I am working on an exercise on ODE and I need the following lemma to solve it.
Lemma: Let two functions $y_1, y_2$: $[x_0, x_1]\to\mathbb{R}$, such that $y_1(x_0) = y_2(x_0)$ and $y_1(x_1) = y_2(x_1)$. Moreover $y_1(x)>y_2(x) \; \forall x\in (x_0, x_1)$. Prove that there exists some point $z\in(x_0,x_1)$, for which $y_1'(z) > y_2'(z)$.
It is geometrically obvious, as every point near enough $x_0$ satisfies the condition, but I couldn't prove it formally. I tried working with local maximums and minimums, but I think there is a more obvious way than considering all options for $max$ and $min$.
EDIT: $y_1, y_2$ are differentiable in $(x_0, x_1)$.
Assuming that both $y_1, y_2$ are continuously differentiable:
Let $f(x) = y_1(x)-y_2(x)$. Then $f(x_0)=f(x_1)=0$ and $f(x) > 0$ for all $x\in(x_0, x_1)$.
Let $x_2 \in(x_0, x_1)$. Then by Mean Value Theorem:
$$\exists z \in (x_0,x_2): f'(z)=\frac {f(x_2)-f(x_0)}{x_2-x_0}=\frac {f(x_2)}{x_2-x_0}>0$$
Hence $y_1'(z)-y_2'(z)>0$.
As a bonus you can also prove there are points where $y_1'(\xi)=y_2'(\xi)$, and even $y_1'(\xi)<y_2'(\xi)$.