Prove $(x+y)^a\leq x^a+y^a$ if $0Prove $(x+y)^a\leq x^a+y^a$ if $0<a\leq1$ and $x,y\geq0$
I need to prove this step for a bigger question. It should be quite basic but I just have no idea...

The equation is trivial if $x=0$ or $y=0$, so assume neither are $0$.
For $z\ge0$, consider
$$
f(z)=(1+z)^a-z^a\tag{1}
$$
Note that $f(0)=1$. Furthermore, since $a-1\le0$,
$$
\begin{align}
f'(z)
&=a\left((1+z)^{a-1}-z^{a-1}\right)\\
&\le0\tag{2}
\end{align}
$$
Thus, for $z\ge0$ and $0\lt a\le1$
$$
(1+z)^a-z^a\le1\tag{3}
$$
Set $z=y/x$ and multiply $(3)$ by $x^a$ to get
$$
(x+y)^a\le x^a+y^a\tag{4}
$$
