Sum of real powers: $\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$ Let $\{x_i\}_{i=1}^{N}$ be positive real numbers and $\beta \in \mathbb{R}$.
Can we say that:
$$ \sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$$
I know that this holds if $\beta \in \mathbb{N}$. Does the above inequality have a name in case it's true?
 A: Note that this amounts to showing that 
$$
\left(\sum_{i=1}^{N}{x_i}^{\beta}\right)^{1/\beta} \leq \sum_{i=1}^N
 x_i
$$
This fact holds for all $\beta \geq 1$, and it is an instance of the Minkowski inequality.
A: If $\beta \ge 1$, this is a consequence of Jensen's inequality applied to the convex function $t \mapsto t^{\beta}$. 
If $\beta < 1$, this is false with $x_i = 1$.
A: The case $0<\beta<1$ we have re reversed inequality
$$\left(\sum_{i=1}^N x_i\right)^\beta\leq \sum_{i=1}^N x_i^\beta$$
Also, this can actually be used to prove the inequality of the present post. 
Here are some details (for the case $\beta>1$ choose $p=1$ and $q=\beta$).
A: If $0<\beta<1$, then the opposite inequality is true, since the function $x\mapsto x^\beta$ is subadditive in this case. See Prove variant of triangle inequality containing p-th power for 0 < p < 1 and An inequality concerning concave functions.
$$\left(\sum_{i=1}^{N}{x_i}\right)^{\beta} \leq \sum_{i=1}^{N}{x_i^{\beta}}.$$
For $\beta=1$ then there is equality
$$\sum_{i=1}^{N}{x_i^{\beta}} = \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}.$$
If $\beta>1$, then $x\mapsto x^\beta$ is superadditive and thus.
$$\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}.$$
In general, concave function $f(x)$ with $f(0)=0$ is subadditive and concave function $f(x)$ with $f(0)=0$ is superadditive. The proof can be found, for example, in:


*

*The Wikipedia article on subadditivity 

*This answer: Prove variant of triangle inequality containing p-th power for 0 < p < 1
See also Exercise 16.6.4, p.480, in M. Kuczma: An introduction to the theory of functional equations and inequalities. 
A: The best part about the inequality $$\sum_{n = 1}^{N}x_{i}^{\beta} \leq \left(\sum_{i = 1}^{N}x_{i}\right)^{\beta}\tag{1}$$ is that it is homogeneous in variables $x_{i}$ and hence we can assume without any loss of generality that $\sum x_{i} = 1$. Since each each $x_{i}$ is positive and they sum to $1$ it follows that each $x_{i}$ lies between $0$ and $1$ and hence $x_{i}^{\beta} < x_{i}$ if $\beta > 1$ and therefore $$\sum x_{i}^{\beta} < \sum x_{i} = 1 = 1^{\beta} = \left(\sum x_{i}\right)^{\beta}$$ If $0 < \beta < 1$ or $\beta < 0$ then we have $x_{i}^{\beta} > x_{i}$ and hence the inequality is reversed.
