Bounding solutions of ODE I want to prove the following lemma:

Let $x'=f(x,t)$ be a ODE and consider a derivable function $y(t)$ such that $y'(t) < f(y(t), t)$. Then if for a certain solution $x$ of the ODE and a certain $t_0$ we have that $x(t_0) \ge y(t_0)$ then we always have that $x(t) > y(t)$ for all $t > t_0$

Intuitively this seems correct to me, but the formal proof eludes me.
Is this true? Is there a simple proof?

Let $t_*>t_0$ be the point of the first intersection of $y=y(t)$ and $x=x(t)$:

We have $x(t_*)=y(t_*)$;
$y(t)$ is below $x(t)$, thus, $\dot y(t_*)\ge \dot x(t_*)=f(x(t_*),t)=f(y(t_*),t)$. It contradicts with the condition $\dot y(t)<f(y(t),t)$, therefore, $y=y(t)$ and $x=x(t)$ do not intersect.
