How does the three chords inequality lead to the bound on the right derivative?
Recall the three chords inequality is that, for $a<x<b$ and $g$ convex, $$ \frac{g(x)-g(a)}{x-a}\leq\frac{g(b)-g(a)}{b-a}\leq \frac{g(b)-g(x)}{b-x} $$
Perhaps we just say that, by rewriting the three chords inequality as , $$ \frac{g(x)-g(a)}{x-y}\leq\frac{g(y)-g(a)}{y-a}\leq \frac{g(z)-g(a)}{z-a} $$ where $x< a < y\leq z$
and the above gives us that the right derivative is decreasing (since it is the formula for right derivative, and it is lower or $y\leq $(
The function $\delta:(0,\infty)\to \Bbb R$, $\delta(h)=\frac{g(a+h)-g(a)}h$ is weakly increasing by (the first half of) the three-chord inequality, and $g_+'(a)=\lim_{h\to 0^+} \delta(h)$. Therefore $g_+'(a)\le \delta(h)$ for any single $h>0$.