I'm trying to prove the following question:
Suppose that $f$ is concave up on an interval $I$ and $x_1, x_2, x_3$ and $x_4$ are numbers in $I$ with $x_1 < x_2 < x_3 < x_4$. Show that the slope of the secant line through the points at $x=x_1$ and $x=x_3$ is less than the slope of the secant line through the points at $x=x_2$ and $x=x_4$.
Now using concavity theorem, I have been able to derive the following equations:
$\dfrac{f(x_3) - f(x_1)}{x_3 - x_1} < \dfrac{f(x_4) - f(x_3)}{x_4 - x_3}$
$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < \dfrac{f(x_4) - f(x_2)}{x_4 - x_2}$
Rewriting the above equations as:
$m_1 < \dfrac{f(x_4) - f(x_3)}{x_4 - x_3}$
$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < m_2$
My objective is to prove that $m_1 < m_2$
I also know that:
$\dfrac{f(x_2) - f(x_1)}{x_2 - x_1} < \dfrac{f(x_3) - f(x_4)}{x_3 - x_4}$
Now i'm stuck at this point. I'm not exactly sure how to proceed so as to prove $m_1 < m_2$.
You can prove:
$$m_1 < m < m_2$$ where $m = \dfrac{f(x_3) - f(x_2)}{x_3-x_2}.$ The most elementary way to proceed would be to make use of the following simple fact: if $c,d > 0$ and $\dfrac ac < \dfrac bd,$ then
$$\dfrac{a}{c} < \dfrac{a+b}{c+d} < \dfrac{b}{d}.$$
You can prove the above easily by expanding and then use this for appropriately chosen $a,b,c,d$ to obtain both sides of the equality you want to prove.