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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!

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    I have tried starting with the Minkovsky inequality..but I don't seem to get anywhere.2012-11-28

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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 $

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    Thank you very much! I have understood your proof, and I agree with it, but I have one problem: I don't seem to understand how the Minkovsky inequality for counting measure holds. For yn = 1 it is trivial..but in the more general case?2012-11-28