Since $f''(x) \geq 0$, $f'(x)$ is monotone increasing (as you already observed). Let $x, y \in (a, b)$ and WLOG $x < y$. Therefore, $x < \frac{x + y}{2} < y$. Now, applying MVT in $\left(x, \frac{x+y}{2}\right)$ and $\left(\frac{x + y}{2}, y\right)$, we get $ f'(c_1) = \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}}\\ f'(c_2) = \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} $ for some $c_1 \in \left(x, \frac{x+y}{2}\right)$ and some $c_2 \in \left(\frac{x + y}{2}, y\right)$.
Now $c_1 \leq c_2$, therefore $f'(c_1) \leq f'(c_2)$. This implies, $ \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}} \leq \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} \\ f\left(\frac{x + y}{2}\right) \leq \frac{f(x) + f(y)}{2} $