How can we prove that slopes increase in a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ from the definition?

Question

The definition I have of a convex function $f: \mathbb{R} \rightarrow \mathbb{R}$ is that for every $x, y \in \mathbb{R}$ and every $\lambda \in [0, 1]$, $$ f(\lambda x + (1-\lambda )y) \leq \lambda f(x) + (1- \lambda )f(y).$$

By proving that slopes increase I mean that for $x \leq y \leq z$, we get $$\frac{f(y) - f(x)}{y-x} \leq \frac{f(z) - f(x)}{z-x} \leq \frac{f(z) - f(y)}{z-y}. $$

Is there a simple proof of this which doesn't assume that such a convex function has a non-negative second derivative? It's difficult to see how the definition gets us here.

Answer 1

Hint.

Note that if $x\leqslant y\leqslant z$, then we can write $$ y=\lambda x+(1-\lambda)z $$ for some $\lambda\in[0,1]$. Explicitly, you can find what is $\lambda$ in terms of $x,y$ and $z$. On the other hand, the definition of convexity tells you $$ f(y)\leqslant \lambda f(x)+(1-\lambda)f(z). $$ Now do some algebra to get what you want.

Answer 2

I like to think of this geometrically. The definition of convexity implies $(y,f(y))$ lies on or below the line $L$ through $(x,f(x))$ and $(z,f(z)).$ Suppose it's below. Then the line through through $(x,f(x))$ and $(y,f(y))$ plainly has slope less than the slope of $L.$ And then to move back up to $(z,f(z)),$ the line through $(y,f(y))$ and $(z,f(z)$ plainly has slope greater than the slope of $L.$ Now interpret all of this with appropriate mind-numbing formulae and algebra.