Prove that line segment $AC$ lays above all points (excluding ends) described by line segments $AB$ and $BC$ if the slope of $AB \lt$ slope of $BC$ Claim to Prove
Suppose there are two line segments $\overline{AB}$ and $\overline{BC}$ on a cartesian coordinate system such that $A=(a,f(a))$, $B=(b,f(b))$, and $C=(c,f(c))$, where $a \lt b \lt c$. Further, suppose that the slope of $\overline{AB} \lt \overline{BC}$...i.e.
$$\frac{f(b)-f(a)}{b-a} \lt \frac{f(c)-f(b)}{c-b}$$
Prove that the line segment $\overline{AC}$ lays above all points described, in total, by the line segments $\overline{AB}$ and $\overline{BC}$, excluding the end points $A$ and $C$.

I'll post my proof below, but I am under the impression that there is a simpler manner to go about this. You'll see that I invoked the Intermediate Value Theorem (IVT), which may be overkill - though, perhaps not.
Throughout this argument, we will make use of the following general formula for a line $L_{m,n}$ which connects the point $(m,f(m))$ to $(n,f(n))$, where $m \lt n$:
$$L_{m,n}(x)=\frac{f(n)-f(m)}{n-m}(x-m)+f(m)$$
Note that when the domain of $L_{A,B}$ is restricted to $[a,b]$, we have the line segment $\overline{AB}$..similarly for $L_{B,C}$ and $\overline{BC}$.

Consider the following 3 lines:
$L_{A,B}(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$
$L_{B,C}(x)=\frac{f(c)-f(b)}{c-b}(x-a)+f(b)$
$L_{A,C}(x)=\frac{f(c)-f(a)}{c-a}(x-a)+f(a)$
It is straightforward to demonstrate that any two lines intersect at no more than 1 point $(*_1)$. By construction, we see that $L_{A,B}$ intersects $L_{A,C}$ at $\left(a,f(a)\right)$ and $L_{B,C}$ intersects $L_{A,C}$ at $\left(c,f(c)\right)$.
Next, let $L'$ represent the following function:
$L'(x)=\begin{cases}L_{A,B}(x) \quad &\text{if } x \leq b \\ L_{B,C}(x) \quad &\text{if }x\gt b \end{cases}$
We can see that $L_{A,C}$ intersects $L'$ at precisely two locations: $\left(a,f(a) \right)$ and $\left(c,f(c) \right)$. By $(*_1)$, this means that $L_{A,C}$ will not intersect $L'$ at any other locations $(*_2)$. Therefore, if we can show that there exists a point $x$ in $(a,c)$ with the following property: $L_{A,C}(x) \gt L'(x)$, then we are done. The reason for this can be understood with an application of IVT.
Suppose there is such an $x$ such that $x \in (a,c)$ and $L_{A,C}(x) \gt L'(x)$. Next, suppose that there is another $x^* \in (a,c)\setminus\{x\}$ such that $L_{A,C}(x^*) \lt L'(x^*)$. Consider the function: $H(z)=L_{A,C}(z)-L'(z)$. Note that $H$ is continuous. Thus, $H(x) \gt 0$ and $H(x^*) \lt 0$. WLG, let $x \lt x^*$. By the IVT, we know that there exists an $x' \in (x,x^*)$ such that $H(x')=0$, or, equivalently, $L_{A,C}(x')=L'(x')$. But $(x,x^*)\subset (a,c)$, which implies that $x' \in (a,c)$, contradicting $(*_2)$.

We now need to show that such an $x$ exists. By assumption, we know that $\frac{f(b)-f(a)}{b-a} \lt \frac{f(c)-f(b)}{c-b}$, which means that:
$$[f(b)-f(a)] (c-b) \lt [f(c)-f(b)] (b-a) \quad (\dagger_1)$$
Next, consider the point $L_{A,B}(b)$, written as previously described:
\begin{align} L_{A,B}(b) &= \frac{f(c)-f(a)}{c-a}(b-a)+f(a) \\ &=\frac{f(c)-f(b)+f(b)-f(a)}{c-a}(b-a)+f(a) \\&=[f(c)-f(b)]\cdot\frac{b-a}{c-a} +[f(b)-f(a)]\cdot\frac{b-a}{c-a}+f(a) \\ &\gt [f(b)-f(a)]\cdot \frac{c-b}{c-a}+[f(b)-f(a)]\cdot\frac{b-a}{c-a}+f(a) \quad \quad \text{by }(\dagger_1) \\ &= [f(b)-f(a)] \cdot \left(\frac{(c-b)+(b-a)}{c-a} \right) + f(a) \\ &= f(b)\end{align}
Finally, noting that $L'(b)=f(b)$, we see that $L_{A,B}(b) \gt L'(b)$. $b$ is our single point in $(a,c)$ that we wanted to find.

 A: The idea is to show first that $\overline{AC}$ lies strictly above the line segments at $x=b$, this is in fact equivalent to the given condition on the slopes. Then conclude that $\overline{AC}$ lies above $\overline{AB}$ on $(a, b)$ and above $\overline{BC}$ on $(b, c)$.
For $m < n$ let us denote more generally with
$$
 L_{m, n; u, v}(x) = \frac{n-x}{n-m} u + \frac{x-m}{n-m}v
$$
the (unique) linear function $L$ satisfying $L(m)=u$ and $L(n) = v$. Note that $L_{m, n; u, v}(x)$ is strictly increasing in both $u$ and $v$ for fixed $x \in (m, n)$.
The condition
$$
\frac{f(b)-f(a)}{b-a} \lt \frac{f(c)-f(b)}{c-b}
$$
is equivalent to
$$
f(b) < \frac{c-b}{c-a}f(a) + \frac{b-a}{c-a}f(c) = L_{a, c; f(a), f(c)}(b) =: y\, .
$$
This proves the desired estimate for $x=b$. For $a < x < b$ is
$$
 L_{a, b; f(a), f(b)}(x) < L_{a, b; f(a), y}(x) = L_{a, c; f(a), f(c)}(x)
$$
where the inequality on the left holds because $f(b) < y$, and the equality on the right holds because both $L_{a, b; f(a), y}$ and $L_{a, c; f(a), f(c)}$ join $(a, f(a))$ with $(b, y)$.
Similarly, for $b < x < c$,
$$
 L_{b, c; f(b), f(c)}(x) < L_{b, c; y, f(c)}(x) = L_{a, c; f(a), f(c)}(x) \, .
$$
