Is $f(a)\leq (1-\delta) f(\frac{1}{\sqrt{C}}) \Rightarrow a\leq \frac{1- \epsilon}{\sqrt{C}} $ for $f(x)=\frac{1}{2} x- \frac{C}{6}x^{3}$?

Question

Define $f:\mathbb R \to \mathbb R$ as $f(x)=\frac{1}{2} x- \frac{C}{6}x^{3}$ (for some fix constnat $C>0$).

Note that $f'(x)= \frac{1}{2}- \frac{C}{2}x^2$ and $f''(x)=-Cx.$ Also note that $f'(x)>0$ iff $\frac{C}{2}x^2<\frac{1}{2}$ and hence $f$ is strictly incersing on $(0, \frac{1}{\sqrt{C}})$

Take $b=\frac{1}{\sqrt{C}}$ (so that $f'(b)=0$ and $f''(b)\neq 0$), and choose $0<a<b$ with the property that $f(a)\leq (1-\delta) f(b)$ for some fix $\delta >0$.

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

Motivation: I am trying to figure out the proof of Lemma 3.4 Equation 3.5, page 9-10. (Note that as per the notations the I am taking, $a=\bar{y}, $ and $b=y_{C}$ and $N=3$(This $C$ is not the same as above)

Answer 1

Be $\frac{a}{2}-\frac{Ca^3}{6}\leq \frac{a}{2}\leq \frac{1-\delta}{3\sqrt{C}}$.

With $1-\epsilon:=2\frac{1-\delta}{3}$ you get your inequality for $a$.