How would you prove this idea?

Question

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.

Note that you drew a picture with a convex ("concave up") function. It might be concave ("concave down") or changing concavity all over the place. — Ted Shifrin, Dec 24 '17 at 01:36
But if it was concave down, the conditions would no longer apply; the slope from AB would be greater than BC for instance. — James, Dec 24 '17 at 01:40
Sorry. I thought you were asking if AB must be less than BC etc. I didn't realize those were a given. — fleablood, Dec 24 '17 at 02:25
The answer to the question depends on the given function. For the function $f$ with the equation $f(x)=x^3-3x$ you'll find points, where your hypothesis isn't true. — Fakemistake, Dec 24 '17 at 10:02

Henry · Accepted Answer · 2017-12-24T09:53:15.873

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

Yea, sorry about that. I hope I cleared that up in the third edit. — James, Dec 24 '17 at 01:04
Could you go over that sentence regarding consequences? What is its relation to the implication? — James, Dec 24 '17 at 01:11
@James - $h,g,j,i$ correspond to the differences between successive co-ordinates. So this corresponds to saying that that long slope is between the short slopes, and so the long same side of one short slope as the other is — Henry, Dec 24 '17 at 09:30

How would you prove this idea?

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