Proof that slopes of a convex function are increasing

Proof

We write the middle point as a convex combination of the outer points and rearrange the convexity inequality.

1. Let $x < y < z$ be points of $I$. Put $t = \dfrac{z - y}{z - x}$. Then $0 < t < 1$ and $1 - t = \dfrac{y - x}{z - x}$.
2. We have $t x + (1 - t) z = \dfrac{(z - y) x + (y - x) z}{z - x} = \dfrac{y (z - x)}{z - x} = y$.
3. By the definition of a convex function, $f(y) \le t f(x) + (1 - t) f(z)$.
4. Multiplying by $z - x > 0$ gives $(z - x) f(y) \le (z - y) f(x) + (y - x) f(z)$. Call this inequality $(\ast)$.
5. Subtract $(z - x) f(x)$ from both sides of $(\ast)$. Since $(z - y) - (z - x) = -(y - x)$, we get $(z - x)(f(y) - f(x)) \le (y - x)(f(z) - f(x))$.
6. Dividing by $(z - x)(y - x) > 0$ gives $\dfrac{f(y) - f(x)}{y - x} \le \dfrac{f(z) - f(x)}{z - x}$.
7. Subtract $(z - x) f(z)$ from both sides of $(\ast)$. Since $(y - x) - (z - x) = -(z - y)$, we get $(z - x)(f(y) - f(z)) \le (z - y)(f(x) - f(z))$.
8. Multiplying by $-1$ reverses the inequality: $(z - x)(f(z) - f(y)) \ge (z - y)(f(z) - f(x))$.
9. Dividing by $(z - x)(z - y) > 0$ gives $\dfrac{f(z) - f(x)}{z - x} \le \dfrac{f(z) - f(y)}{z - y}$.
10. Steps 6 and 9 together are the two claimed inequalities.