Proving a function is convex From the Defintion of convex:

Theorem to be proven:

If $f$ is differentiable and $f'$ is increasing, then $f$ is convex. Use Proof by Contradiction.

Consider, $I = (a, b)$ with $a < x < b$. 
If $f(x)$ is convex then,
$$\frac{f(x) - f(a)}{x-a} < \frac{f(b) - f(a)}{b-a}, \space \forall x$$
Suppose, 
$$\frac{f(x) - f(a)}{x-a} > \frac{f(b) - f(a)}{b-a} \space \mathrm{for \space some} \space x$$
From MVT< $\exists x_1$: 
$$\exists x_1 \in (a, b) \implies f'(x_1) = \frac{f(b) - f(a)}{b-a}$$
We have the inequality (from the assumptions) , 
$$f'(x) > f'(x_1)$$ 
If $x_1 > x$ then contradiction for $f'$ is increasing. 
If $x_1 < x$ <---- I cant come up with a contradiction? 
 A: First of all, I want to note that your definition of convexity corresponds to strict convexity. Thus, I'll have to assume that $f'$ is strictly increasing.
Let's have a look at the convexity condition first and reformulate it a little. So, for $a<x<b$ the convexity condition gives
$$
\begin{align}
& & \frac{f(x)-f(a)}{x-a} &< \frac{f(b)-f(a)}{b-a}
\\ \Leftrightarrow & & (f(x)-f(a))(b-a) &< (f(b)-f(a))(x-a)
\\ \Leftrightarrow & & (f(x)-f(a))(b-a) + (f(x)-f(a))(a-x) &< ((f(b)-f(a))(x-a) + (f(x)-f(a))(a-x)
\\ \Leftrightarrow & & (f(x)-f(a))(b-x) &< ((f(b)-f(a))(x-a) + (f(a)-f(x))(x-a)
\\ \Leftrightarrow & & (f(x)-f(a))(b-x) &< ((f(b)-f(x))(x-a)
\\ \Leftrightarrow & & \frac{f(x)-f(a)}{x-a} &< \frac{f(b)-f(x)}{b-x}.
\end{align}
$$
Now assume that $f$ is not convex. Then there exist $a<x<b$ such that
$$
\frac{f(x)-f(a)}{x-a} \geq \frac{f(b)-f(a)}{b-a}
$$
or, as we just computed, equivalently
$$
\frac{f(x)-f(a)}{x-a} \geq \frac{f(b)-f(x)}{b-x}.
$$
By the mean value theorem we get $\xi_1 \in (a,x)$ and $\xi_2 \in (x,b)$ with
$$
\begin{align}
f'(\xi_1) = \frac{f(x)-f(a)}{x-a} && \text{and} && f'(\xi_2) = \frac{f(b)-f(x)}{b-x}.
\end{align}
$$
If we combine this with the above inequality we have $\xi_1 < \xi_2$ with
$$
f'(\xi_1) = \frac{f(x)-f(a)}{x-a} \geq \frac{f(b)-f(x)}{b-x} = f'(\xi_2).
$$
But $f'(\xi_1) \geq f'(\xi_2)$ for $\xi_1 < \xi_2$ is a contradiction to $f'$ strictly increasing. Hence, the contradiction yields that $f$ must be convex.
