Proving a Simple Fact about Slopes of Lines The following problem is a detail from a proof I wrote recently -- a detail that I left unproven, and would like to prove.
Let there be three points $a$, $b$, and $c = \frac{a+b}{2}$, with $a<b$. We have a function $f$ such that $f(a) < f(c) \leq \frac{f(a)+f(b)}{2} < f(b)$.
So we can draw a line from $(a,f(a))$ to $(b,f(b))$ and the point $(c,f(c))$ will be on or below that line. 
We then claim that the slope of the line from $(a,f(a))$ to $(c,f(c))$ is less than or equal to the slope of the line from $(c,f(c))$ to $(b,f(b))$. Why is this true? The intuition is clear to me, but I don't know how to formalize this.
A diagram  below for clarity.

No matter how we place $f(c)$, the point $(c,f(c))$ will clearly always be such that the slope from $(a,f(a))$ to $(c,f(c))$ is less than the slope from $(c,f(c))$ to $(b,f(b))$ -- if this were not the case, then the point $(c,f(c))$ would have to lie above the line from $(a,f(a))$ to $(b,f(b))$. But how do I prove this?
 A: Edit: I misunderstood the question and will prove the actual question while leaving what I initially understood to be the question below.
We want to show that $(f(b) - f(c))/(b-c) \geq (f(c) - f(a))/(c-a)$.
Note: $(b-c) = (c-a) = (b-a)/2$
The inequality holds as $f(c)$ is at or below the midpoint of $f(b) - f(a)$

We also show that the slope of the line from $(a, f(a))$ to $(b, f(b))$ is larger  or equal to the slope from $(a, f(a))$ to $(c, f(c))$.
To do this we must show that $(f(b) - f(a))/(b-a) \geq (f(c) - f(a))/(c - a)$.
Note that $c = (a + b)/2$
Substituting into the right hand side of the inequality yields $(f(b) - f(a))/(b-a) \geq 2(f(c) - f(a))/(b - a)$
We know the inequality holds as $f(c) - f(a)$ is strictly less than $f(b) - f(a)$ and $f(c)$ is at or below the midpoint of $f(a)$ and $f(b)$ so $2(f(c)-f(a)) \leq (f(b) - f(a))$ 
A: Note that this property implies that for any $t=\dfrac m{2^n}$, we have $f(ta+(1-t)b)\le tf(a)+(1-t)f(b)$. If $f$ is continuous, this implies $f$ in convex, and it is easy to prove the slope of chords that have one extremity at $A$, $\dfrac{f(x)-f(a)}{x-a}\,$ is a non-decreasing function of $x$. 
So your assertion results from the fact that $f$, ig continuous, is a convex function.
