I do not believe there is a name for your specific inequality (which I rewrite as following):
$$
|x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ }
$$
However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably many other inequalities after appropriate substitution and/or change of variables.
The closes call would probably be the generalized mean inequality:
$$
M_j\left( x_1, \dots, x_n \right) \leq M_i\left( x_1, \dots, x_n \right)
\quad \text{ whenever } \quad j<i.
\label{*} \tag{*}
$$
Here $M_k \left( x_1, \dots, x_n \right)$ is so-called power mean, which is defined as
$$
M_k(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^k \right)^{\frac{1}{k}}.
$$
! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\left(x_1, \dots, x_n \right) := \left(\,\chi, \gamma \right)$, we get
$$
\begin{aligned}
M_1\big(\left|\,\chi\right|, \left|\gamma\right|\big)
& = \dfrac{1 }{2}\big(\left|\,\chi\right| + \left|\gamma\right|\big)
= \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2},
\\
M_p\big(\left|\,\chi\right|,\left|\gamma\right|\big)
& = \left( \dfrac{1}{2} \left(\left|\,\chi\right|^p+\left|\gamma\right|^p\right)\right)^{\frac{1}{p}}
=
\Bigg(
\left(\frac{\left|\,\chi\right|}{2}\right)^p +
\left(\frac{\left|\gamma\right|}{2}\right)^p
\Bigg)^{\frac{1}{p}}.\\
\end{aligned}
$$
By $\eqref{*}$, we have
$$
\dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2} \le
\left(
\bigg(\frac{\left|\,\chi\right|}{2^{\frac{1}{p}}}\bigg)^p +
\bigg(\frac{\left|\gamma\right|}{2^{\frac{1}{p}}}\bigg)^p
\right)^{\frac{1}{p}} .
\label{**} \tag{**}
$$
Denoting $ x := 2^{-\frac{1}{p}}\chi, \ \ y := 2^{-\frac{1}{p}}\gamma$ and raising both sides of $\eqref{**}$ to the power $p$, we get
$$
\left|x\right|^{p}+\left|y\right|^{p} \le
\big(\left|x\right|+\left|y\right|\big)^{p}.
$$
To summarize, I believe that (strictly speaking) there is probably no name for your inequality.
However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.