Prove $\liminf_{n\to\infty} \left(n^2(4a_n(1-a_{n-1})-1)\right)\leq\frac14$ for any non-negative seqence $\{a_n\}$.

Question

Let $\{a_n\}_{n=1}^\infty$ be a sequence of non-negative numbers. Prove that $$\liminf_{n\to\infty} \left(n^2\left(4a_n(1-a_{n-1})-1\right)\right)\leq\frac14.$$

I found this problem on one of my old notebooks, and clearly I forgot where it came from. I searched using Approach0 and it led me to a post on AOPS without solutions.

I have to admit that I don't know how to start. First of all, it is easy to notice that if we have a subsequence $\{a_{n_k}\}$ with $a_{n_k}\geq 1$ for all $k$, then we must have $\liminf_{n\to\infty} \left(n^2(4a_n(1-a_{n-1})-1)\right)\leq0\leq \frac14$; also, the case where there are infinitely many $0$'s is easy to handle. So, WLOG we can assume that $a_n\in(0,1)$ for all $n$.

Secondly, notice that $4x(1-x)-1=-(2x-1)^2\leq 0$, with equality holds iff $x=\frac12$. It may suggest that $\frac12$ is an important number here. Now it seems natural to write $a_n=\frac12+b_n$, so we have $b_n\in(-1/2,1/2)$ and we want to prove that $$\liminf_{n\to\infty}\left(n^2(b_n-b_{n-1}-2b_nb_{n-1})\right)\leq\frac18.$$ But this is not obvious, either.

Remark. The bound $1/4$ (or $1/8$ in terms of $b_n$) is sharp. If we choose $b_n=-\frac1{4n}$, or equivalently $a_n=\frac12-\frac1{4n}$ for $n\geq1$, then $$n^2(b_n-b_{n-1}-2b_nb_{n-1})=\frac{n^2}{8n(n-1)}\to\frac18,\qquad n\to\infty.$$

The above are simplest (kind of silly) observations. Any help would be appreciated!

Answer 1

Finally, I come up with this proof. Thanks to @Jack D'Aurizio, whose second thought is the main body of this proof; also thanks to @Adam Rubinson, some ideas formed during comunicating with whom also play an important role in this proof.

I'll divide the proof into $6$ steps, which may get readers startled at first glance. I just want to write things clearly, to make sure there is no mistake in every step. I hope that every line you'll see below is easy to understand. If there are mistakes somewhere, please let me know.

We are going to prove

Let $\{a_n\}_{n=1}^\infty$ be a sequence of non-negative numbers. Prove that $$\liminf_{n\to\infty} \left(n^2\left(4a_n(1-a_{n-1})-1\right)\right)\leq\frac14.\tag{1}$$

Step 1. In the first step, I'll reduce the choices of $\{a_n\}$ to $a_n\in(0,1)$ for all $n\geq 1$, as I did in OP. If we have a subsequence $\{a_{n_k}\}$ with $a_{n_k}\geq 1$ for all $k$, then $4a_{n_k+1}(1-a_{n_k})-1\leq 0$ for all $k$ and thus $(1)$ holds; also, the case where there are infinitely many $0$'s is easy to handle. So, WLOG we can assume that $a_n\in(0,1)$ for all $n\geq 1$.

Step 2. In this step we rewrite $(1)$ in terms of $b_n$, which is more convenient to deal with. Let $a_n=\frac12+b_n$, then we have $b_n\in(-1/2,1/2)$ and our aim is to prove $$\liminf_{n\to\infty}\left(n^2\left(b_n-b_{n-1}-2b_nb_{n-1}\right)\right)\leq\frac18.\tag{2}$$ Fix an arbitrary $a>\frac18$, we will prove $$\liminf_{n\to\infty}\left(n^2\left(b_n-b_{n-1}-2b_nb_{n-1}\right)\right)\leq a.\tag{3}$$ Indeed, if $(3)$ holds for all $a>\frac18$, then $(2)$ follows easily.

Step 3. Suppose that $(3)$ is not true for some $a>\frac18$ (we will get a contradiction in the end of the proof,) then there exists $N>0$ such that $$b_n-b_{n-1}-2b_nb_{n-1}>\frac a{n^2},\qquad \forall n>N.\tag{4}$$ We may assume WLOG that $N=1$, so we have $$b_n>\frac{b_{n-1}+\frac a{n^2}}{1-2b_{n-1}},\qquad \forall n\geq 2.\tag{5}$$ We claim that $b_n\leq 0$ for all $n\geq1$. Suppose $b_{n_0}>0$ for some $n_0\geq 1$, then it follows from $(5)$ that $$b_n>\frac{b_{n-1}}{1-2b_{n-1}},\qquad \forall n\geq n_0+1.$$ By induction we can easily show that $b_n>0$ for all $n\geq n_0$ and $\{b_n\}_{n=n_0}^\infty$ is increasing. Recalling that $b_n<\frac12$ for all $n\geq1$, the limit $L:=\lim_{n\to\infty} b_n$ exists and satisfies $0<b_{n_0}\leq L\leq \frac12$. Letting $n\to\infty$ in $(4)$, we get $-2L^2\geq 0$, which implies that $L=0$, a contradiction. This proves our claim that $b_n\leq 0$ for all $n\geq1$.

Step 4. Let $c_n=-2b_n$, then $c_n\in[0,1)$ for all $n\geq 1$ and $$c_n<\frac{c_{n-1}-\frac{2a}{n^2}}{1+c_{n-1}},\qquad \forall n\geq2. $$ And we also have $c_n>\frac{2a}{(n+1)^2}$ for all $n\geq 1$. Define $\{d_n\}_{n=1}^\infty$ by $d_1=c_1$ and $$d_n=\frac{d_{n-1}-\frac{2a}{n^2}}{1+d_{n-1}},\qquad \forall n\geq2. \tag{6}$$ Since the map $x\mapsto\frac{x-\frac{2a}{n^2}}{1+x}=1-\frac{1+\frac{2a}{n^2}}{1+x}$ is increasing for $x>0$, we can easily prove by induction that $$\frac{2a}{(n+1)^2}<c_n\leq d_n,\qquad n\geq1.\tag{7}$$

Step 5. In this step we prove the limit $\lim_{n\to\infty}nd_n$ exists and is finite. By $(7)$, we have $nd_n>0$ for all $n\geq1$. We prove that $\{nd_n\}$ is decreasing after finite terms. Indeed, for $n\ge 1$, \begin{align*} &\frac{(n+1)d_{n+1}}{nd_n}=\frac{(n+1)d_{n+1}-\frac{2a}{n+1}}{nd_n(1+d_n)}<1\\ \Longleftrightarrow& (n+1)d_{n+1}-\frac{2a}{n+1}<nd_n+nd_n^2\\ \Longleftrightarrow& nd_n^2-d_n+\frac{2a}{n+1}>0\\ \Longleftrightarrow& \left(d_n-\frac{1}{2n}\right)^2+\frac{2a}{n(n+1)}-\frac{1}{4n^2}=\left(d_n-\frac{1}{2n}\right)^2+\frac{(8a-1)n-1}{4n^2(n+1)}>0. \end{align*} Since $a>\frac18$, there eixsts $N>0$ such that $$0<(n+1)d_{n+1}<nd_n,\qquad n>N.$$ Therefore $\lim_{n\to\infty} nd_n$ exists and is finite.

Step 6. In the final step, we deduce a contradiction. By $(6)$, we have $$-n^2(d_{n+1}-d_n) = \frac{1}{1+d_n}\left(2a\frac{n^2}{(n+1)^2}+(nd_n)^2\right),\qquad n\geq 1.$$ Assume $A:=\lim_{n\to\infty} nd_n$, then $0\leq A<\infty$ and $\lim_{n\to\infty}d_n=0$. By Stolz-Cesàro theorem, \begin{align*} A&=\lim_{n\to\infty} nd_n=\lim_{n\to\infty} \frac{d_n}{\frac1n}=\lim_{n\to\infty}\frac{d_{n+1}-d_n}{\frac1{n+1}-\frac1n}\\ &=-\lim_{n\to\infty}n^2(d_{n+1}-d_n)=\lim_{n\to\infty}\frac{1}{1+d_n}\left(2a\frac{n^2}{(n+1)^2}+(nd_n)^2\right)\\ &=2a+A^2. \end{align*} However, since $a>\frac18$, there is no $A\in\mathbb R$ such that $A=2a+A^2$. $\color{red}{\text{This is a contradiction!}}$

So, the proof is completed now. I would like to express my gratitude to anyone who are reading this long post and answer. Comments or remarks on a easier proof, or improved results, or anything else related to this problem, are very welcome.

Answer 2

Just some ideas.

Given $a_n\in(0,1]$, let $a_n=\sin^2\theta_n$ for $\theta_n\in\left(0,\frac{\pi}{2}\right]$. We want to show that

$$ \liminf_{n\to +\infty} n^2\left(4\sin^2\theta_n\cos^2\theta_{n-1}-1\right)\leq\frac{1}{4} $$

i.e. that $\sin\theta_n\cos\theta_{n-1}=\sin\theta_n\sin\left(\frac{\pi}{2}-\theta_{n-1}\right)$ is frequently close to $\frac{1}{2}$, precisely $$ \sin\theta_n\sin\left(\frac{\pi}{2}-\theta_{n-1}\right)\leq \frac{1}{2}\sqrt{1+\frac{1}{4n^2}}. \tag{1}$$

If $\theta_n\leq\frac{\pi}{6}$ or $\theta_{n-1}\geq \frac{\pi}{3}$ then $(1)$ is automatically fulfilled. It follows that we may assume $\theta_{n-1}<\frac{\pi}{3}$ and $\theta_n>\frac{\pi}{6}$, so from some point on all the $\theta_n$s lie in the central part of $\left(0,\frac{\pi}{2}\right]$. Once this is settled, we have $\sin\theta_n \leq \frac{\sqrt{3}}{2}$, hence we may further restrict the $\theta_n$s to the interval $\left[\arcsin\frac{1}{\sqrt{3}},\arccos\frac{1}{\sqrt{3}}\right]$. Here the sine function is bounded by $\sqrt{\frac{2}{3}}$, allowing a further restriction to $\left[\arcsin\sqrt{\frac{3}{8}},\arccos\sqrt{\frac{3}{8}}\right]$. Here the sine function is bounded by $\sqrt{\frac{5}{8}}$, allowing a further restriction to $\left[\arcsin\sqrt{\frac{2}{5}},\arccos\sqrt{\frac{2}{5}}\right]$.

I guess it is clear where I am headed here. We just have to prove that by performing this restriction process (bringing $\left[\arcsin\sqrt{a},\arccos\sqrt{a}\right]$ into $\left[\arcsin\sqrt{\frac{1}{4(1-a)}},\arccos\sqrt{\frac{1}{4(1-a)}}\right]$) $N$ times we end up in a neighbourhood of $\frac{\pi}{4}$ with a certain diameter, granting $(1)$ anyhow. More precisely, by studying the iterations of $f:a\mapsto\frac{1}{4(1-a)}$ we have the following: if $\sin\theta_n\cos\theta_{n-1}>\frac{1}{2}$ holds for any $n\leq N$, then for any $n>N+1$ we have that $\theta_n$ lies in the interval

$$\left[\arcsin\sqrt{\frac{1}{2}-\frac{1}{2N+2}},\arccos\sqrt{\frac{1}{2}-\frac{1}{2N+2}}\right] $$ so $\sin\theta_{N+2}\cos\theta_{N+1}$ is at most $\frac{1}{2}+\frac{1}{2N+2}$. Unluckily, this only proves (since $N$ is arbitrary) $$ \liminf_{n\to +\infty} \color{red}{n}(4a_n(1-a_{n-1})-1)\leq \color{red}{2}.$$

---

Alternative approach. Let us assume that $$ b_n-b_{n-1}-2 b_n b_{n-1} > \frac{1}{8n^2} \tag{2}$$ holds for all values of $n$ from some point on: up to reindexing we may as well assume that $(2)$ holds for all $n\geq 1$. Let us study the solutions of

$$ b_n = \frac{b_{n-1}+\frac{1}{8n^2}}{1-2 b_{n-1}} \tag{3}$$ for $n\geq 1$ and $b_0\in\left(-\frac{1}{2},\frac{1}{2}\right)$. If $b_0>0$ then $b_n > \frac{b_{n-1}}{1-2 b_{n-1}}$ and $b_n\to +\infty$ violating the boundedness of $\{b_n\}_{n\geq 0}$. Let us set $c_n=-b_n$ and consider $$ c_n = \frac{c_{n-1}-\frac{1}{8n^2}}{1+2 c_{n-1}}\tag{4} $$ for $c_0\in\left(0,\frac{1}{2}\right)$. Letting $2c_n=d_n$, we may as well study $$ d_n = \frac{d_{n-1}-\frac{1}{4n^2}}{1+d_{n-1}}\tag{5} $$ for $d_0\in(0,1)$. If at some point $d_n$ becomes negative, the destiny of the sequence is to blow up. We want to show that the only chances are a blow up or $\lim n d_n=\frac{1}{2}$. Luckily, the recurrence $$ \widetilde{d}_n = \frac{\widetilde{d}_{n-1}}{1+\widetilde{d}_{n-1}}$$ with $\widetilde{d}_0=d_0$ has a simple solution, $\widetilde{d}_n=\frac{1}{n+\frac{1}{d_0}}$, leading to $d_n\leq \frac{1}{n+\frac{1}{d_0}}$. It follows that as long as the sequence stays positive, $nd_n$ is bounded by one. By plugging these inequalities back into $(5)$ it should be possible to achieve a good understanding of the asymptotic behaviour of the solutions of $(5)$. We may also notice that if the limit $\lim nd_n$ exists it is necessarily $\frac{1}{2}$ by Cesaro Stolz, since $$ -n^2(d_{n+1}-d_n) = \frac{1}{1+d_n}\left(\frac{n^2}{4(n+1)^2}+(nd_n)^2\right).$$

Answer 3

Partial answer. As discussed in the comments, here is a proof that $\liminf_{n\to\infty} \left(4a_n(1-a_{n-1})-1\right)\leq\frac14:$

Suppose by way of contradiction that $\liminf_{n\to\infty} \left(4a_n(1-a_{n-1})-1\right) > \frac14$. Then there are finitely many $a_n$ such that $4a_n(1-a_{n-1})-1 \leq \frac14,\ $ i.e., $\ a_n(1-a_{n-1}) \leq \frac{5}{16}.$ This implies there are six consecutive members of $(a_n)$ which satisfy $\ a_n(1-a_{n-1}) > \frac{5}{16}.$ But it is not too difficult to show that this is impossible.

Suppose WLOG that for all $n\in\{2,3,4,5,6,7\}*,\ a_n(1-a_{n-1}) > \frac{5}{16}.$ In order for $a_2(1-a_1) > \frac{5}{16},$ we must have $a_2 > \frac{5}{16}.$ Since $a_3(1-a_2) > \frac{5}{16}$ and $a_2 > \frac{5}{16},$ we must have $a_3 > \frac{5/16}{11/16} = \frac{5}{11}.$ Since $a_4(1-a_3) > \frac{5}{16}$ and $a_3 > \frac{5}{11},$ we must have $a_4 > \frac{5/16}{6/11} = \frac{55}{96}.$ Since $a_5(1-a_4) > \frac{5}{16}$ and $a_4 > \frac{55}{96},$ we must have $a_5 > \frac{5/16}{41/96} = \frac{30}{41}.$ Since $a_6(1-a_5) > \frac{5}{16}$ and $a_5 > \frac{30}{41},$ we must have $a_6 > \frac{5/16}{11/41} = \frac{205}{176}.$ Since $\ 1-a_6<0\ $ and $\ a_7\geq 0,\ a_7(1-a_6) \leq 0 < \frac{5}{16},$ a contradiction. Therefore $\liminf_{n\to\infty} \left(4a_n(1-a_{n-1})-1\right) \leq \frac14.$

*It may not be $2,3,4,5,6,7,$ but my proof is easier to follow than if I wrote $N, N+1, N+2, N+3, N+4, N+5.$