Solution
= Solution
Let
$$
m=\frac{f(b)-f(a)}{b-a},\qquad
h(x)=f(x)-f(a)-m(x-a).
$$
Then $h(a)=h(b)=0$. Since $f$ is not linear, $h(c)\ne0$ for some $c\in(a,b)$. If $h(c)>0$, the <mean value theorem> on $[a,c]$ gives a point $\xi$ with
$$
h'(\xi)=\frac{h(c)-h(a)}{c-a}>0.
$$
If $h(c)<0$, apply it on $[c,b]$ to obtain
$$
h'(\xi)=\frac{h(b)-h(c)}{b-c}>0.
$$
In either case $h'=f'-m$, and therefore
$$
\boxed{f'(\xi)>\frac{f(b)-f(a)}{b-a}}.
$$