How would you prove this idea? Is it true that if the slope of the secant line between A and B is less than the slope of the secant line between B and C, then the slope of the secant line between A and B is less than the slope of A to C? Is this true and if so, how would you prove it?
Edit: A, B, and C are all points on a function
Edit2: A possible diagram for the question 
Edit3: Sorry, I realized I was missing information thanks to the comment and answer. The x values of A, B, and C are in increasing order i.e. A's x < B's x < C's x.
 A: Triatticus's question in the comment matters: for example 
Try $A$ being $(2,0)$ and $B$ being $(0,0)$ and C being $(0,1)$: the the slope of $AB$ is $0$ less than the slope of $BC$ being $1$, but the slope of $AB$ being $0$ is greater than the slope of $AC$ being $-1$
But with  $A$ being $(x_a,y_a)$ and $B$ being $(x_b,y_b)$ and C being $(x_c,y_c)$ then if $x_a \lt x_b \lt x_c$ it will be true that $\dfrac{y_b-y_a}{x_b-x_a} \lt \dfrac{y_c-y_b}{x_c-x_b} \implies \dfrac{y_b-y_a}{x_b-x_a} \lt \dfrac{y_c-y_a}{x_c-x_a}$.  This is a consequence of $\dfrac{g+i}{h+j}$ being between $\dfrac{g}{h}$ and $\dfrac{i}{j}$ when $h$ and $j$ are positive
You would be saying  $\left( \frac{y_c-y_a}{x_c-x_a}-\frac{y_b-y_a}{x_b-x_a}\right) \left( \frac{y_c-y_b}{x_c-x_b}-\frac{y_c-y_a}{x_c-x_a}\right) = \frac{{{\left( x_b y_c-x_a y_c-x_c y_b+x_a y_b+x_c y_a-x_b y_a\right) }^{2}}}{\left( x_b-x_a\right)  {{\left( x_c-x_a\right) }^{2}} \left( x_c-x_b\right) }$ where the numerator of the right hand side is a square and so non-negative and usually positive, while the denominator of the right hand side is a product of positive terms positive when $x_a \lt x_b \lt x_c$, implying that the two terms on the left hand side (the differences in slopes) have the same sign
