Lower bound of $\frac{f(x)}{x^{n+1}}$

Question

Let $f:[0,a]\to \Bbb{R}_{\geq 0}$ be real analytic, $a<1$. Furthermore, $f(0) = 0$ and $f$ is strictly increasing on $[0,a]$. Let $n\in \Bbb{N}$ be the smallest positive integer such that $f^{(n)}(0)\ne 0$.

$\textbf{Claim.}$ Let $y\leq a$, then $$ \min_{x\leq y}\frac{f(x)}{x^{n+1}}\geq f(y). $$

$\textbf{Ideas.}$ My first idea was to consider a proof by contradiction. Suppose there exists some $x_0 < y$ such that $$ \frac{f(x_0)}{x_0^{n+1}}< f(y) \implies f(x_0) < x_0^{n+1}f(y). $$ By Taylor's Remainder Theorem, $$ f(x_0) = \frac{f^{(n)}(0)}{n!}x_0^n + \frac{f^{(n+1)}(c_{x_0})}{(n+1)!}x_0^{n+1}, $$ where $c_{x_0}$ is dependent on $x_0$. Hence, it follows that \begin{equation} 0<x_0^{n+1}f(y) - f(x_0) = -\frac{f^{(n)}(0)}{n!}x_0^n +\left(f(y) -\frac{f^{(n+1)}(c_{x_0})}{(n+1)!}\right)x_0^{n+1}. \quad (1) \end{equation} Let $$ g(x,y,c) = -\frac{f^{(n)}(0)}{n!}x^n +\left(f(y) -\frac{f^{(n+1)}(c)}{(n+1)!}\right)x^{n+1}. $$ For fixed $y,c$, the zeros of $g$ are $$ x = 0, \frac{\frac{f^{(n)}(0)}{n!}}{f(y)-\frac{f^{(n+1)}(c)}{(n+1)!}}. $$ At this point I want to argue that for $0<c<x<y$, $$ \frac{\frac{f^{(n)}(0)}{n!}}{f(y)-\frac{f^{(n+1)}(c)}{(n+1)!}} > y $$ contradicting (1), (it just needs to be greater than $x$, but in testing it seems to be greater than $y)$. However, this seems dubious. Furthermore, I haven't utilized the fact that $f$ is strictly increasing, so this attempt doesn't seem to be the correct way. Any advice would be appreciate.

Answer 1

A counterexample: $n=1$, $f(u)=e^{pu}-1$, $x=1/2$, $y=3/4$, and $p\in(0,\infty)$. Then $$\frac{f(x)}{x^{n+1}}\Big/f(y)=4\frac{e^{p/2}-1}{e^{3p/4}-1}<1$$ if $p$ is large enough.