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!
 A: 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
$$ \| \sum_n v(n) \|_p \le \sum_n \|v(n)\|_p $$
A: This is kind of the complementary of the question here.
So, I adapt the answer of Saulspatz to your case.
I will adapt the notation, instead of writing $|X_n|$ for the sum elements, I will write $x_i$ so that $\forall i\; x_i>0$. Then, your question states:
\begin{equation}
\left(\sum{x_i}\right)^k \leq\sum{x_i^k}\quad \text{if } x_i>0\;\forall i\ \text{and } 0 \leq k\leq1
\end{equation}
Proof:
For the cases $k=0,1$ the proof is trivial, let us see now the cases $ 0 < k < 1$
If you write $$f(x_1,x_2,\dots,x_n)=\left(\sum{x_i}\right)^k-\sum{x_i^k},$$
then $f(0,0,...,0)=0$ and it's easy to show that all the first-order partial derivatives $\frac{\partial f}{\partial x_j}$ are strictly negative when $k>1.$  Indeed,$$
\frac{\partial f}{\partial x_j}=k\left(\left(\sum{x_i}\right)^{k-1}-x_j^{k-1}\right)<0$$ when $x_i>0 \forall i$.
Because $g(x)=x^{k-1}$ for $0<k<1, x>0$ is a decreasing function and $\sum x_i > x_j\; \forall j$ since $x_i>0\;\forall i$
