that for $\epsilon, \delta > 0$, $X$ a random variable,
$$E[\vert X \vert^2 ; \vert X \vert > \epsilon] \leq \frac{1}{\epsilon^\delta} E \vert X \vert^{2+\delta}$$
By Chebyshev's inequality we know that
$$P( \vert X \vert > \epsilon) \leq \frac{E \vert X \vert ^\delta}{\epsilon^\delta}$$
I'm guessing that's used here somewhere but now sure how?
It follows from the inequality $$ \varepsilon^\delta |X|^2I(|X|>\varepsilon)\leq |X|^{2+\delta} $$ which can be easily verified and taking expected values (Here $I$ is the indicator function).