Prove that for $0<p<1$, $|x-y|^p$ is a metric space on $\mathbb R^{n}$

Question

Define the function $f_p : \mathbb R^{n} \to \mathbb R^{n}$ for $n ≥ 2$ by $f_p(x) = \sum_{k=1}^{n} \lvert x\rvert^{p}$.

Show that for $0<p<1$, we get $d_b(x,y) = f_p(x-y)$ is a metric on $\mathbb R^{n}$.

I tried to use Minkowski's inequality to prove the triangle inequality: $$d_b(x,y) \leq d_b(x,z) + d_b(z,y)$$ by having $|x-y| \leq |x-z|+|z-y|$, which leads to $$|x-y|^p \leq (|x-z|+|z-y|)^{p},$$ but how do I show that $(|x-z|+|z-y|)^{p}\leq |x-z|^{p} + |z-y|^{p}$ (by raising both sides by $\frac{1}{p}$, but then doesn't make sense to do the binomial expansion if $\frac{1}{p}$ is not integer).

Answer 1

It suffices to show that, for every $a,b> 0$, $$ (a+b)^p\le a^p+b^p, $$ or equivalently $$ 1\le \left(\frac{a}{a+b}\right)^p+\left(\frac{b}{a+b}\right)^p. $$ But $$ \left(\frac{a}{a+b}\right)^p\ge \frac{a}{a+b} \quad\text{and}\quad \left(\frac{b}{a+b}\right)^p\ge \frac{b}{a+b}, $$ as $0<p<1$. Thus $$ 1= \frac{a}{a+b}+\frac{b}{a+b} \le \left(\frac{a}{a+b}\right)^p+\left(\frac{b}{a+b}\right)^p. $$