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Proving that the second derivative of a convex function is nonnegative

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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.


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