We know that Chebyshev's inequality is:
$\Pr[|X-\mathbb E(X)| \ge t] \le \frac{\mathrm Var (X)}{t^2}$
And for all $r>0$ and $t>0$,
$\Pr[|X-\mathbb E(X)| \ge t]= \Pr[|X-\mathbb E(X)|^r \ge t^r]$
So by Markov's inequality,
$\Pr[|X-\mathbb E(X)| \ge t] \le \frac{\mathbb E[|X-\mathbb E(X)|^r] }{t^r}$
My first question is: in what situation, we can get a better bound that the Chebyshev's bound using the above bound? The second question is: how large should we I choose $r$ to get a best bound?