Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.
A hint is given that for $0 \le \alpha \le \beta$ there exist a number $\xi \in (\beta, \alpha + \beta)$ s.t. $$(\alpha + \beta)^p - \beta^p = p \xi^{p-1}\alpha \le \alpha^{p}$$.
I see that by proving the above hint the result holds for positive numbers.
In order to prove the above hint I've considered the Intermediate value theorem, however I've not yet succeeded.
We want to show $$1\le \left(\frac{|x|}{|x|+|y|}\right)^p+\left(\frac{|y|}{|x|+|y|}\right)^p,$$ or $$1\le a^p+b^p\tag{1}$$ for $a,b\in [0,1]$, $a+b=1$, and $0< p\le1$. But (1) is clear since $$a\le a^p,b\le b^p.$$
We want to show that for non-negative $s$ and $t$ we have $(s+t)^p\le s^p+t^p$. Fix $s$, and let $$f(t)=s^p+t^p-(s+t)^p.$$ Note that $f(0)=0$. For $t\gt 0$ we have $f'(t)=p(t^{p-1}-(s+t)^{p-1})\ge 0$, so $f$ is non-decreasing.
