Suppose $f''(y) > 0$ for all $y\in(x,z)$. Prove that g is strictly increasing on $(x,z]$. The problem, word for word: Let $a\lt b$ and $f:[a,b]\rightarrow\mathbb R$ a twice differentiable function. Let $x,z \in [a,b]$ with $x\lt z$ and define another function $g:(x,z]\rightarrow\mathbb R$ by
$$g(y) = \frac{f(y)-f(x)}{y-x}$$
Suppose $f''(y) > 0$ for all $y\in(x,z)$. Prove that g is strictly increasing on $(x,z]$. 
My attempt
Given $f$ is twice differentiable and $f$ acts on the interval $[x,y]$ then by the mean value theorem we can deduce that there exists a $c\in(x,y)$ such that 
$$f'(c) = \frac{f(y) - f(x)}{y-x}$$
Now I wonder if that implies that $f'(c) = g(y) \implies f''(c) = g'(y) \implies g'(y) \gt 0$ (by hypothesis). If this derivation is correct, I have yet to figure out how that relates back to the overall question.
Any help and pointers are very much appreciated, thank you.
 A: I think you are on the right track by noticing that according to MVT, there is some $c \in (x, y)$ such that
$$f'(c) = \frac{f(y) - f(x)}{y - x}$$
Next, in order to show $g$ is strictly increasing on $(x, z]$, it suffices to show $g' > 0$ on $(x, z)$. (We can see from the definition of $g$ that it is continuous on $(x, z]$ and differentiable on $(x, z)$, so we can apply this criterion.)
Observe that
$$g'(y) = \frac{(y - x)f'(y) - (f(y) - f(x))}{(y - x)^2}$$
Hence
$$\begin{align}
g'(y) > 0 &\iff (y - x)f'(y) - (f(y) - f(x)) > 0 \\
&\iff (y - x)f'(y) > f(y) - f(x) \\
&\iff f'(y) > \frac{f(y) - f(x)}{y - x} \\
&\iff f'(y) > f'(c)
\end{align}$$
and this is true since $y > c$ and $f'' > 0$ on $(x, z)$ implies that $f'$ is strictly increasing on $(x, z]$. (Again we use the fact that $f'$ is continuous on $(x, z]$ and differentiable on $(x, z)$ so that we can apply the criterion). We are done!
A: Since $f''(y)>0$, $f$ is strictly convex on $(x,z)$. Take $y_1$ and $y_2$ with $x < y_1 < y_2 < z$. Then there exists a $t\in (0,1)$ such that $y_1 = (1-t)y_2 + tx $. By strict convexity we get
$$ \begin{align*}
 g(y_1) &= \frac{f(y_1)-f(x)}{y_1 - x } \\
        &= \frac{f\left((1-t)y_2 + tx \right)-f(x)}{(1-t)y_2 + tx - x }\\
&< \frac{(1-t)f(y_2)+(t-1)f(x)}{(1-t)y_2 + (t-1)x} = \frac{f(y_2)-f(x)}{y_2 - x} = g(y_2).
\end{align*}
$$

Edit: I also found an alternative answer. Note that proving $g'(y)>0$ is equivalent with showing that 
$$
  k(y)=  f'(y) - \frac{f(y)-f(x)}{y-x} > 0.
$$
We will prove by contradiction that $k(y)>0$. 
Let us assume that $k(y)\leq 0$.
First note that $\lim_{y\to x}k(y) = 0$ by the definition of a derivative.
Next, deriving $k(y)$ gives
$$ k'(y)= f''(y) - \frac{f'(y)(y-x)-(f(y)-f(x))}{(y-x)^2} = f''(y) - \frac{k(y)}{y-x}\tag{*}.$$
By the given assumption that $f''(y)>0$ and our assumption that $k(y)\leq 0$, it follows that $k'(y)>0$. But that would mean that $k(y)$ is an increasing function. Since $\lim_{y\to x }k(y)=0$, this implies that $k(y)>0$. Contradiction. 
