(Problem Solving) Proving $|x|^p +|y|^p \geq |x+y|^p$

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I have been given a problem as practice that I do not understand how to solve. Here is the question:

Given $x$,$p$ and $y$ are real numbers, and $0\leq p\leq 1$, prove that $|x|^p +|y|^p \geq |x+y|^p$

I have tried writing down everything I know that could help, which lead me to these inequalities:

$|a|^p + |b|^p \leq |a| + |b|$

$|a+b|^p \leq |a+b| $

$|a+b| \leq |a| + |b|$

However, these don't seem to be able to lead to the desired result. If anyone could tell me what I'm missing, that would be greatly appreciated!

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As suggested by one of the comments, use Minkowski inequality: That is, if $q\geq 1$ then we have \begin{align} (|x|^q+|y|^q)^{1/q} \leq |x|+|y|. \end{align} To show the inequality for $0<p<1$, it suffices to show \begin{align} |x|+|y| \leq (|x|^p+|y|^p)^{1/p}. \end{align}

Set $s = |x|^{p}$ and $t=|y|^{p}$. Then observe that \begin{align} |x|+|y| = t^q+s^q \end{align} where $q=1/p>1$. Then by Minkowski inequality, we have \begin{align} t^q+s^q \leq (t+s)^{q} \ \ \Longleftrightarrow \ \ \ |x|+|y| \leq (|x|^p+|y|^p)^{1/p}. \end{align}