I also overcome with a different approach. Assume initially $g(t)\geq 0$.
We will consider the auxiliary function $\int_{a'}^x g(t)dt$ to that this function satisfies the convexity conditions in $a', that is,
$\int_{a'}^x g(t)dt\leq \frac{x-a'}{b'-a'}\left(\int_{a'}^x g(t)dt+\int_x^{b'} g(t)dt\right).$
To prove that observe
$\int_{a'}^x g(x)dt (1-\frac{x-a'}{b'-a'})=\frac{x-a'}{b'-a'}\int_{x}^{b}g(x)dt$
(Note that the integration is in t and g(x) is constant in that
Because $g$ is increasing we have $\int_{a'}^x g(t)dt\leq \int_{a'}^x g(x)dt$ and $\int_{x}^{b'} g(x)dt\leq \int_{x}^{b'} g(t)dt$
Then $\int_{a'}^x g(t)dt (1-\frac{x-a'}{b'-a'})\leq\frac{x-a'}{b'-a'}\int_{x}^{b}g(t)dt$
Rearranging it is the desired result.