Suppose that $g(x)$ is strictly increasing; $g,h$ are smooth functions. Both $g,h$ are defined on a closed interval $I$.
Can we claim that $f(x)=g(x)+ah(x)$ is also strictly increasing for some small positive constant $a$?
Here is my "proof":
Since $g,h$ are smooth and $I$ is closed, then $g'(x),h'(x)$ are bounded variation functions.
Define: $$\min_{x\in I} h'(x)=c$$ $$\min g'(x)=b>0$$
By setting $a<b/|c|$, it is obvious that $f'(x)>0$?
Thank you for the counterexample; the claim was wrong. I think in order to have the claim correct, we must further require that $g'(x)>0$ for that closed interval $I$, rather than just being "strictly increase"?
It's not true that $\min g'(x) > 0$.
Hint: try $g(x) = x^2$ and $h(x) = -x$ on $[0,1]$.