We will prove \begin{equation}\label{3} \displaystyle \frac{f(t)-f(s)}{t-s}\leq\frac{f(u)-f(s)}{u-s}. \end{equation}The proof of the right-most inequality is similar.

Suppose () does not hold. Then for some $s, t, u$ , \begin{equation}\label{2} \displaystyle \frac{f(t)-f(s)}{t-s}>\frac{f(u)-f(s)}{u-s}. \end{equation}This inequality is just the statement of the slope of the line segment $\overline{AB}, A=(t, f(t)), B=(s, f(s))$ , being larger than the slope of the segment $\overline{CB}, C=(u, f(u))$ . Since $t$ is between $s$ and $u$ , and $f$ is continuous, this implies \begin{equation} f(t)>h(x)=\frac{f(u)-f(s)}{u-s}(x-s)+f(s), \end{equation}$s<x<u$ . This contradicts convexity of $f$ on $(a, b)$ . Hence, () is false and () 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 $\lambda$ .