Q:

If $x'(t)+f(x(t))\leq y'(t)+f(y(t))$ and x$(0)\leq y(0)$ then $x(t)\leq y(t)$.

Let $f:\mathbb{R}\to\mathbb{R}$ a non-increasing function. And let $x,y:[0,T]\to \mathbb{R}$ two function of class $C'$. If $$x'(t)+f(x(t))\leq y'(t)+f(y(t)),\ \forall t\in [0,T]$$ 
and $x(0)\leq y(0)$ then we have to prove that $x(t)\leq y(t)$ for all $t\in[0,T]$. 
Note:  I'm trying to solve this problem, but without success! 
Is this problem easy?
Any hint or comment is welcome. Thanks.

A:

For contradiction's sake, suppose there exists a time $t$ such that $x(t) > y(t)$. By continuity and differentiability of $x$ and $y$, there exists some time $s < t$ such that $x(s) = f(s)$ and $x'(s) > y'(s)$. However, that violates the first condition, since $x'(s) \leq y'(s)$. (We have that $f(x(s)) = f(y(s))$.)
I'm a little worried about this proof since I didn't need the fact $f$ is non-increasing, but I think the proof is legitimate.

