How to prove for $s<1,|a+b|^s\le|a|^s+|b|^s$
for positive $a,b$ and $s \in (0,1)$ $$ a^s +b^s = s\left(\int_0^a x^{s-1}dx+\int_0^b x^{s-1}dx \right) \\ \ge s\left(\int_0^a x^{s-1}dx+\int_0^b (x+a)^{s-1}dx \right) \\ = s\left(\int_0^a x^{s-1}dx+\int_a^{a+b} x^{s-1}dx \right) \\ = s\left(\int_0^{a+b}x^{s-1}dx \right) \\ = (a+b)^s $$
Hint: Consider the function $f(x)=(1+x)^s-1-x^s$ for $x\geq 0$