Problem. Let $f:D\to\mathbb R,$ be continuous, where $D\subset\mathbb R^2$, open, and let $\varphi,\psi :[\tau:T]\to\mathbb R,$ differentiable functions. If $\varphi(\tau)=\psi(\tau),$ and $$ \varphi'(t)=f\big(t,\varphi(t)\big) \quad\text{and}\quad \psi'(t)<f\big(t,\psi(t)\big), $$ for every $\,t\in [\tau,T],$ then show that $\psi(t)<\varphi(t),$ for all $t\in (\tau,T]$.
This is an exercise in an undergraduate ODEs class of rather elementary level. I have produced a solution (which I post), but I am afraid that it is a bit too technical. It assumes the knowledge of Calculus of M. Spivak's level. In particular, it uses the fact that the zero set of a continuous function is closed, and if it is bounded it contains a minimum. I was wondering if there is a simpler solution, more suitable for the level of that class.
I am providing a sketch of proof.
Assume that $\varphi(t_1)\le \psi(t_1),$ for some $t_1\in (\tau,T]$.
Step 1. Since $\varphi(\tau)=\psi(\tau),\,$ and $\varphi'(\tau)>\psi'(\tau),\,$ then there exists a $\delta>0$, such that $\varphi(t)>\psi(t),\,$ for $t\in (\tau,\tau+\delta)$.
Step 2. Combining the assumption that $\varphi(t_1)\le \psi(t_1),$ for some $t_1\in (\tau,T]$, with the result of Step 1, we obtain that (due to the Intermediate Value Theorem) the set $\,S=\{t\in [\tau+\delta,T]:\varphi(t)=\psi(t)\},\,$ is non-empty.
Step 3. Let $s=\min\{t\in [\tau+\delta,T] : \varphi(t)=\psi(t)\}$. As $\varphi(t)>\psi(t),\,$ for $t\in (\tau,s)$, then $$ \varphi'(s)=\lim_{t\to s^-}\frac{\varphi(t)-\varphi(s)}{t-s} \le\lim_{t\to s^-}\frac{\psi(t)-\psi(s)}{t-s} =\psi'(s)<f\big(s,\psi(s)\big)\\ =f\big(s,\varphi(s)\big)=\varphi'(s). $$ Contradiction.