Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while:
$\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n \;\rho(d_i)\;$ with $\rho(d_k) := |d_k|^p, \quad 0 < p < 1 $ and $ d_i \in \mathbb{R}$.
I encountered this situation while dealing with an energy-based regularization approach. While I do not expect this inequality to hold for $p > 1$, I basically believe that it does so for $0 < p < 1$, implying that the energy $\rho$ for the sum of all $d_i$ is lower than the sum of energies of the individual $d_i$, under such a quasi-norm. I attempted to show this for the less general case where $p := \frac{q}{r}$ is a rational number, by raising both sides to the $r$-th power:
$\mid\sum\limits_i^n d_i\mid^p \leq \sum\limits_i^n \mid d_i \mid^p \; \Leftrightarrow \; \mid\sum\limits_i^n d_i\mid^q \leq \left( \sum\limits_i^n \mid d_i \mid^p \right)^r = \sum\limits_i^n \mid d_i \mid^{pr} + u = \sum\limits_i^n \mid d_i \mid^{q} + u$,
where $u$ contains a sum of mixed terms. Next, I tried to get rid of the $q$ exponent in order to somehow make use of the triangle inequality. However, it seems like I am going in circles here.
Could somebody give me a hint on how to better approach this? Is there an existing inequality that I could use in proving this? I guess these are basics that have been studied before, I am just not sure what keywords to search for. Thank you very much!