Suppose $f$ is strictly convex on $(a,b)$, let $x_1<x_2<x_3<x_4<x_5$, $x_i\in(a,b)$
By strictly convex, we have $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}<\frac{f(x_4)-f(x_3)}{x_4-x_3}<\frac{f(x_5)-f(x_4)}{x_5-x_4}$$
Let $x_2\to x_1^+, x_4\to x_5^-$, we have $f'(x_1)<f'(x_5)$. So $f'$ strictly increasing.
The other side is true by using Mean value theorem.
For $x_1<x_2<x_3$, $x_i\in(a,b)$ since $f$ is differentiable,
$\exists c_1\in(x_1,x_2)$, s.t. $\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c_1)$,
$\exists c_2\in(x_2,x_3)$, s.t. $\frac{f(x_3)-f(x_2)}{x_3-x_2}=f'(c_2)$,
Since $f'$ is strictly increasing, $f'(c_1)<f'(c_2)$, hence $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$
Lemma. For $\forall \,u \lt v \lt w \in (a,b)\,$:
$$
\frac{f(v)-f(u)}{v-u} \le \cfrac{f(w)-f(v)}{w-v}
$$
This simply formalizes the intuition that adjacent chords on the graph of a convex function have non-decreasing slopes, and follows directly from the definition of convexity with $\lambda = \frac{w-v}{w-u} \in [0,1]$ and $1-\lambda=\frac{v-u}{w-u}\,$:
$$
\require{cancel}
\begin{align}
& f\big(\lambda u + (1-\lambda)w)\big) \le \lambda f(u) + (1-\lambda)f(w) \\
\iff \quad & f\left(\frac{(\cancel{w}-v)u+(v-\cancel{u})w}{w-u}\right) \le \frac{w-v}{w-u} f(u) + \frac{v-u}{w-u} f(w) \\
\iff \quad & f(v) \big((w-v)+(v-u)\big) \le (w-v)f(u) + (v-u)f(w) \\
\iff \quad & \big(f(v)-f(u)\big)(w-v) \le \big(f(w)-f(v)\big)(v-u)
\end{align}
$$
Proof. Since $f$ is increasing and not constant there exists $\exists \,c \in (a,b)$ such that $f(c) \gt f(a)$. For $\forall \,x_1,x_2$ such that $a\lt c\lt x_1\lt x_2\lt b$ applying the previous lemma twice gives:
$$
\frac{f(x_2)-f(x_1)}{x_2-x_1} \ge \frac{f(x_1)-f(c)}{x_1-c} \ge \frac{f(c)-f(a)}{c-a} \gt 0
$$
The above proves that $f$ is strictly increasing on $[c,b)$ since $x_1 \gt c \implies f(x_1) \gt f(c)$ and $x_2 \gt x_1 \gt c \implies f(x_2) \gt f(x_1)$.
Best Answer
Take $$ f(x) = \log(1+x) $$ Then $f(0)=0$, $f'(x) = 1/(1+x)>0$, but $f''<0$, so $f$ is concave but satisfies all the hypotheses.