Using Mean Value Theorem to Prove Derivative Greater than Zero I'm working on a problem where at one point I have to show that for $x\ge a$,  
$$g (x) = \int_a^x f - (x-a) f \left({a+x \over 2} \right)$$, $g'(x) \ge 0$. 
Additional information: I know that $f''(x)\gt0$, $f'(x)<0$, and $f(x)\gt0$. 
Here is what I have so far:
$g'(x) = f(x) - f(a) - f({a+x \over 2}) - (x-a) f'({a+x \over 2})({1 \over 2})$
Using the Mean Value Theorem, I replaced $f(x) - f(a)$ with $(x-a) f'(x_0)$ where $x_0 \in [a,x]$, resulting in: $\displaystyle g'(x) = (x-a)\left[f'(x_0) - {1 \over 2} f'\left({a+x \over 2}\right)\right] - f\left({a+x \over 2}\right)$.
I am having trouble, however, going from this step (assuming I've taken the right steps thus far, and getting the result that $g'(x) \ge 0$.

if $f''(x)>0,x\in [a,b]$, then we know
$$\dfrac{1}{b-a}\int_{a}^{b}f(x)dx\ge f\left(\dfrac{a+b}{2}\right)\tag{1}$$
I think you want prove this well know inequality?
take $b\to x$,then
$$\int_{a}^{x}f(t)dt-(x-a)f\left(\dfrac{x+a}{2}\right)\ge 0$$
Indeed.for $(1)$ inequality we can use 
$$f(x)\ge f\left(\dfrac{a+b}{2}\right)+f'\left(\dfrac{a+b}{2}\right)(x-a)$$
solve it 
because $f''(x)>0$
so
$$\int_{a}^{b}f(x)dx\ge \int_{a}^{b}\left(f\left(\dfrac{a+b}{2}\right)+f'\left(\dfrac{a+b}{2}\right)(x-a)\right)dx=(b-a)f\left(\dfrac{a+b}{2}\right)$$
