Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.
Solution 1:
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 $$a\le a^p,b\le b^p.$$
Solution 2:
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.