proof of $\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le \frac{f(u)-f(t)}{u-t}$ for convex $f$

We will prove

$$\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}.$$(1)

The proof of the right-most inequality is similar.

Suppose (1) does not hold. Then for some $s,t,u$,

$$\frac{f(t)-f(s)}{t-s}>\frac{f(u)-f(s)}{u-s}.$$(2)

This inequality is just the statement of the slope of the linesegment $\overline{AB},A=(t,f(t)),B=(s,f(s))$, being larger thanthe slope of the segment $\overline{CB},C=(u,f(u))$. Since $t$ isbetween $s$ and $u$, and $f$ is continuous, this implies

$$f(t)>h(x)=\frac{f(u)-f(s)}{u-s}(x-s)+f(s),$$(3)

. This contradicts convexity of $f$ on $(a,b)$. Hence,(2) is false and (1) follows.

Note that we have tacitly use the fact that $x=\lambda u+(1-\lambda )s$ and $h(x)=\lambda f(u)+(1-\lambda )f(s)$ for some $\lambda$.