Given $X_i$ are (not necessarily independent) and $\max_{j \leq n} (E|X_j|^p)^{1/p} = \sigma_p < \infty$, $p>1$
Prove that : $E\ \max_{j \leq n} |X_j| \leq n^{1/p} \sigma_p$
Approach:
$$ E\ \max|X_i| \leq E\ \max|X_i|^P \leq \sum_{j \leq n} E|X_j|^p \leq n\ \max E|X_j|^p = n\sigma_p^p $$
By Hölder's inequality (or Lyapunov's inequality), $$ \operatorname E|X|\le(\operatorname E|X|^p)^{1/p}. $$ Hence, $$ \operatorname E\max_{i\le n}|X_i| \le(\operatorname E\max_{i\le n}|X_i|^p)^{1/p} \le\biggl(\operatorname E\biggl[\sum_{i=1}^n|X_i|^p\biggr]\biggr)^{1/p} \le (n\sigma_p^p)^{1/p} =n^{1/p}\sigma_p. $$