A function $f(x)$ in some domain $a\leq x \leq b$ is convex if and only if for any $x_1 < x_2 < x_3$ from domain $[a,b]$,
$$\frac{(f(x_2)-f(x_1))}{(x_2-x_1)} \leq \frac{(f(x_3)-f(x_1))}{(x_3-x_1)} \leq \frac{(f(x_3)-f(x_2))}{(x_3-x_2)}.$$
I am aware that for any $(x_1,x_2)$ for a convex function $f(x_1) \geq f(x_1)+f'(x_1)(x_2-x_1)$. How do I proceed further to prove the above inequality?
There are two equivalent definitions of convexity that will be used to prove this.
Let $f$ be a real-valued function that is convex on some interval $I$. Then:
$\dagger_1$: For any $a,b \in I$ and for any $\alpha \in [0,1]$, $f(\alpha \cdot b+(1-\alpha) \cdot a) \leq \alpha f(b)+(1-\alpha)f(a)$
$\dagger_2$: For any $x_1 \lt x_2 \lt x_3 \in I$, $\frac{f(x_3)-f(x_1)}{x_3-x_1} \geq \frac{f(x_2)-f(x_1)}{x_2-x_1}$
Consider any $x_1 \lt x_2 \lt x_3 \in I$. Next, let $\alpha=\frac{x_2-x_1}{x_3-x_1}$ noting that $\alpha \in (0,1)$. Then $\alpha-1=\frac{x_3-x_2}{x_3-x_1}$.
Next, we have that $x_2=\alpha \cdot x_3 + (1-\alpha)\cdot x_1$. We can now apply $\dagger_1$, which gives us the following inequality:
$$f(x_2) \leq \alpha f(x_3)+(1-\alpha)f(x_1) \quad (*)$$
From this inequality, we can infer from $\dagger_2$ the following inequality:
$$\frac{\alpha f(x_3)+(1-\alpha)f(x_1)-f(x_1)}{x_2-x_1} \geq \frac {f(x_2)-f(x_1)}{x_2-x_1} \quad \color{red}{(\dagger_3)}$$
From $(*)$ and some rules of inequalities, we have the following chain of inferences:
\begin{align} f(x_2) &\leq \alpha f(x_3)+(1-\alpha)f(x_1) \implies\\ -f(x_2) &\geq - \left[ \alpha f(x_3)+(1-\alpha)f(x_1)\right] \implies \\ \frac{f(x_3)-f(x_2)}{x_3-x_2} &\geq \frac{f(x_3)- \left[ \alpha f(x_3)+(1-\alpha)f(x_1)\right]}{x_3-x_2} \quad \color{red}{(\dagger_4)} \end{align}
If we can show that $\frac{f(x_3)- \left[ \alpha f(x_3)+(1-\alpha)f(x_1)\right]}{x_3-x_2}=\frac{\left[\alpha f(x_3)+(1-\alpha)f(x_1)\right]-f(x_1)}{x_2-x_1}$, we are done. Although referring to a simple drawing would show this to be true, we can arrive at the answer through a written argument.
Consider the line $L_{x_1,x_3}$ that connects $\left(x_1,f(x_1)\right)$ to $\left(x_3,f(x_3) \right)$.
One possible way of describing the slope of this line is $\frac{f(x_3)-f(x_1)}{x_3-x_1}$. However, it is straightforward to show that for any two points $\left(a,f(a) \right), \left(b,f(b) \right)$ that are part of this line, the slope can be equivalently described as $\frac{f(b)-f(a)}{b-a}=\frac{f(x_3)-f(x_1)}{x_3-x_1}$. To show that $\frac{f(x_3)- \left[ \alpha f(x_3)+(1-\alpha)f(x_1)\right]}{x_3-x_2}=\frac{\left[\alpha f(x_3)+(1-\alpha)f(x_1)\right]-f(x_1)}{x_2-x_1}$, we simply need to demonstrate that the point $\left(x_2,\alpha f(x_3)+(1-\alpha)f(x_1) \right)$ sits on our line.
Applying the traditional definition of a line, the second component of our collection of ordered pairs that populate $L_{x_1,x_3}$ can be described using the following equation: $y=\frac{f(x_3)-f(x_1)}{x_3-x_1}\cdot(x-x_1)+f(x_1)$. Plug in $x=x_2$, and you will precisely get $\frac{x_2-x_1}{x_3-x_1}f(x_3)+(1-\frac{x_2-x_1}{x_3-x_1})f(x_1)=\alpha f(x_3)+(1-\alpha)f(x_1)$.
This means that $\left(x_2,\alpha f(x_3)+(1-\alpha)f(x_1) \right) \in L_{x_1,x_3}$. Therefore, we can conclude that: $$\frac{f(x_3)- \left[ \alpha f(x_3)+(1-\alpha)f(x_1)\right]}{x_3-x_2}=\frac{\left[\alpha f(x_3)+(1-\alpha)f(x_1)\right]-f(x_1)}{x_2-x_1}=\frac{f(x_3)-f(x_1)}{x_3-x_1}$$
This means that we can rewrite $\color{red}{(\dagger_3)}$ and $\color{red}{(\dagger_4)}$ as:
$$\frac{f(x_3)-f(x_1)}{x_3-x_1} \geq \frac {f(x_2)-f(x_1)}{x_2-x_1}$$
and
$$\frac{f(x_3)-f(x_2)}{x_3-x_2} \geq \frac {f(x_3)-f(x_1)}{x_3-x_1}$$, respectively...the combination of which yields us our desired result:
$$\frac{f(x_3)-f(x_2)}{x_3-x_2} \geq \frac{f(x_3)-f(x_1)}{x_3-x_1} \geq \frac {f(x_2)-f(x_1)}{x_2-x_1}$$
It is easier than it seems! $x_{2}=tx_{1}+(1-t)x_{3}$ for some $t$ in $(0,1)$. Then the first inequality becomes:
$f(tx_{1}+(1-t)x_{3})-f(x_{1})/(1-t)(x_{3}-x_{1})\leq f(x_{3})-f(x_{1})/x_{3}-x_{1}$
which is equivalent to :
$f(tx_{1}+(1-t)x_{3})$$\leq\,tf(x_{1})+(1-t)f(x_{3})\,\,$ (J)
which is the definition of convexity!! Likewise the second inequality is again the
definition of convexity! If$ f$ is differentiable and you
define convexity to be : "non-decreasing $f'$", then all you
have to prove is that Jenssen's inequality (J) holds (which is
very easy, using the mean value theorem) and go to the above proof!!