Consequence of concavity

This should be easy, but I just can not solve this problem: Assume we have $a, b \in \mathbb{R}$ with $a< b$. The condition $$ f(\theta b + (1-\theta)a) \geq \theta f(b) + (1 - \theta)f(a)~\forall \theta \in (0, 1) $$ for some $f: \mathbb{R} \rightarrow \mathbb{R}$ holds. I want to show that $$ \frac{f(x) - f(a)}{x - a} - \frac{f(b) - f(x)}{b - x} \geq 0 $$ for all $x \in (a, b)$. I set $x = \theta b+(1-\theta)a$ for some appropriate value of $\theta \in (0, 1)$. Then I just computed a little and used concavity once: $$ \frac{f(x) - f(a)}{x - a} - \frac{f(b) - f(x)}{b - x} = \frac{f(x)(b-x)-f(a)(b-x)-f(b)(x-a)}{(x-a)(b-x)} \geq \frac{\theta(f(b)-f(a))(b-x)-f(b)(x-a)}{(x-a)(b-x)} $$ The last expression looks like a huge mess. Can someone help me out?

$\theta=\frac {x-a} {b-a}$ and $1-\theta=\frac {b-x} {b-a}$. So $f(x)\geq \frac {x-a} {b-a}f(a)+\frac {b-x} {b-a} f(b)$. Subtract $f(a)$ from both sides and divide by $x-a$. It should now be easy for you to simpilfy the right side and get the desired inequality.