Given that $f(x)\geq x^2,\;\;\forall\;x\in(0,\infty),\;\;a_{n+1}\leq a_n -2f(a_n),\;\;\forall n\geq 1,$ prove that $\liminf\limits_{n\to \infty}a_n=0$

This question looks to me, very difficult but I believe there are people here who can help me, by providing a very detailed answer. Here is the question

Let $\{a_n\}$ be a nonnegative bounded sequence and $f:(0,\infty)\to(0,\infty)$ be a monotone increasing function which satisfies the following condition: $$f(x)\geq x^2,\;\;\forall\;x\in(0,\infty).\;\;(*)$$ Suppose the following inequality holds: $$a_{n+1}\leq a_n -2f(a_n),\;\;\forall n\geq1.\;\;(**)$$ Prove that $\liminf\limits_{n\to \infty}a_n=0.$

Here are useful hints.

Hint

Since $\{a_n\}$ is bounded and nonnegative, it follows that $\liminf\limits_{n\to \infty}$ exists and is nonnegative. Let $\liminf\limits_{n\to \infty}a_n\geq 0.$ To prove that $\liminf\limits_{n\to \infty}=0$, we proceed by contradiction. So, assume $a\neq 0.$ Then, it must be strictly positive, i.e., $a>0.$

(a.) Complete the following statement. $\liminf\limits_{n\to \infty}=a$ implies that, given $\epsilon>0$, there exists $n(\epsilon)\in \Bbb{N}$ such that $$a-\epsilon<a_n,\;\;\forall\;n\geq n(\epsilon).$$

(b.) Since $a>0$, we can take $\epsilon =\frac{a}{2}>0.$ Put this value for $\epsilon$ in your statement in part (a).

(c.) Use the hypothesis that $f$ is monotone increasing to get, from part (b), that $$f\left(\frac{a}{2}\right)<f(a_n),\;\;\forall\;n\geq n(\epsilon).\;\;\;\;\;(***)$$

(d.) Use inequality $(***)$ in $(**)$ and use condition $(*)$ to get that $$0<a^2\leq2(a_n-a_{n+1}),\;\;\forall\;n\geq n(\epsilon).\;\;\;\;\;(****).$$

(e.) Use condition $(**)$ to justify that the sequence $\{a_n\}$ is monotone non-increasing (and it is bounded below by $0$) and so converges.

(f.) Now take limit in $(****)$ to obtain your contradiction. Explain this contradiction.

Any help will be highly appreciated!

By definition (of $\liminf$), there exists $N \in \mathbb{Z}_+$ so that $a - \frac{a}{2} < a_n, \ \forall n > N$. Thus, $\frac{a}{2} < a_n, \ \forall n>N$. Since $f$ is a monotone increasing function, we have $f\left(\frac{a}{2}\right) < f(a_n), \ \forall n>N$. Therefore, \begin{equation} \label{star} \tag{*} a_{n+1} \leq a_n - 2f(a_n) < a_n - 2f\left(\frac{a}{2}\right) < a_n - \frac{a^2}{2}, \ \forall n>N. \end{equation} Rearranging gives, for $n>N$, \begin{equation*} \label{double star} \tag{**} 0<a^2<2(a_n-a_{n+1}). \end{equation*} Additionally, by \eqref{star} we know that $\{a_n\}$ is non-increasing (at least for $n>N$). Since $\{a_n\}$ is bounded it follows that it converges (to $a$). Letting $n \to \infty$ in \eqref{double star} gives \begin{equation*} a^2\leq 2(a-a)=0. \end{equation*}