$f$ is strictly monotonic on interval $I$ $\iff$ $\forall~x_1,x_2,x_3 \in I$, if $x_10$ Prove: $f$ is strictly monotonic on interval $I$ $\iff$ $\forall~x_1,x_2,x_3 \in I$, if $x_1<x_2<x_3$, then
$$
(f(x_1)-f(x_2))(f(x_2)-f(x_3))>0.
$$
My (false) attempt:
strictly monotonic $\iff$ $f(x_1)<f(x_2)<f(x_3)$ or $f(x_1)>f(x_2)>f(x_3)$, then the statement simply follows.
However, then I notice that there might be two set of numbers $x_1,x_2,x_3$ and $y_1,y_2,y_3$, satisfying $f(x_1)<f(x_2)<f(x_3)$ and $f(y_1)>f(y_2)>f(y_3)$ at the same time. So the "$\iff$" should be $\Rightarrow$ in fact.
Can you help me show that there shouldn't be such two set of numbers, or prove this proposition using another approach? Thanks!
 A: Proof of $(\Leftarrow)$. The condition implies that for $x_1\neq x_2, x_2\neq x_3, x_3\neq x_1$ we have
$$\text{sgn}\left(\frac{f(x_1)-f(x_2)}{x_1-x_2}\right)=\text{sgn}\left(\frac{f(x_2)-f(x_3)}{x_2-x_3}\right)=\text{sgn}\left(\frac{f(x_3)-f(x_1)}{x_3-x_1}\right). \tag{1}$$
Indeed, to see $(1)$, we note that $(1)$ is cyclically symmetric on $x_1, x_2, x_3$, so we can assume WLOG that $x_1<x_2<x_3$ and then $(1)$ follows from the condition.
Using $(1)$ to another triple $x_2, x_3, x_4$, we know that if $x_2\neq x_3, x_3\neq x_4, x_4\neq x_2$, then
$$\text{sgn}\left(\frac{f(x_2)-f(x_3)}{x_2-x_3}\right)=\text{sgn}\left(\frac{f(x_3)-f(x_4)}{x_3-x_4}\right)=\text{sgn}\left(\frac{f(x_4)-f(x_2)}{x_4-x_2}\right). \tag{2}$$
Combining $(1)$ and $(2)$, we have
$$\text{sgn}\left(\frac{f(x_1)-f(x_2)}{x_1-x_2}\right)=\text{sgn}\left(\frac{f(x_3)-f(x_4)}{x_3-x_4}\right), \tag{3}$$
if $x_1\neq x_2, x_2\neq x_3, x_3\neq x_1$ and $x_2\neq x_3, x_3\neq x_4, x_4\neq x_2$. Indeed, we can easily check that $(3)$ holds for all $x_1, x_2, x_3, x_4$ with $x_1\neq x_2, x_3\neq x_4$. (The details are omitted here, you just need to check several cases, not too much.) Now we have
$$\text{sgn}\left(\frac{f(x_1)-f(x_2)}{x_1-x_2}\right)=\text{sgn}\left(\frac{f(x_3)-f(x_4)}{x_3-x_4}\right),\qquad x_1\neq x_2, x_3\neq x_4. \tag{4}$$
Now we choose $x_3, x_4\in I$ with $x_3\neq x_4$ in $(4)$ and then fix them. The condition implies that $f(x_3)\neq f(x_4)$, so $\text{sgn}\left(\frac{f(x_3)-f(x_4)}{x_3-x_4}\right)$ is a definite number, which is $1$ or $-1$. If it is $1$, then $(4)$ implies that $f$ is strictly increasing; if it is $-1$, then $(4)$ implies that $f$ is strictly decreasing.
