Problem 11A.9(c) in Spivak's Calculus (4th edition) asks the following (I'm paraphrasing):
Suppose $f$ is convex. Show that $f'(a)$ exists iff $f_+'(x)$ is continuous at $a$.
($f_+'(x)$ is the right-derivative at $a$.) I worked out a proof of this fact, and found the $\implies$ direction particularly laborious. Then I consulted the answer book (to the third edition, where the problem also appears), and found the line of reasoning in the picture below. It appears to be a big mistake: they have two values $a>b$, and they've written $\frac{f(a)-f(b)}{a-b} < f_+'(b)$. Surely the inequality should be the other direction, right? Or am I very confused?
Normally I would assume it's a simple mistake, but usually Spivak's book is pretty careful, and here it seems the whole idea of the proof is wrong.
You are correct. Since the slopes are increasing on a convex function then:
$$\frac{f(a)-f(b)}{a-b}\ge\frac{f(b+h)-f(b)}{h}$$
where $h>0$ and small enough that $b+h<a$. Letting $h$ tend to $0$ gives the inequality that you describe. However, the inequality may not be strict unless the function is strictly convex.