Proving that the second derivative of a convex function is nonnegative My task is as follows:

Let $f:\mathbb{R}\to\mathbb{R}$ be a twice-differentiable function,
  and let $f$'s second derivative be continuous. Let $f$ be convex with
  the following definition of convexity: for any $a<b \in \mathbb{R}$:
  $$f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2}$$ Prove that
  $f'' \geq 0$ everywhere.

I've thought of trying to show that there exists a $c$ in every $[a,b] \subset \mathbb{R}$ such that $f''(c) \geq 0$, and then just generalizing that, but I haven't been able to actually do it -- I don't know how to approach this. I'm thinking that I should use the mean-value theorem. I've also thought about picking $a < v < w < b$ and then using the MVT on $[a,v]$ and $[w,b]$ to identify points in these intervals and then to take the second derivative between them, and showing that it's nonnegative.
However I'm really having trouble even formalizing any of these thoughts: I can't even get to a any statements about $f'$. I've looked at a few proofs of similar statements, but they used different definitions of convexity, and I haven't really been able to bend them to my situation.
I'd appreciate any help/hints/sketches of proofs or directions.

I would set up a proof by contradiction.  Assuming a single point where $f''(x) < 0$, you can use the continuity of $f''(x)$ to find an interval $[a,b]$, where $f''(x) < 0$ throughout.  The intuition is then clear, in the sense that if you draw a concave down segment, then any secant line lies below your curve.  I will leave it to you to fill in the details from there.
