How to show that $(a+b)^p\leq a^p + b^p$ for $a,b\geqslant 0$ and $0

Question

Rudin-Functional Analysis p.37

Let $a,b$ be nonnegative reals and $0<p<1$.

Why is it that $(a+b)^p\leq a^p + b^p$?

Since $x^p$ is not convex, I'm not sure why this inequality must hold.

Answer 1

$f(a) = (a+b)^p - a^p - b^p$ on $[0,\infty)$. We have $f'(a) = p((a+b)^{p-1} - a^{p-1}) < 0$. Thus $f(a) \leq f(0) = 0 \Rightarrow (a+b)^p \leq a^p+b^p$.