When one can say $f(a)\leq (1-\delta) f(b) \Rightarrow a\leq (1-\epsilon) b $?

Question

Fix $a, b \in (0, \infty)$ such that $a<b$. Let $f:\mathbb R \to \mathbb R$ be twice differentiable function. Assume that $f$ is nonnegative and strictly increasing on $(a,b)$ and $f'(x)=0 $ if and only if $f(b)=0.$ And $f''(b)\neq 0,$ $f(a)\leq (1-\delta) f(b)$ for a fix $\delta >0.$ Also $f(0)=0.$

My Question: Can we expect to find $\epsilon >0$(depending on $\delta$ and $C$) such that $a\leq (1-\epsilon) b$?

Edit: Let $f(x)=\frac{1}{2} x- \frac{C}{6}x^{3}$ (for some fix constnat $C>0$) Then what can we say?

Answer 1

Certainly not. If $f$ has the property for some $a,b$, then for any $c>0$, the function $x\mapsto f(x-c)$ has the property for $a+c,b+c$. And $\frac{a+c}{b+c}\to 1$ as $c\to\infty$.

Edit: I just noticed that I ignored the condition $f(0)=0$, which seems to destroy this simple form of translation. But that is easily mended by working piece-wise.