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The question is as follows.

Let $f$ be differentiable real function defined in $(a,b)$. Prove that $f$ is convex iff $f'$ is monotonically increasing. Assume next that $f″(x)$ exists for every $x\in (a,b)$, and prove that $f$ is convex iff $f″(x)\geq 0$ for all $x\in (a,b)$.

For the second part of the question, can we directly say that $f'(x)$ is monotonically increasing if and only if $f''(x)\geq 0$? I have this question because the theorem derived from the Mean Value Theorem only directly implies the backwards direction (i.e. if the derivative is larger than 0, the function is monotonically increasing).

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The forward direction is simple: If $g$ is increasing ($x < y \implies g(x) \le g(y)$), then for $h > 0, g(x + h) \ge g(x)$, and so $$\frac{g(x+ h) - g(x)}{h} \ge 0$$ If $g'(x)$ exists, then $$g'(x) = \lim_{h \to 0+} \frac{g(x+ h) - g(x)}{h} \ge 0$$ (Note that existance of the two-sided limit implies that the one-sided limit exists and has the same value. Alternatively, you can show that the same inequality also holds for $h < 0$.)

Let $g = f'$ to apply this to your question.

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If $f$ is convex in $[a,b]$ we have that

$$\frac{f(x)-f(a)}{x-a}\le\frac{f(b)-f(a)}{b-a}\le \frac{f(b)-f(x)}{b-x},\quad\forall x\in(a,b)$$

Now apply the MVT above and you are done.