Strictly convex if and only if derivative strictly increasing?

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 $$\frac{f(x_2)-f(x_1)}{x_2-x_1}<\frac{f(x_3)-f(x_2)}{x_3-x_2}$$