calculus - 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 $$f\left(\frac{a+b}{2}\right) \leq \frac{f(a)+f(b)}{2}$$ 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.

Answer

Here's how I ended up solving it. I made reference to this answer for the general principle, and this answer for the part about proving slopes of lines.

Let $m = \frac{a+b}{2}$. Suppose we draw a line (on a two-dimensional Cartesian grid) from $(a,f(a))$ to $(b,f(b))$. Then the point $(m,f(m))$ lies on or below this line because $f$ is midpoint convex: the height of the line at $m$ is given $\frac{f(a)+f(b)}{2}$, which is $ \geq f(m)$ by definition.

Since $(m,f(m))$ lies on or below this line, it follows that the slope of the line from $(a,f(a))$ to $(m,f(m))$ is less than or equal to the slope of the line from $(m,f(m))$ to $(b,f(b))$. We briefly show this:

We want to show that $\frac{f(b)-f(m)}{b-m} \geq \frac{f(m)-f(a)}{m-a}$. Remark that $b-m = \frac{b-a}{2} = m - a$, so the inequality simplifies to $f(b)-f(m) \geq f(m)-f(a)$, which is true since $f(m) \leq \frac{f(m)+f(a)}{2}$.

Having shown that the slope of $((a,f(a)),(m,(f(m))$ is less than or equal to the slope of $((m,f(m)),(b,(f(b))$, we apply the MVT to both of these two lines:
$$\exists c \in (a,m) \text{ s.t. } \frac{f(m)-f(a)}{m-a} = f'(c)$$
$$\exists d \in (m,b) \text{ s.t. } \frac{f(b)-f(m)}{b-m} = f'(d)$$
And knowing that the slope of the line joining $((a,f(a)),(m,f(m)))$ is less than or equal to that of the slope of the line joining $((m,f(m)),(b,f(b)))$, we have that $f'(c) \leq f'(d)$. We then apply the MVT once more:

$$\exists z \in (c,d) \text{ s.t. } \frac{f'(d)-f'(c)}{d-c} = f''(z)$$

So $f'(d)-f'(c) = f''(z)(d-c)$. We know that $f'(d) \geq f'(c)$, so $f'(d)-f'(c) \geq 0$. We also know that $d-c > 0$ because $c \in (a,m)$ and $d \in (m,b)$, where $a < b$. So our statement reduces to this inequality:
$$f''(z) \geq 0$$
As desired.