The key point is that since $f''(x)\neq 0$ for all $x\in (a,b)$ and $(a,b)$ is an open interval, the function $f'$ has neither a maximum, nor a minimum on $(a,b)$. In other words, for any $x\in (a,b)$, one can find $u,v\in (a,b)$ such that $f'(u)<f'(x)<f'(v)$. (This is the only way in which the assumption will be used).
It follows the the set $J:=f'((a,b))$ is an open interval. Indeed, this is an interval by the intermediate value theorem (no need to use the more subtle Darboux theorem here), and this interval is open by the above remark.
Consider the triangle $\Delta=\{ (x_1,x_2)\in (a,b)\times (a,b);\; x_1<x_2\}$, and the map $\Phi:\Delta\to \mathbb R$ defined by
$$\Phi(x_1,x_2):=\frac{f(x_2)-f(x_1)}{x_2-x_1}\cdot $$
The map $\Phi$ is continuous. Since $\Delta$ is a connected set, it follows that $I:=\Phi(\Delta)$ is a connected subset of $\mathbb R$, i.e. an interval.
Now, observe that $J=f'((a,b))$ is contained in the closure of $I=\Phi(\Delta)$. Indeed, since $(a,b)$ is right-open, we may write $f'(x)=\lim_{z\to x^+}\frac{f(z)-f(x)}{z-a}$ for any $ x\in (a,b)$, so that $f'(x)\in \overline{\Phi(\Delta)}$.
Since $J$ is an open set and $I$ is an interval, it follows that in fact $J\subset I$. So we have $$f'((a,b))\subset \Phi(\Delta)\, ,$$ which gives the required result.
Edit Note that the result may be wrong is $f'$ has a maximum or a minimum on $(a,b)$. For example consider $f(x)=\sin x$ on $(a,b)=(-\frac\pi2,\frac\pi2)$. Then $f'(0)=1$, but $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \cos t\, dt< x_2-x_1$ whenever $x_1<x_2$, because $\cos t<1$ for all but one $t\in (a,b)$. More generally, the result fails if $f'$ has an isolated maximum or an isolated minimum at some point $x\in (a,b)$.
You have shown that $\sin x$ is strictly concave downwards on $(0,\pi)$. Now for $x\in (0,\pi/2)$, note that:
$\begin{align}\sin x= &\sin (\frac 2\pi x (\frac \pi 2))\\=&
\lim_{n\to \infty} \sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)) \end {align}$
Noting that $\{2x/\pi,1-2x/\pi\}\subset (0,1)$ and by by strictly concave downward nature of $\sin $ on $(0,\pi)$, we have $\sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)\gt (1-\frac 2\pi x) \sin (\frac 1n) + \frac 2\pi x \sin (\frac \pi 2)$
It follows that: $\lim _{n\to \infty} \sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)\ge \lim_{n\to \infty}(1-\frac 2\pi x) \sin (\frac 1n) + \frac 2\pi x \sin (\frac \pi 2)=\frac 2\pi x$
So it follows that $\sin x\ge \frac 2 \pi x$ for $x\in (0,\pi/2)$.
Now it remains to rule out the equality. If $\sin t=\frac 2\pi t$ for some $t\in (0,\pi/2)$, then define $g:[0,\pi/2]\to \mathbb R$ as follows:
$g(x)=\begin{cases} 0; \text{ when }x=0\\ \sin x-\frac 2\pi x; \text{ when } x\in (0,\pi/2)\\0;\text{ when } x=\pi/2\end{cases}$
By LMVT on $g,\exists c_1\in (0,t)\land c_2\in (t,\pi/2)$ such that $g'(c_1)=\cos c_1=\frac 2\pi=g'(c_2)=\cos c_2$, which is a contradiction as $\cos $ is a decreasing function in $(0,\pi/2)$ in strict sense.
Best Answer
Equivalently you can replace (1) with (2): $$f(x+h) +f(x-h) - 2f(x) \geq 0 \tag{2}$$ for all $x,x+h,x-h\in (a,b)$.
As the OP correctly notes, every convex function is midpoint convex, so it is sufficient to rely on any theorem that asserts that a function is convex.
Or...one might prove directly since (as observed in a comment) that defeats the spirit of the problem.
The simplest proof maybe. Consider the function $$t \to \frac{f(x+t) + f(x-t) -2 f(x)}{t}$$ and apply the Cauchy Mean Value theorem [see below] on the interval $[0,h]$ to obtain $\tau \in (0,h)$ with $$ \frac{f(x+h) + f(x-h) -2 f(x)}{h} = \frac{f'(x+\tau) - f'(x-\tau)}{1} \geq 0.$$ QED.
But I jumped in here with a totally different motive.
What about this notion of midpoint convexity. Is that a thing? If every convex function is midpoint convex, then is every midpoint convex function really just convex?
No. Not every midpoint convex function is convex.
But every continuous midpoint convex function is convex.
If a midpoint convex function is not convex then it is pretty weird. Blumberg (1919) and Sierpiński (1920) independently proved that every measurable midpoint convex function must be convex.
But there are plenty of nonmeasurable functions that are midpoint convex. All of them are unbounded in every open subinterval, so not your ordinary everyday function.
There is a big literature on midpoint convex functions which I can only encourage interested parties to consult.
Notes:
Calculus students are not provided with many tools. About the only reliable and often-used one is the mean-value theorem. I would suggest that you memorize a small upgrade. The weak version is easy to remember; this is almost as easy.
Cauchy's Mean Value Theorem: Let $F,G:R→R$ be continuous on $[a, b] $ and differentiable on $(a, b)$. Suppose that $G(b)≠G(a)$. Then there exists $c∈(a, b)$ such that $G′(c)≠0$ and such that $$\frac{F(b) - F(a)}{G(b) - G(a)} = \frac{F'(c)}{G'(c)}.$$