(convex function) Let $f: [a,b] \to \mathbb R$ twice differentiable, and $f''(x) \ge 0$, $\forall x \in [a,b]$.... 
Let $f: [a,b] \to \mathbb R$ twice differentiable, and $f''(x) \ge 0$, $\forall x \in [a,b]$.Prove that $f(\frac{x_1 + x_2} {2}) \le \frac{1}{2}[f(x_1) + f(x_2)], \forall x_1, x_2 \in [a,b].$

My attempt:
As a hypothesis we assume that $f''(x) \ge 0$ and we know that .If $f: (a,b) \to \mathbb R$ has second order derivatives on $(a, b)$ the function $f$ is convex if and only if $f''(x) \ge 0$. If $f’’(x) \gt 0$, the function $f$ is strictly convex. So $f$ is convex. And by the definition of convex, A function $f: (a,b) \to \mathbb R$ is calling convex on (a,b) if for any $x_1, x_2 \in (a,b)$ and for any pair of real numbers $\alpha_1 \ge 0, \alpha_2 \ge 0$ such that $\alpha_1 + \alpha_2 = 1$ the following inequality $f(\alpha_1x_1 + \alpha_2x_2) \le \alpha_1f(x_1) + \alpha_2f( x_2)$.
So we have:
$f(\frac{x_1 + x_2}{2}) \le \frac{1}{2}[f(x_1) + f(x_2)] = f(\frac{1}{2}[x_1]+\frac{1}{2}[x_2])  \le \frac{1}{2}f(x_1) + \frac{1}{2}f(x_2) \to f(\alpha_1x_1 + \alpha_2x_2) \le \alpha_1f(x_1) + \alpha_2f(x_2)$
How is my answer?
Thank's for any help
 A: 
Definition.  a function $f$ on an interval $(a,b)$ is said to be
midpoint convex if $$f\left(\frac{x_1 + x_2} {2}\right) \le \frac{1}{2}[f(x_1) + f(x_2)] \tag{1}$$  for  all $x_1, x_2 \in (a,b)$.

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.

Problem.  Show [directly] that a differentiable function $f$ with a nondecreasing derivative $f'$ is midpoint convex.

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
$$\frac{F(b) - F(a)}{G(b) - G(a)} = \frac{F'(c)}{G'(c)}.$$
A: Let $x_0 := (x_1+x_2)/2$ and let $h = (x_2-x_1)/2$.
The desired inequality is $f(x_0) \le \frac{1}{2}(f(x_0-h) + f(x_0+h))$,
which can be rearranged as
$$f(x_0) - f(x_0-h) \le f(x_0 + h) - f(x_0).\tag{$*$}$$
By Taylor's theorem, there exist $\xi_-$ and $\xi_+$ in $[0, h]$ such that
$$f(x_0-h) = f(x_0) - h f'(x_0) + \frac{h^2}{2} f''(x_0-\xi_-)$$
and
$$f(x_0+h) = f(x_0) + h f'(x_0) + \frac{h^2}{2} f''(x_0+\xi_+).$$
Can use these two equations to prove the above inequality ($*$)?
