I can't seem to finish the proof that for all $x \in \mathbb{R}_{++}$ (strictly positive reals) and $\{a \in \mathbb{R} | a \leq 0 \text{ or } a \geq 1\}$, $f(x) = x^a$ is convex using the first order convexity condition *.
My work so far:
Assume $x, y \in \mathbb{R}_{++}$ and $x < y$. Then we need \begin{align} y^a &\geq x^a + ax^{a-1}(y-x)\\ y^a &\geq x^a + ayx^{a-1} - ax^a\\ (y/x)^a &\geq 1 + a(y/x) - a && (\text{$x$ is positive})\\ r^a &\geq 1 + a(r - 1) \end{align} with $r > 1$ because $y > x$.
Where I need help:
If $a = 0$ or $a = 1$, there's equality. For $a > 1$, $r^a$ is exponential in $a$ with $r > 1$ while $1 + a(r-1)$ is linear in $a$. Thus, as $a$ increases, it should be the case that $exponential > linear$. How can I make this intuition more rigorous?
I'm not sure how to handle the $a < 0$ case.
*First order convexity condition (in one dimension) that I'm working with is that a function $f$ is convex iff its domain is convex and the following inequality holds for all $x, y \in dom(f)$: $f(y) \geq f(x) + f'(x)(y - x)$.