Proof of the power of sums smaller than sum of powers I have stumbled upon this problem which keeps me from finishing a proof:
$(\sum_{n} {|X_n|})^a \leq \sum_{n} {|X_n|}^a$,
where $n \in \mathbb{N}$ and $ 0 \leq a \leq 1 $
I have no idea how to prove this. It is something like the Cauchy-Schwarz inequality which applies in case $0 \leq a \leq 1$?
Any tip is welcome.
Thanks!

I'll leave the case $a=0$ to you.  Otherwise, let $a = 1/b$, $b \ge 1$.  If $y_n = |X_n|^a$, we have
$|X_n| = y_n^b$, and your inequality says
$$ \left(\sum_n y_n^b\right)^{1/b} \le \sum_n y_n$$
which is essentially Minkowski's inequality for counting measure: if 
$v(n)$ is the vector with $v(n)_n = y_n$, $v(n)_j = 0$ otherwise, then $\|v(n)\|_p = y_n$, and your inequality becomes
$$ \| \sum_n v(n) \|_p \le \sum_n \|v(n)\|_p $$
